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<h2 class="hd hd-2 unit-title">1. Motivation</h2>
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<h3 class="hd hd-2">Infinite series</h3>
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<h2 class="hd hd-2 unit-title">2. Infinite series</h2>
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<p><b class="bfseries">Objectives</b></p><ul class="itemize"><li><p>
Know the definition of <b class="bf"><span style="color:#27408C">partial sums</span></b> and <b class="bf"><span style="color:#27408C">series</span></b>. </p></li><li><p>
Determine when a <b class="bf"><span style="color:#27408C">geometric series</span></b> is <b class="bf"><span style="color:#27408C">convergent</span></b> or <b class="bf"><span style="color:#27408C">divergent</span></b>, and evaluate it when it is convergent.<br/></p></li><li><p>
Use <b class="bf"><span style="color:#27408C">the divergence test</span></b>, <b class="bf"><span style="color:#27408C">integral comparison</span></b>, <b class="bf"><span style="color:#27408C">direct comparison</span></b>, <b class="bf"><span style="color:#27408C">limit comparison</span></b>, <b class="bf"><span style="color:#27408C">the ratio test</span></b>, <b class="bf"><span style="color:#27408C">the root test</span></b> to determine if a positive series is convergent or divergent when possible. </p></li><li><p>
Know the difference between <b class="bf"><span style="color:#27408C">absolute convergence</span></b> and <b class="bf"><span style="color:#27408C">conditional convergence</span></b>. </p></li><li><p>
For <b class="bf"><span style="color:#27408C">alternating series</span></b>, use the <b class="bf"><span style="color:#27408C">alternating series test</span></b> to determine convergence. </p></li></ul><p><b class="bfseries">Higher goals</b></p><ul class="itemize"><li><p>
Understand how <b class="bf"><span style="color:#27408C">mathematical induction</span></b> works. </p></li></ul><p><b class="bfseries">Contents: 23 pages</b></p><p>
15 videos (108 minutes 1x speed) 40 questions </p>
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<h2 class="hd hd-2 unit-title">3. Introduction</h2>
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A geometric series
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<p>
Let </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000000965" style="table-layout:auto" width="100%">
<tr id="a0000000966">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle S_0[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle 1[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.107)</td>
</tr>
<tr id="a0000000967">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle S_1[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle 1+\frac{1}{2}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.108)</td>
</tr>
<tr id="a0000000968">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle S_2[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle 1+\frac{1}{2}+\frac{1}{4}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.109)</td>
</tr>
<tr id="a0000000969">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle S_3[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.110)</td>
</tr>
<tr id="a0000000970">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle S_4[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.111)</td>
</tr>
<tr id="a0000000971">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \vdots[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \vdots[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.112)</td>
</tr>
<tr id="a0000000972">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle S_ N[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle 1+\frac{1}{2}+\frac{1}{4}+\cdots +\left(\frac{1}{2}\right)^{N}.[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.113)</td>
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<p>
Mark [mathjaxinline]\, S_0,[/mathjaxinline] [mathjaxinline]\, S_1,[/mathjaxinline] [mathjaxinline]\, S_2,[/mathjaxinline] [mathjaxinline]\, S_3,[/mathjaxinline] [mathjaxinline]\, S_4,[/mathjaxinline] on the real number line. <br/>(Scroll down to submit.)<br/></p>
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A geometric series continued
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Is it possible that [mathjaxinline]\, S_ N\geq 2\,[/mathjaxinline] for some very large [mathjaxinline]\, N[/mathjaxinline]? </p>
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<h2 class="hd hd-2 unit-title">4. The geometric series and formulas</h2>
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A <b class="bf"><span style="color:#27408C">geometric series</span></b> is an infinite sum </p><table id="a0000000973" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000974"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle 1+a+a^2+a^3+\cdots =a^0+a^1+a^2+a^3+\cdots \qquad (a\, \text { is any number}).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
In other words, it is the limit </p><table id="a0000000975" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000976"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \lim _{N\rightarrow \infty } \left( 1+a+a^2+a^3+\cdots +a^{N} \right) \qquad (a\, \text { is any number}).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
Notice that each term (except the first) is [mathjaxinline]a[/mathjaxinline] times the previous term. In other words, [mathjaxinline]\, a\,[/mathjaxinline] is the ratio of consecutive terms.<br/></p><p>
When [mathjaxinline]\, \displaystyle |a|<1,\,[/mathjaxinline] the infinite sum is finite, and is given by </p><table id="a0000000977" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000978"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \left(1+a+a^2+a^3+\cdots \right)\,[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{1}{1-a} \qquad (|a|<1).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.114)</td></tr></table><p>
We will derive this formula in the exercises below. </p>
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Finite sums of the geometric series
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Let [mathjaxinline]\, a\,[/mathjaxinline] be any number. Define the finite sum </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000000979" style="table-layout:auto" width="100%">
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[mathjaxinline]\displaystyle \displaystyle S_ N[/mathjaxinline]
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[mathjaxinline]\displaystyle = 1+a+a^2+\cdots +a^ N.[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.115)</td>
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<p>
(This gives [mathjaxinline]\, \displaystyle S_{N+1}\, =\, 1+a+a^2+\cdots +a^{N}+a^{N+1}[/mathjaxinline].)<br/></p>
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Which of the following are equal to [mathjaxinline]S_{N+1}[/mathjaxinline]?<br/>(Check all that apply.)<br/><div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_series-tab4-problem1_2_1">
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Equate the different expressions of [mathjaxinline]\, S_{N+1}\,[/mathjaxinline] to find a formula for [mathjaxinline]\, S_ N[/mathjaxinline] in terms of [mathjaxinline]\, a\,[/mathjaxinline] and [mathjaxinline]\, N[/mathjaxinline].<br/>(Note that the answer is case sensitive: type N not n.) <p style="display:inline">[mathjaxinline]S_ N=\,[/mathjaxinline]</p><div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="formulaequationinput_series-tab4-problem1_3_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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Evaluate the finite sums
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Use the formula for [mathjaxinline]\, S_ N\,[/mathjaxinline] from the problem above to evaluate the following finite sums:<br/>(You must enter your answers as math expressions).<br/></p>
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<p style="display:inline">[mathjaxinline]\displaystyle 1+3+3^2+\cdots +3^{10} =\,[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\displaystyle \frac{1}{16} +\frac{1}{32} + \cdots +\frac{1}{2^{12}} = \,[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\displaystyle -1+\frac{1}{2}-\frac{1}{4}+\frac{1}{8} -\cdots -\frac{1}{2^{8}} = \,[/mathjaxinline]</p>
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Formula for the series
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As above, let [mathjaxinline]\, S_ N= 1+a+a^2+\cdots +a^ N,\,[/mathjaxinline] but now restrict to [mathjaxinline]\, -1&lt;a&lt;1[/mathjaxinline].<br/></p>
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Evaluate the limit [mathjaxinline]\, \displaystyle \lim _{N\rightarrow \infty } S_ N\,[/mathjaxinline] using the formula for [mathjaxinline]\, S_ N\,[/mathjaxinline] found above.<br/>(Your answer should be in terms of [mathjaxinline]\, a\,[/mathjaxinline] only.)<br/></p>
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<p style="display:inline">[mathjaxinline]\displaystyle \lim _{N\rightarrow \infty } S_ N=\,[/mathjaxinline]</p>
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Evaluate the series
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Evaluate the following infinite series:<br/>(Enter <b class="bf">infty</b> if the answer is infinite.)<br/></p>
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<p style="display:inline">[mathjaxinline]\displaystyle 1+3+3^2+\cdots =\,[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\displaystyle \frac{1}{16} +\frac{1}{32} +\frac{1}{64} + \cdots = \,[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\displaystyle -1+\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\cdots = \,[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">5. Divergence of the geometric series</h2>
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We are concerned with the different behaviors of the geometric series for different values of the ratio [mathjaxinline]\, a[/mathjaxinline]. <br/></p><p>
If the geometric series tends to a finite number, as when [mathjaxinline]\, |a|<1,\,[/mathjaxinline] we say it is <b class="bf">convergent</b>. Otherwise, we say it is <b class="bf">divergent</b>.<br/></p>
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Testing for divergence
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Consider the infinite series </p>
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[mathjaxinline]\displaystyle \displaystyle \left(\frac13 +10^{-9} \right)+\left(\frac19 +10^{-9}\right)+\left(\frac1{27} +10^{-9}\right)+\cdots .[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.136)</td>
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Determine if the series above approaches a finite number or tends to [mathjaxinline]\, \infty \,[/mathjaxinline].</p>
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<text> The series approaches [mathjaxinline]\, \displaystyle \frac{3}{2}+10^{-9}[/mathjaxinline].</text>
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<text> The series approaches [mathjaxinline]\, \displaystyle \frac{1}{2}+10^{-9}[/mathjaxinline].</text>
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<text> The series tends to [mathjaxinline]\, \displaystyle \infty[/mathjaxinline] and so diverges.</text>
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Comparison with a divergent series
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Let [mathjaxinline]\, a_ i&gt;0\,[/mathjaxinline] for all [mathjaxinline]\, i.\, \,[/mathjaxinline] And suppose </p>
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[mathjaxinline]\displaystyle \displaystyle \, b_ i\geq a_ i\, \qquad \text {for all}\, \, i&gt;100.[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.141)</td>
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If [mathjaxinline]\, a_1+a_2+a_3+\cdots \,[/mathjaxinline] diverges to [mathjaxinline]\, \infty ,[/mathjaxinline] then what can you conclude about the series [mathjaxinline]\, b_1+b_2+b_3+\cdots \,[/mathjaxinline]? </p>
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<text> The series can either diverge or converge since we have no information about the first 100 terms of the series.</text>
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<text> The series approaches a finite number and so converges.</text>
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<text> The series tends to [mathjaxinline]\, \displaystyle \infty[/mathjaxinline] and so diverges.</text>
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<h2 class="hd hd-2 unit-title">6. Notation</h2>
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<p><b class="bfseries">Divergence test</b></p><p>
One of the first and simplest tests on a series is the <b class="bf"><span style="color:#27408C">divergence test</span></b>:<br/></p><p>
If the sequence of numbers [mathjaxinline]\, a_1, a_2, a_3,\ldots \,[/mathjaxinline] does not tend to [mathjaxinline]\, 0,\,[/mathjaxinline] that is, if [mathjaxinline]\, \displaystyle \lim _{n\rightarrow \infty } a_ n\neq 0,\,[/mathjaxinline] then the series [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } a_ n[/mathjaxinline] diverges. </p><p>
This is very intuitive. For an infinite sum to approach a finite number, the terms being added had better approach [mathjaxinline]\, 0[/mathjaxinline]. </p>
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Review: summation notation
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Write the following <b class="bf">finite</b> sum in the [mathjaxinline]\, \displaystyle \sum[/mathjaxinline] notation:<br/></p>
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[mathjaxinline]\displaystyle \displaystyle -\frac{5}{(2)2^2}+\frac{5}{(3)2^3}-\frac{5}{(4)2^4}+\frac{5}{(5)2^5}-\frac{5}{(6)2^6}[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.148)</td>
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(Enter the lower limit, the summand, and the upper limit of the sum. All three boxes below are graded together. They will be either all correct or all wrong.)<br/></p>
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<p><b class="bfseries">Notation for series</b></p><p>
A <b class="bf"><span style="color:#27408C">partial sum</span></b> [mathjaxinline]\, \displaystyle S_ N\,[/mathjaxinline] is the <b class="bf">finite</b> sum </p><table id="a0000001031" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001032"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle S_ N[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{N} a_ n.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.150)</td></tr></table><p>
A <b class="bf"><span style="color:#27408C">series</span></b> [mathjaxinline]\, S\,[/mathjaxinline] is the <b class="bf">infinite sum</b> [mathjaxinline]\, \displaystyle \sum _{n=0}^{\infty } a_ n[/mathjaxinline].<br/></p><p>
If the limit of the partial sum [mathjaxinline]\displaystyle \lim _{N\rightarrow \infty } \, S_ N\,[/mathjaxinline] exists, then </p><table id="a0000001033" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001034"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle S[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \lim _{N\rightarrow \infty }S_ N[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.151)</td></tr></table><p>
and we say that the series [mathjaxinline]\, S\,[/mathjaxinline] <b class="bf"><span style="color:#27408C">converges</span></b>.<br/></p><p>
If the limit does not exist, we say the series [mathjaxinline]\, S\,[/mathjaxinline] <b class="bf"><span style="color:#27408C">diverges</span></b>. </p><p>
Note that a divergent series does not have to tend to [mathjaxinline]\, \infty ,\,[/mathjaxinline] and we have already seen the different divergent behaviors of the geometric series.<br/></p>
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Determine if each of the following series is convergent or divergent. If it is convergent, evaluate it.<br/>(Enter <b class="bf">div</b> if the series diverges. Enter numerical answers as math expressions or to 3 decimal places.)<br/></p>
<p>
<p style="display:inline">[mathjaxinline]\displaystyle \sum _{k=5}^{\infty } \frac{(-2)^{k-2}}{3^ k } \, =\,[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } \sin \left(\frac{n\pi }{8}\right) \, =\,[/mathjaxinline]</p>
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Review: partial fraction
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Find [mathjaxinline]\, A\,[/mathjaxinline] and [mathjaxinline]\, B\,[/mathjaxinline] such that </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001039" style="table-layout:auto" width="100%">
<tr id="a0000001040">
<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \displaystyle \, \frac{1}{(n+1)(n+2)}\,[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle =\frac{A}{n+1}+\frac{B}{n+2}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.155)</td>
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<p style="display:inline">[mathjaxinline]\displaystyle A = \,[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\displaystyle B = \,[/mathjaxinline]</p>
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<h3 class="hd hd-2">Worked example: evaluate a series</h3>
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Evaluate the series </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001043" style="table-layout:auto" width="100%">
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \displaystyle S[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle \sum _{n=1}^{\infty } \frac{1}{1+2+\cdots +n}[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle \frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\cdots[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.157)</td>
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<p><i class="itshape">Hint</i>: Use [mathjaxinline]\, \displaystyle 1+2+\cdots + n=\frac{n(n+1)}{2}[/mathjaxinline].<br/>(Enter <b class="bf">div</b> if the series diverges.)<br/></p>
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<p style="display:inline">[mathjaxinline]\displaystyle S = \,[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">8. Mathematical induction</h2>
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<p>
Sometimes we can guess a formula for the partial sum [mathjaxinline]\, S_ N,\,[/mathjaxinline] but how do we know that our guess is correct for all [mathjaxinline]\, N\,[/mathjaxinline]? </p><p>
One way to show that the formula indeed works for all [mathjaxinline]\, N\,[/mathjaxinline] is by <b class="bf"><span style="color:#27408C">mathematical induction</span></b>. <br/></p><p><b class="bf"><span style="color:#27408C">Mathematical induction </span></b> consists of two steps:<br/></p><dl class="description"><dt><b class="bf"><span style="color:#27408C">Base case</span></b>: </dt><dd><p>
Show the formula is true for the [mathjaxinline]\, N=1\,[/mathjaxinline], </p></dd><dt><b class="bf"><span style="color:#27408C">Induction step</span></b>:</dt><dd><p>
Show that <b class="bf">if </b> the formula is true for [mathjaxinline]S_ N\,[/mathjaxinline], then formula would also be true for [mathjaxinline]\displaystyle \, S_{N+1}[/mathjaxinline]. </p></dd></dl><p>
If both statements are true, then the formula works for all [mathjaxinline]S_ N[/mathjaxinline]. <br/></p><div id="a0000001049" class="figure"><center><img src="/assets/courseware/v1/baadd4af5239f10c3b58c9f05efb4f32/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/images_series_bluedomino.svg" width="450px" alt="See caption" style="margin: 10px 25px 25px 25px"/><div class="caption"><b>Figure 35</b>: <span> All tiles in a domino falls with a push on the first tile if each tile is placed close enough to the one before.</span></div></center></div><br/><p>
Mathematical induction works like the domino. Showing a formula is true for [mathjaxinline]\, N\,[/mathjaxinline] is analogous to having the [mathjaxinline]N^{\text {th}}\,[/mathjaxinline] tile fall. The base case is analogous to the push on the first tile. The induction step is analogous to making sure that if one tile falls, it pushes the next one down. </p>
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<p><b class="bfseries">A partial sum formula</b></p><p>
Use mathematical induction to show the formula: </p><table id="a0000001050" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001051"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle S_ N[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle 1+2+\cdots + N[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle \frac{N(N+1)}{2}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.161)</td></tr></table><p>
is true for all [mathjaxinline]\, N[/mathjaxinline]. We used this formula in the previous problem. <br/></p><p>
Follow the two steps of induction:<br/><b class="bf">Base case:</b> Show [mathjaxinline]\displaystyle \, S_ N=\frac{N(N+1)}{2}[/mathjaxinline] for [mathjaxinline]\, N=1[/mathjaxinline].<br/><b class="bf">Induction step:</b> Assume [mathjaxinline]\displaystyle \, S_ N=\frac{N(N+1)}{2}\,[/mathjaxinline] and compute [mathjaxinline]\, S_{N+1}\,[/mathjaxinline] using [mathjaxinline]\, \displaystyle S_{N+1}=S_{N}+(N+1)[/mathjaxinline]. </p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Proof of formula<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
We follow the two steps for induction:<br/><b class="bf">Base case:</b> [mathjaxinline]\, N=1[/mathjaxinline]. </p><table id="a0000001052" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001053"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle S_1[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle 1[/mathjaxinline]
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[mathjaxinline]\displaystyle =\frac{1(1+1)}{2}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.162)</td></tr></table><p><b class="bf">Induction step: </b> Show that [mathjaxinline]\displaystyle \, S_ N= \frac{N(N+1)}{2} \, \Longrightarrow S_{N+1}= \frac{(N+1)\left((N+1)+1\right)}{2}[/mathjaxinline]. </p><table id="a0000001054" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001055"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle S_{N+1}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle S_ N+(N+1)\qquad \left(\text {by definition}\right)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.163)</td></tr><tr id="a0000001056"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{N(N+1)}{2}+ (N+1)\qquad \left(\text {assume}\, \, S_ N= \frac{N(N+1)}{2}\right)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.164)</td></tr><tr id="a0000001057"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{N(N+1)+2(N+1)}{2}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.165)</td></tr><tr id="a0000001058"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{(N+1)(N+2)}{2}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.166)</td></tr><tr id="a0000001059"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{(N+1)\left((N+1)+1\right)}{2}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.167)</td></tr></table><p>
Since both steps of the induction work, [mathjaxinline]\, \displaystyle S_ N=\frac{N(N+1)}{2}\,[/mathjaxinline] for all [mathjaxinline]\, N[/mathjaxinline]. </p></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><SCRIPT src="/assets/courseware/v1/9567be3f4e91361f26a8beaadf749715/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/latex2edx.js" type="text/javascript"/><LINK href="/assets/courseware/v1/daf81af0af57b85a105e0ed27b7873a0/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/latex2edx.css" rel="stylesheet" type="text/css"/>
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Guess a formula for the partial sum </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001060" style="table-layout:auto" width="100%">
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \displaystyle S_ N= \sum _{n=1}^{N} \left(\frac{1}{\sqrt {n}}-\frac{1}{\sqrt {n+1}}\right)[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.168)</td>
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<p>
and use mathematical induction to show the formula works for all [mathjaxinline]\, N[/mathjaxinline].<br/></p>
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First, evaluate two cases before guessing the formula: <p style="display:inline">[mathjaxinline]\displaystyle S_1 = \,[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="formulaequationinput_series-tab8-problem1_2_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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<p style="display:inline">[mathjaxinline]\displaystyle S_2 = \,[/mathjaxinline]</p>
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Now, guess a formula for [mathjaxinline]\, \displaystyle S_ N[/mathjaxinline].<br/>(Note that the answer is case sensitive: <b class="bf">N</b> and <b class="bf">n</b> are different.)<br/><p style="display:inline">[mathjaxinline]\displaystyle S_ N = \,[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 3" role="group"><div id="formulaequationinput_series-tab8-problem1_4_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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Verify your formula for [mathjaxinline]\, S_ N\,[/mathjaxinline] is correct for all [mathjaxinline]\, N\,[/mathjaxinline] by induction.<br/></p>
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Evaluate the series
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Use the partial sum from the previous problem to evaluate the following series.<br/></p>
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \displaystyle S[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=1}^{\infty } \left(\frac{1}{\sqrt {n}}-\frac{1}{\sqrt {n+1}}\right)[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.178)</td>
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<p style="display:inline">[mathjaxinline]S \, =[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">9. Integral comparison</h2>
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Review of Riemann sum
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Find the expression of the right Riemann sum of [mathjaxinline]\displaystyle \, \int _{1}^{100} \frac{1}{x^2}\, dx\,[/mathjaxinline] with [mathjaxinline]\, \Delta x =1.\, \,[/mathjaxinline]<br/>(Recall the right Riemann sum uses the integrand values at the right endpoints of the intervals.)<br/>(All three answer boxes are graded together. They are either all correct or all wrong.)<br/></p>
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Which of the following inequality is true?<br/><div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_series-tab9-problem1_3_1">
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<h3 class="hd hd-2">Integral comparison: 1 over n</h3>
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How fast does the harmonic series diverge?
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Let us get a feeling of how fast [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty }\frac{1}{n}\,[/mathjaxinline] diverges. <br/></p>
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Let [mathjaxinline]\, \displaystyle S_ N=\sum _{n=1}^{N}\frac{1}{n}[/mathjaxinline].<br/></p>
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Find an estimate for the number [mathjaxinline]\, N\,[/mathjaxinline] for which [mathjaxinline]\, 23&lt; S_ N&lt;24\,[/mathjaxinline] by comparing the series with [mathjaxinline]\, \displaystyle \int _1^{N}\frac{1}{x}\, dx[/mathjaxinline].<br/>(Enter the number as a math expression.)<br/></p>
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<p style="display:inline">[mathjaxinline]23\, &lt;\, S_ N\, &lt;\, 24\,[/mathjaxinline] for [mathjaxinline]\, N=\,[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">10. Integral comparison</h2>
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<p><b class="bfseries">Integral comparison test</b></p><p>
If [mathjaxinline]\, f(x)>0\,[/mathjaxinline] and is decreasing, then [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } f(n)\,[/mathjaxinline] and [mathjaxinline]\, \displaystyle \int _1^{\infty } f(x)\, dx\,[/mathjaxinline] either <b class="bf">both converge</b> or <b class="bf">both diverge</b>.<br/>Moreoever, we have the following inequality </p><table id="a0000001087" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001088"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } f(n)- \int _1^{\infty } f(x)\, dx[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle <[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle f(1).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.186)</td></tr></table><p>
This inequality is equivalent to the pair of inequalities shown in the figure below.<br/></p><table class="tabular" cellspacing="0" style="table-layout:auto"><tr><td style="text-align:center; border:none"><img src="/assets/courseware/v1/840e019aa437eaee8fc567003407af97/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/images_series_integraltest.svg" width="380px" alt="" style="margin: 10px 25px 25px 25px"/></td><td style="text-align:center; border:none"> </td><td style="text-align:center; border:none"><img src="/assets/courseware/v1/fdcc0a0d1f26a343c7ce7b45785b4858/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/images_series_integraltest2.svg" width="380px" alt="" style="margin: 10px 25px 25px 25px"/></td></tr><tr><td style="text-align:center; border:none">
[mathjaxinline]\, \displaystyle \int _1^\infty f(x) \, dx > \sum _{n=1}^{\infty } f(n)- f(1)[/mathjaxinline]</td><td style="text-align:center; border:none"> </td><td style="text-align:center; border:none">
[mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } f(n)> \int _1^\infty f(x) \, dx[/mathjaxinline]</td></tr></table>
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Use integral comparison to determine all powers [mathjaxinline]\, p\,[/mathjaxinline] for which [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \frac{1}{n^ p}\,[/mathjaxinline] converges. This series is also known as the <b class="bf"><span style="color:#27408C">p-series</span></b>.<br/>(Check all that apply.)<br/></p>
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True or false:<br/></p>
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<p style="display:inline">If [mathjaxinline]\, \displaystyle \lim _{n\rightarrow \infty } a_ n = 0,\,[/mathjaxinline] then [mathjaxinline]\, \displaystyle \sum _{n=n_0}^{\infty } a_ n\,[/mathjaxinline] converges.</p>
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<p style="display:inline">If [mathjaxinline]\, \displaystyle \lim _{n\rightarrow \infty } a_ n \neq 0,\,[/mathjaxinline] then [mathjaxinline]\, \displaystyle \sum _{n=n_0}^{\infty } a_ n\,[/mathjaxinline] diverges.</p>
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<h2 class="hd hd-2 unit-title">11. Harder examples</h2>
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<h3 class="hd hd-2">Worked example: Integral comparison test for convergence</h3>
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<p>
In the video above, the final answer to the second series should be [mathjaxinline]\, \displaystyle \frac{1}{\ln (2)}\,[/mathjaxinline] not [mathjaxinline]\, \displaystyle \ln (2)[/mathjaxinline]. </p>
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Harder example 1
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Use integral comparison to find all values of [mathjaxinline]\, p\,[/mathjaxinline] for which [mathjaxinline]\, \displaystyle \sum _{n=2}^{\infty } \frac{1}{n\left(\ln (n)\right)^ p}[/mathjaxinline] converges.<br/>(Check all that apply.)<br/></p>
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<text>[mathjaxinline]p\leq 0[/mathjaxinline]</text>
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<text>[mathjaxinline]0&lt;p&lt;1[/mathjaxinline]</text>
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Harder examples 2
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Determine if each of the following series is convergent or divergent. <br/></p>
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<p style="display:inline">[mathjaxinline]\, \displaystyle \sum _{n=3}^{\infty } \frac{1}{n\ln (n)\ln (\ln (n))}\qquad[/mathjaxinline]</p>
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<p style="display:inline"> [mathjaxinline]\, \displaystyle \sum _{n=3}^{\infty } \frac{1}{n\ln (n)\left(\ln (\ln (n))\right)^{1.0001}}\, \qquad[/mathjaxinline]</p>
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<option value="converges."> converges.</option>
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<h2 class="hd hd-2 unit-title">12. Using the integral test for estimation</h2>
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<p><b class="bf">Video Errata:</b></p><ul class="itemize"><li><p>
At time 7:40, Joel has the wrong inequality, he has an extra [mathjaxinline]N+1[/mathjaxinline] in the upper limit of the right hand term. It should be </p><table id="a0000001108" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\int _3^{N+1}\frac{\ln x}{x}dx \lt S_ N \lt \frac{\ln 3}{3} + \int _3^ N \frac{\ln x}{x}dx[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></li><li><p>
At the end of the video, Joel mentions the computation time would be 50,000 times the age of the universe. He lost a power of ten, this should be 500,000 times the age of the universe! </p></li></ul>
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<h2 class="hd hd-2 unit-title">13. Direct comparison</h2>
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Let [mathjaxinline]\, \displaystyle a_ n, b_ n&gt;0,\,[/mathjaxinline] and suppose[mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } b_ n[/mathjaxinline] converges.<br/></p>
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<p style="display:inline">If [mathjaxinline]\displaystyle \, a_ n&lt; b_ n\,[/mathjaxinline] for all [mathjaxinline]\, n&gt;N[/mathjaxinline], <br/>then [mathjaxinline]\, \, \displaystyle \sum _{n=1}^{\infty } a_ n\,[/mathjaxinline]</p>
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<p style="display:inline">On the other hand, if [mathjaxinline]\displaystyle \, a_ n &gt; b_ n\,[/mathjaxinline] for all [mathjaxinline]\, n&gt;N[/mathjaxinline],<br/>then [mathjaxinline]\, \, \displaystyle \sum _{n=1}^{\infty } a_ n\,[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">14. Direct comparison test</h2>
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Let [mathjaxinline]\, \displaystyle 0< a_ n <b_ n\,[/mathjaxinline] for all [mathjaxinline]\, n\geq N[/mathjaxinline].<br/></p><p>
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if [mathjaxinline]\, \displaystyle \sum _{n=N}^{\infty } b_ n\,[/mathjaxinline] converges, then [mathjaxinline]\, \displaystyle \sum _{n=N}^{\infty } a_ n\,[/mathjaxinline] also converges; </p></li><li><p>
if [mathjaxinline]\, \displaystyle \sum _{n=N}^{\infty } a_ n\,[/mathjaxinline] diverges, then [mathjaxinline]\, \displaystyle \sum _{n=N}^{\infty } b_ n[/mathjaxinline] also diverges. </p></li></ul>
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Let [mathjaxinline]f(n)\,[/mathjaxinline] be a function of [mathjaxinline]\, n\,[/mathjaxinline] such that [mathjaxinline]\, f(n)&gt;0[/mathjaxinline] for all [mathjaxinline]n[/mathjaxinline]. </p>
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Suppose [mathjaxinline]\displaystyle \, \sum _{n=1}^{\infty } \frac{1}{\left(f(n)\right)^{2}}\,[/mathjaxinline] converges, what can you conclude about [mathjaxinline]\displaystyle \, \sum _{n=1}^{\infty } \frac{1}{\left(f(n)\right)^{2.01}}[/mathjaxinline]? </p>
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<text> [mathjaxinline]\displaystyle \, \sum _{n=1}^{\infty } \frac{1}{\left(f(n)\right)^{2.01}}[/mathjaxinline] converges.</text>
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<text> [mathjaxinline]\displaystyle \, \sum _{n=1}^{\infty } \frac{1}{\left(f(n)\right)^{2.01}}[/mathjaxinline] diverges.</text>
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<text> We cannot conclude convergence of [mathjaxinline]\displaystyle \, \sum _{n=1}^{\infty } \frac{1}{\left(f(n)\right)^{2.01}}\,[/mathjaxinline] from the convergence of [mathjaxinline]\displaystyle \, \sum _{n=1}^{\infty } \frac{1}{\left(f(n)\right)^{2}}[/mathjaxinline].</text>
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Use direct comparison along with results from integral comparison to determine whether the series below converges or diverges.<br/></p>
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[mathjaxinline]\displaystyle \displaystyle \sum _{n=0}^{\infty } \frac{1}{2n+1}[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle 1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.207)</td>
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<text> The series converges.</text>
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[mathjaxinline]\displaystyle \displaystyle \displaystyle \sum _{n=33}^{\infty } \frac{1}{\sqrt {n+2}+\sqrt {n}}[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.211)</td>
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<h2 class="hd hd-2 unit-title">15. Limit comparison</h2>
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<p><b class="bfseries">Limit comparison</b></p><p>
If </p><ol class="enumerate"><li value="1"><p>
[mathjaxinline]\, \displaystyle \frac{f(n)}{g(n)}\rightarrow c\,[/mathjaxinline] where [mathjaxinline]\, \, c\neq 0\,[/mathjaxinline] is finite, </p></li><li value="2"><p>
[mathjaxinline]\, \displaystyle g(n)>0\,[/mathjaxinline] for all [mathjaxinline]\, n>N\,[/mathjaxinline] for some [mathjaxinline]\, N>0[/mathjaxinline], </p></li></ol><p>
then [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } f(n)\,[/mathjaxinline] and [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } g(n)\,[/mathjaxinline] either <b class="bf">both converge</b> or <b class="bf">both diverge</b>.<br/></p><p>
In other words, if [mathjaxinline]\, f(n)\,[/mathjaxinline] and [mathjaxinline]\, g(n)\,[/mathjaxinline] decay at the same rate as [mathjaxinline]\, n\,[/mathjaxinline] tends to [mathjaxinline]\, \infty ,\,[/mathjaxinline] then the series [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } f(n)\,[/mathjaxinline] and [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } g(n)\,[/mathjaxinline] converge or diverge together.<br/></p><p>
This is analogous to limit comparison for improper integrals.<br/></p><p><b class="bf">Note:</b> The condition [mathjaxinline]\, \displaystyle \frac{f(n)}{g(n)}\longrightarrow c\neq 0\,[/mathjaxinline] as [mathjaxinline]\, \displaystyle n \longrightarrow \, \infty \,[/mathjaxinline] is equivalent to </p><table id="a0000001129" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001130"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(n)\sim c\, g(n),\qquad \text {that is,}\quad \frac{f(n)}{c\, g(n)}\longrightarrow 1\, \, \text {as }\, \, n\, \longrightarrow \, \infty .[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.214)</td></tr></table>
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Limit comparison 1
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Use limit comparison to determine if the following series converges or diverges.<br/></p>
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[mathjaxinline]\displaystyle \displaystyle \sum _{k=1}^{\infty }\frac{3k^6-2}{5k^7+3k^6+2k^3+k^2+4}[/mathjaxinline]
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<text> The series tends to [mathjaxinline]\, \displaystyle \infty[/mathjaxinline] and so diverges.</text>
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Limit comparison 2
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Use limit comparison to determine if the following series converges or diverges.<br/></p>
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[mathjaxinline]\displaystyle \displaystyle \sum _{k=5}^{\infty }\frac{1}{\sqrt {2k^3+10}}[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.220)</td>
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<text> The series converges.</text>
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<text> The series tends to [mathjaxinline]\, \displaystyle \infty[/mathjaxinline] and so diverges.</text>
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Limit comparison 3
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Use limit comparison to determine if the following series converges or diverges.<br/></p>
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[mathjaxinline]\displaystyle \displaystyle \sum _{n=10}^{\infty }\sin \left(\frac{1}{n^2}\right)[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.223)</td>
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<h2 class="hd hd-2 unit-title">16. Worked examples of comparison tests</h2>
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<h3 class="hd hd-2">Worked examples: review and more on limit comparison</h3>
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<p><b class="bfseries">More on limit comparison</b></p><p>
We can also use limit comparison in the following way. <br/></p><p>
Suppose [mathjaxinline]\, f(n), g(n)> 0[/mathjaxinline] for all [mathjaxinline]n\geq N[/mathjaxinline] for some large [mathjaxinline]\, N[/mathjaxinline].<br/></p><p>
If [mathjaxinline]\, \displaystyle \frac{f(n)}{g(n)}\rightarrow 0\,[/mathjaxinline] as [mathjaxinline]\, \displaystyle n\rightarrow \infty ,\,[/mathjaxinline] that is, [mathjaxinline]\, f(n)\,[/mathjaxinline] decays faster than [mathjaxinline]\, g(n),\,[/mathjaxinline] then </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle \, \sum _{n=N}^{\infty } g(n)\,[/mathjaxinline] converges implies [mathjaxinline]\displaystyle \, \sum _{n=N}^{\infty } f(n)\,[/mathjaxinline] converges; </p></li><li><p>
[mathjaxinline]\displaystyle \, \sum _{n=N}^{\infty } f(n)\,[/mathjaxinline] diverges implies [mathjaxinline]\displaystyle \sum _{N}^{\infty } g(n)\,[/mathjaxinline] diverges. </p></li></ul>
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More practice on limit comparison
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Use limit comparison to find all values of [mathjaxinline]\, p\,[/mathjaxinline] for which [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \frac{\ln n }{n^ p}[/mathjaxinline] converges.<br/>(Check all that apply.)<br/></p>
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<text>[mathjaxinline]p\leq 0[/mathjaxinline]</text>
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Using both direct and limit comparisons
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Determine if the following series converges or diverges. </p>
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<p style="display:inline">[mathjaxinline]\, \displaystyle \sum _{n=10}^{\infty } \frac{\cos ^2\left(n^3+n^2+1\right)}{n\sqrt {n-2}}\qquad[/mathjaxinline]</p>
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<option value="converges."> converges.</option>
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<h2 class="hd hd-2 unit-title">17. Ratio tests</h2>
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<p><b class="bfseries">Ratio test</b></p><p>
The <b class="bf"><span style="color:#27408C">ratio test</span></b> is another way to determine convergence of a series. </p><p>
Consider [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } a_ n\,[/mathjaxinline] where [mathjaxinline]\, a_ n>0[/mathjaxinline] for all [mathjaxinline]\, n>N[/mathjaxinline]. </p><p>
Define </p><table id="a0000001159" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001160"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle L=\lim _{n\rightarrow \infty } \frac{a_{n+1}}{a_ n}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.232)</td></tr></table><p>
There are three cases: </p><ol class="enumerate"><li value="1"><p>
If [mathjaxinline]L<1,\,[/mathjaxinline] then the series <b class="bf">converges</b>; </p></li><li value="2"><p>
If [mathjaxinline]L>1,\,[/mathjaxinline] then the series <b class="bf">diverges</b>; </p></li><li value="3"><p>
If [mathjaxinline]L=1,\,[/mathjaxinline] then there is <b class="bf">no conclusion</b>, i.e. the series can diverge or converge. </p></li></ol><p>
Here, we focus on series [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } a_ n\,[/mathjaxinline] where [mathjaxinline]\, a_ n>0[/mathjaxinline] for all [mathjaxinline]\, n[/mathjaxinline] large. You will see shortly that you can use the ratio test on series whose tail is not always positive.<br/></p>
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Ratio test practice 1
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Use the ratio test to determine if the following series converges or diverges.<br/></p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001161" style="table-layout:auto" width="100%">
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty }\frac{2^ n}{n!} \qquad \text {where}\, \, n!=n(n-1)(n-2)\cdots (3)(2)(1)[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.233)</td>
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First, compute [mathjaxinline]\, \displaystyle L=\lim _{n\rightarrow \infty } \frac{a_{n+1}}{a_ n}[/mathjaxinline].<br/></p>
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<p style="display:inline">[mathjaxinline]L\, =\, \quad[/mathjaxinline]</p>
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Now, use [mathjaxinline]\, L\,[/mathjaxinline] to determine if the series converges or diverges.<br/></p>
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<text> The series converges.</text>
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<text> The series tends to [mathjaxinline]\, \displaystyle \infty[/mathjaxinline] and so diverges.</text>
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<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
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<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
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<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
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<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
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<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
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<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
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<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
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<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
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<th class="formulainput" rowspan="3" scope="row">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
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<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
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<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
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<th class="formulainput" scope="row">Differentials</th>
<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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Ratio test practice 2
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Use the ratio test to find all values of [mathjaxinline]\, x&gt;0\,[/mathjaxinline] for which the following series converge. </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001167" style="table-layout:auto" width="100%">
<tr id="a0000001168">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \frac{x^ n}{n^2}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.237)</td>
</tr>
</table>
<p>
First, compute [mathjaxinline]\, \displaystyle L=\lim _{n\rightarrow \infty } \frac{a_{n+1}}{a_ n}[/mathjaxinline] in terms of [mathjaxinline]\, x[/mathjaxinline].<br/></p>
<p>
<p style="display:inline">[mathjaxinline]L\, =\, \quad[/mathjaxinline]</p>
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For which [mathjaxinline]\, x&gt;0\,[/mathjaxinline] does the series converge?<br/>(Check all that apply).<br/><div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_series-tab17-problem2_3_1">
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<input type="checkbox" name="input_series-tab17-problem2_3_1[]" id="input_series-tab17-problem2_3_1_choice_1" class="field-input input-checkbox" value="choice_1"/><label id="series-tab17-problem2_3_1-choice_1-label" for="input_series-tab17-problem2_3_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_series-tab17-problem2_3_1"> <text>[mathjaxinline]x=1[/mathjaxinline]</text>
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<input type="checkbox" name="input_series-tab17-problem2_3_1[]" id="input_series-tab17-problem2_3_1_choice_2" class="field-input input-checkbox" value="choice_2"/><label id="series-tab17-problem2_3_1-choice_2-label" for="input_series-tab17-problem2_3_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_series-tab17-problem2_3_1"> <text>[mathjaxinline]1&lt;x&lt;2[/mathjaxinline]</text>
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<input type="checkbox" name="input_series-tab17-problem2_3_1[]" id="input_series-tab17-problem2_3_1_choice_3" class="field-input input-checkbox" value="choice_3"/><label id="series-tab17-problem2_3_1-choice_3-label" for="input_series-tab17-problem2_3_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_series-tab17-problem2_3_1"> <text>[mathjaxinline]x=2[/mathjaxinline]</text>
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<input type="checkbox" name="input_series-tab17-problem2_3_1[]" id="input_series-tab17-problem2_3_1_choice_4" class="field-input input-checkbox" value="choice_4"/><label id="series-tab17-problem2_3_1-choice_4-label" for="input_series-tab17-problem2_3_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_series-tab17-problem2_3_1"> <text>[mathjaxinline]x&gt;2[/mathjaxinline]</text>
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<th class="formulainput" scope="col">Example Entries</th>
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<th class="formulainput" rowspan="3" scope="row">Numbers</th>
<td class="formulainput">Integers</td>
<td class="formulainput">
<font color="#0078b0">2520</font>
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<font color="#0078b0">2/3</font>
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<td class="formulainput">Decimals </td>
<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
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<tr class="formulainput">
<th class="formulainput" rowspan="4" scope="row">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
</tr>
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<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
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<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
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<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
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<th class="formulainput" rowspan="3" scope="row">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
</tr>
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<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Differentials</th>
<td class="formulainput">dx, dy</td>
<td class="formulainput">Enter a function followed by differential. You must multiply by the differential. <br/> Enter <font color="#0078b0">e^x*dx </font> for \( e^xdx \)<br/>
Enter <font color="#0078b0">(2*pi+y)*dy </font> for \( (2\pi+y)dy \)
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Ratio test practice 3 (*)
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Use the ratio test to determine if the following series converges or diverges.<br/></p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001174" style="table-layout:auto" width="100%">
<tr id="a0000001175">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \frac{n!}{n^ n}[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.242)</td>
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<p>
First, compute [mathjaxinline]\, \displaystyle L=\lim _{n\rightarrow \infty } \frac{a_{n+1}}{a_ n}[/mathjaxinline].<br/><i class="itshape">Hint</i>: Recall that since [mathjaxinline]\, \ln \,[/mathjaxinline] is continuous, </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001176" style="table-layout:auto" width="100%">
<tr id="a0000001177">
<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \displaystyle \ln \left(\lim _{n\rightarrow \infty } b(n)\right)= \lim _{n\rightarrow \infty } \ln (b(n))[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.243)</td>
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provided the [mathjaxinline]\, \displaystyle \lim _{n\rightarrow \infty } b(n)\,[/mathjaxinline] is positive or [mathjaxinline]\, +\infty[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]L\, =\, \quad[/mathjaxinline]</p>
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Now, use [mathjaxinline]\, L\,[/mathjaxinline] to determine if the series converges or diverges.<br/></p>
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<text> The series converges.</text>
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<h2 class="hd hd-2 unit-title">18. Root tests</h2>
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<p>
The <b class="bf"><span style="color:#27408C">root test</span></b> is yet another test to determine the convergence of a series. </p><p>
Consider [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } a_ n\,[/mathjaxinline] where [mathjaxinline]\, a_ n>0[/mathjaxinline] for all [mathjaxinline]\, n>N[/mathjaxinline].<br/></p><p>
Define </p><table id="a0000001189" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001190"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle L=\lim _{n\rightarrow \infty } \sqrt [n]{a_ n} =\lim _{n\rightarrow \infty } (a_ n)^{\frac{1}{n}}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.253)</td></tr></table><p>
The conclusions are the same as for the ratio test, that is, </p><ol class="enumerate"><li value="1"><p>
If [mathjaxinline]L<1,\,[/mathjaxinline] then the series <b class="bf">converges</b>; </p></li><li value="2"><p>
If [mathjaxinline]L>1,\,[/mathjaxinline] then the series <b class="bf">diverges</b>; </p></li><li value="3"><p>
If [mathjaxinline]L=1,\,[/mathjaxinline] then there is <b class="bf">no conclusion</b>, that is, the series can either diverge or converge. </p></li></ol><p>
The root test can also be used for series whose tail consist of negative terms. You will see this shortly.<br/></p><p>
We are not going to discuss the proof of the root test.<br/></p>
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Review: limits
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When using the root test, we often need to evaluate the [mathjaxinline]n[/mathjaxinline]-th root of expression. Here are some examples. <br/></p>
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Evaluate the following limits.<br/>(Enter <b class="bf">infty</b> if the limit is infinite.)<br/></p>
<p>
<p style="display:inline">[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } 2^{\frac{1}{n}}\, =\,[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } (n^ p)^{\frac{1}{n}}\, =\,[/mathjaxinline]</p>
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<p><i class="itshape">Hint</i>: Recall that since [mathjaxinline]\, \ln \,[/mathjaxinline] is continuous, </p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001191" style="table-layout:auto" width="100%">
<tr id="a0000001192">
<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \displaystyle \ln \left(\lim _{n\rightarrow \infty } b(n)\right)= \lim _{n\rightarrow \infty } \ln (b(n))[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.254)</td>
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provided the [mathjaxinline]\, \displaystyle \lim _{n\rightarrow \infty } b(n)\,[/mathjaxinline] is positive or [mathjaxinline]\, +\infty[/mathjaxinline]. </p>
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<p><b class="bfseries">A more difficult limit</b></p><p>
When using the root test, you may also encounter [mathjaxinline]\, \displaystyle \lim _{n\rightarrow \infty } (n!)^\frac {1}{n}.\, \,[/mathjaxinline] This limit is harder to evaluate. In fact, </p><table id="a0000001203" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001204"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \lim _{n\rightarrow \infty } (n!)^\frac {1}{n}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \infty .[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.263)</td></tr></table><p>
We leave the evaluation of this limit as an exercise for you. Try it out and take a look at our computation in the tab below.<br/></p><p><i class="itshape">Hint</i>: Obtain a lower bound on [mathjaxinline]\, \displaystyle \sum _{n=1}^{N} \ln (n)\,[/mathjaxinline] by comparing it with [mathjaxinline]\, \displaystyle \int _1^ N \ln (x)\, dx[/mathjaxinline]. <br/></p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Computation of the limit<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
As in the previous problem, will first compute [mathjaxinline]\, \displaystyle \ln[/mathjaxinline] of the limit and then exponentiate. </p><table id="a0000001205" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001206"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \ln \left( \lim _{n\rightarrow \infty } (n!)^{\frac{1}{n}}\right)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty }\frac{\ln (n!)}{n}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.264)</td></tr><tr id="a0000001207"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } \frac{ \sum _{k=2}^{n} \ln (k)}{n}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.265)</td></tr></table><p>
To show the limit is infinite, we can compare [mathjaxinline]\, \displaystyle \frac{ \sum _{k=2}^{n} \ln (k)}{n}\,[/mathjaxinline] with a smaller function whose limit is infinite. Let us proceed.<br/>The partial sum [mathjaxinline]\displaystyle \, \sum _{k=2}^{n} \ln (k) \,[/mathjaxinline] is the right Riemann sum of [mathjaxinline]\, \displaystyle \int _{1}^{n} \ln (x) \, dx\,[/mathjaxinline] with [mathjaxinline]\, \displaystyle \Delta x=1.\, \,[/mathjaxinline] Since [mathjaxinline]\, \ln (x)\,[/mathjaxinline] is an increasing function, </p><table id="a0000001208" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001209"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{k=2}^{n} \ln (k)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle >[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \int ^{n}_1 \ln (x)\, dx.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.266)</td></tr></table><p>
Now, we evaluate the integral using integration by parts: </p><table id="a0000001210" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001211"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \int \ln (x)\, dx[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle x \ln x-\int \frac{x}{x}\, dx \qquad \left(u=\ln (x), dv= dx\right)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.267)</td></tr><tr id="a0000001212"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle x\ln (x)-x[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.268)</td></tr><tr id="a0000001213"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \Longrightarrow[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \int _1^ n \ln (x)\, dx[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle n \ln (n)-n+1.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.269)</td></tr></table><p>
Therefore, we have a lower bound for [mathjaxinline]\displaystyle \, \sum _{k=1}^{n} \ln (k)[/mathjaxinline]: </p><table id="a0000001214" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001215"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{k=1}^{n} \ln (k)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle >[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle n \ln n-n+1.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.270)</td></tr></table><p>
Let us get back to evaluating the limit. </p><table id="a0000001216" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001217"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \ln \left( \lim _{n\rightarrow \infty } (n!)^{\frac{1}{n}}\right)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } \frac{ \sum _{k=2}^{n} \ln (k)}{n}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.271)</td></tr><tr id="a0000001218"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle >[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } \frac{n \ln n-n+1}{n}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.272)</td></tr><tr id="a0000001219"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } \left(\ln (n)-1 +\frac{1}{n}\right)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.273)</td></tr><tr id="a0000001220"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \infty[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.274)</td></tr></table><p>
This implies </p><table id="a0000001221" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001222"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \lim _{n\rightarrow \infty } \left(n!\right)^{\frac{1}{n}}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle e^{\infty } \, =\, \infty .[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.275)</td></tr></table></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><SCRIPT src="/assets/courseware/v1/9567be3f4e91361f26a8beaadf749715/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/latex2edx.js" type="text/javascript"/><LINK href="/assets/courseware/v1/daf81af0af57b85a105e0ed27b7873a0/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/latex2edx.css" rel="stylesheet" type="text/css"/>
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<h2 class="hd hd-2 unit-title">19. Root test continued</h2>
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<p><b class="bfseries">Worked examples: root test</b></p><p><b class="bfseries">Example 1: the geometric series</b></p><p>
The root test is another way to determine convergence of a geometric series. Consider </p><table id="a0000001223" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001224"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } x^ n \qquad \text {where} \, \, x>0.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.276)</td></tr></table><p>
To use the root test, we first compute [mathjaxinline]\, \displaystyle L[/mathjaxinline]. </p><table id="a0000001225" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001226"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle L[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } \sqrt [n]{x^ n}\,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.277)</td></tr><tr id="a0000001227"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } x[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.278)</td></tr><tr id="a0000001228"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle x \qquad (x>0).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.279)</td></tr></table><p>
Therefore, the root test tells us that [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } x^ n\, \, (x>0)\,[/mathjaxinline] converges when [mathjaxinline]\, L=x<1[/mathjaxinline], and diverges when [mathjaxinline]L=x>1[/mathjaxinline]. The root test does not give any information when [mathjaxinline]L=x=1,\,[/mathjaxinline] even though we already know by the divergence test that [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } 1^ n\,[/mathjaxinline] diverges.<br/></p><p><b class="bfseries">Example 2</b></p><p>
We will use the root test to determine if the following series converges or diverges.<br/></p><table id="a0000001229" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001230"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \frac{ n^2 }{2^ n}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.280)</td></tr></table><p>
First, we compute [mathjaxinline]\, \displaystyle L=\lim _{n\rightarrow \infty } (a_ n)^{\frac{1}{n}}[/mathjaxinline]. </p><table id="a0000001231" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001232"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle L[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } \left(\frac{ n^2 }{2^ n}\right)^{\frac{1}{n}}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.281)</td></tr><tr id="a0000001233"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } \frac{ \left(n^2\right)^{\frac{1}{n}} }{2}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.282)</td></tr><tr id="a0000001234"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{1}{2} \, \lim _{n\rightarrow \infty } \left(n^2\right)^{\frac{1}{n}}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.283)</td></tr><tr id="a0000001235"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{1}{2} \qquad \left( \lim _{n\rightarrow \infty } \left(n^2\right)^{\frac{1}{n}}=1\, \, \text {from previous page}\right).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.284)</td></tr></table><p>
Therefore, since [mathjaxinline]\, \displaystyle L<1,\,[/mathjaxinline] we conclude from the root test that the series [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \frac{ n^2 }{2^ n}\,[/mathjaxinline] converges. </p>
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Root test practice 1
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Use the root test to determine if the following series converges or diverges.<br/></p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001236" style="table-layout:auto" width="100%">
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[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \frac{3^ n (n)^4 }{4^ n}[/mathjaxinline]
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First, compute [mathjaxinline]\, \displaystyle \, L=\lim _{n\rightarrow \infty } (a_ n)^{\frac{1}{n}}[/mathjaxinline].<br/></p>
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<p style="display:inline">[mathjaxinline]L\, =\, \quad[/mathjaxinline]</p>
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<text> The series tends to [mathjaxinline]\, \displaystyle \infty[/mathjaxinline] and so diverges.</text>
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Root test practice 2
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Use the root test to determine for which [mathjaxinline]\, x&gt;0\,[/mathjaxinline] the following series converges.<br/></p>
<table cellpadding="7" cellspacing="0" class="eqnarray" id="a0000001246" style="table-layout:auto" width="100%">
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[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \left(\frac{x}{n}\right)^ n[/mathjaxinline]
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<td class="eqnnum" style="width:20%; border:none;text-align:right">(2.292)</td>
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First, compute [mathjaxinline]\, \displaystyle \, L=\lim _{n\rightarrow \infty } (a_ n)^{\frac{1}{n}}[/mathjaxinline].<br/></p>
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<p style="display:inline">[mathjaxinline]L\, =\, \quad[/mathjaxinline]</p>
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For which [mathjaxinline]\, x&gt;0\,[/mathjaxinline] does the series converge?<br/>(Check all that apply).<br/><div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_series-tab19-problem2_3_1">
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Use the root test to determine if each of the following series converges.<br/></p>
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<p style="display:inline">[mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \left(\frac{n-2}{n}\right)^ n\, \qquad[/mathjaxinline]</p>
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<option value="inconclusive by the root test."> inconclusive by the root test.</option>
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<p style="display:inline">[mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \left(\frac{n-2}{n}\right)^{n^2}\, \qquad[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">20. General series</h2>
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<p><b class="bfseries">Absolute convergence versus conditional convergence</b></p><p>
So far, we have focused on series whose terms are positive (with the exception of the geometric series and the divergence test). For general series, including series with both positive and negative terms, there are two notions of convergence.<br/></p><p>
Consider the series </p><table id="a0000001266" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001267"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle S = \sum _{n=1}^{\infty } a_ n \qquad (a_ n \, \text {can be positive or negative}).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.307)</td></tr></table><p>
The series [mathjaxinline]S[/mathjaxinline] is <b class="bf"><span style="color:#27408C">absolutely convergent</span></b> if [mathjaxinline]\, \sum _{n=1}^{\infty } |a_ n |[/mathjaxinline] converges.<br/></p><p>
The series [mathjaxinline]S[/mathjaxinline] is <b class="bf"><span style="color:#27408C">conditionally convergent</span></b> if it converges but is <b class="bf">not absolutely convergent</b>. <br/></p><p>
Strange to say, an absolute convergent series is <b class="bf">not</b> conditionally convergent.<br/></p><p>
For series with only positive terms, absolute convergence is the same as convergence. <br/></p><p>
In general, absolute convergence of a series implies convergence. <br/></p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Absolute convergence implies convergence<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
Convergence of a series means [mathjaxinline]\, \displaystyle \lim _{M\rightarrow \infty } \left|S-\sum _{n=1}^{M} a_ n\right|=0,\,[/mathjaxinline] and the reason that absolute convergence of a series implies convergence is that </p><table id="a0000001268" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001269"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \left| \sum _{n=M+1}^{N} a_ n \right|[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \leq[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=M+1}^{N}\left| a_ n\right| \longrightarrow 0 \qquad \text {as both}\, \, M,N\longrightarrow \infty ,[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.308)</td></tr></table></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><p>
Because absolute convergence concerns the convergence of [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \left|a_ n\right|,\,[/mathjaxinline] we can apply all of the techniques we have learned to determine absolute convergence. </p><p><b class="bfseries">Example</b></p><p>
Consider the series </p><table id="a0000001270" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001271"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \, \sum _{n=1}^{\infty } \frac{(-1)^ n}{n^ p}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.309)</td></tr></table><p>
Note that </p><table id="a0000001272" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001273"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \left|\frac{(-1)^ n}{n^ p}\right|[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=1}^{\infty } \frac{1}{n^ p}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.310)</td></tr></table><p>
Since [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \frac{1}{n^ p}\,[/mathjaxinline] converges for all [mathjaxinline]\, p>1,\,[/mathjaxinline] [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \frac{(-1)^ n}{n^ p}\,[/mathjaxinline] is absolutely convergent for all [mathjaxinline]\, p>1.[/mathjaxinline]<br/></p><p>
Since [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \frac{1}{n^ p}\,[/mathjaxinline] diverges for all [mathjaxinline]\, p\leq 1,\,[/mathjaxinline] [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \frac{(-1)^ n}{n^ p}\,[/mathjaxinline] is not absolutely convergent for all [mathjaxinline]\, p\leq 1.\, \,[/mathjaxinline] We will see shortly that [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \frac{(-1)^ n}{n^ p}\,[/mathjaxinline] is conditionally convergent for all [mathjaxinline]\, p\leq 1[/mathjaxinline].<br/></p><SCRIPT src="/assets/courseware/v1/9567be3f4e91361f26a8beaadf749715/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/latex2edx.js" type="text/javascript"/><LINK href="/assets/courseware/v1/daf81af0af57b85a105e0ed27b7873a0/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/latex2edx.css" rel="stylesheet" type="text/css"/>
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Determine if the following series absolutely converges, conditionally converges, diverges. </p>
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<p style="display:inline">[mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \left(\frac{-14}{15} \right)^ n\, \qquad[/mathjaxinline]</p>
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<text> absolutely converges.</text>
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<text> conditionally converges.</text>
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<text> diverges.</text>
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<p><b class="bfseries">A note on the ratio test and root test</b></p><p>
We can apply the ratio test and root test to series with negative terms by simply defining [mathjaxinline]\, L\,[/mathjaxinline] with absolute values. </p><p>
For a series [mathjaxinline]\, \displaystyle \sum _{n=0}^{\infty } a_ n,\,[/mathjaxinline] let </p><table id="a0000001274" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001275"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle L=\begin{cases} \lim _{n\rightarrow \infty }\left| \frac{a_{n+1}}{a_ n}\right|, & \mbox{for the ratio test} \\ \lim _{n\rightarrow \infty } \left|a_ n\right|^{\frac{1}{n}}& \mbox{for the root test}. \end{cases}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.311)</td></tr></table><p>
Then for both tests, the conclusions are the same as before except stated in terms of absolute convergence: </p><ol class="enumerate"><li value="1"><p>
If [mathjaxinline]L<1,\,[/mathjaxinline] then the series <b class="bf">absolutely converges</b>; </p></li><li value="2"><p>
If [mathjaxinline]L>1,\,[/mathjaxinline] then the series <b class="bf">diverges</b>; </p></li><li value="3"><p>
If [mathjaxinline]L=1,\,[/mathjaxinline] then there is <b class="bf">no conclusion</b> </p></li></ol><p>
The additional information here is that when [mathjaxinline]L>1,\,[/mathjaxinline] not only does [mathjaxinline]\, \displaystyle \sum _{n=0}^{\infty } \left|a_ n\right|\,[/mathjaxinline] diverge, the series without absolute signs [mathjaxinline]\, \displaystyle \sum _{n=0}^{\infty } a_ n\,[/mathjaxinline] also diverges. This is because when [mathjaxinline]\displaystyle L>1, \, \lim _{n\rightarrow \infty } a_ n \neq 0,\,[/mathjaxinline] so by the divergence test, [mathjaxinline]\, \displaystyle \sum _{n=0}^{\infty } a_ n\,[/mathjaxinline] diverges.<br/></p><p>
When we talk about Taylor series, we will use the ratio test (or the root test) to find what is known as the radius of convergence.<br/></p>
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<h2 class="hd hd-2 unit-title">21. Alternating series</h2>
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An <b class="bf"><span style="color:#27408C">alternating series</span></b> is a series whose terms alternate in signs. That is, an alternating series takes the form </p><table id="a0000001276" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001277"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \pm \sum _{n=1}^{\infty } (-1)^ n c_ n \qquad \text {where}\, \, \, c_ n\geq 0.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.312)</td></tr></table><p><b class="bfseries">The alternating series test</b></p><p>
There is a simple test for convergence of an alternating series.<br/></p><p>
If for all [mathjaxinline]\, n\,[/mathjaxinline] large enough, </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } c_ n =0[/mathjaxinline],<br/></p></li><li><p>
[mathjaxinline]\displaystyle c_ n\,[/mathjaxinline] decreases as [mathjaxinline]\, n\,[/mathjaxinline] increases,<br/></p></li></ul><p>
then [mathjaxinline]\, \displaystyle \pm \sum _{n=1}^{\infty } (-1)^ n c_ n\, \, (c_ n\geq 0)\,[/mathjaxinline] converges.<br/></p><p><b class="bfseries">Example: a conditionally convergent series</b></p><p>
The alternating series </p><table id="a0000001278" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001279"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \frac{(-1)^{n-1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.313)</td></tr></table><p>
is conditionally convergent (and not absolutely convergent).<br/></p><p>
To show this, first check for absolute convergence. The series is not absolutely convergent since </p><table id="a0000001280" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001281"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \left|\frac{(-1)^{n-1}}{n}\right|[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=1}^{\infty } \frac{1}{n}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.314)</td></tr></table><p>
diverges (by integral comparison as discussed before).<br/></p><p>
Since the series is not absolutely convergent, we need to check for conditional convergence. The series satisfies both conditions of the alternating series test: </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } \frac{1}{n} =0[/mathjaxinline], </p></li><li><p>
[mathjaxinline]\displaystyle \frac{1}{n}\,[/mathjaxinline] decreases as [mathjaxinline]n[/mathjaxinline] increases. </p></li></ul><p>
Hence,[mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \frac{(-1)^{n-1}}{n}\,[/mathjaxinline] converges. Since the series converges but is not absolutely convergent, it is conditionally convergent. The same reasoning applies to [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \frac{(-1)^ n}{n^ p}\,[/mathjaxinline] for [mathjaxinline]\, 0<p<1[/mathjaxinline]. </p>
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Alternating series 1
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Determine whether the series below converges absolutely, conditionally, or diverges.<br/></p>
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[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \frac{(-1)^ n}{5n-1}[/mathjaxinline]
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Alternating series 2
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Determine whether the series below converges absolutely, conditionally, or diverges.<br/></p>
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[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \frac{(-1)^ n}{\sqrt {2n^3-n^2+1}}[/mathjaxinline]
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Alternating series 3
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Determine whether the series below converges absolutely, conditionally, or diverges.<br/></p>
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[mathjaxinline]\displaystyle \displaystyle \sum _{n=1}^{\infty } \frac{(-3)^{n+2}}{(n+4)!}[/mathjaxinline]
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Alternating series 4
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Determine whether the series below converges absolutely, conditionally, or diverges.<br/></p>
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[mathjaxinline]\displaystyle \displaystyle \sum _{n=2}^{\infty } \frac{(-1)^ n}{n+\ln n}[/mathjaxinline]
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<text> The series converges absolutely.</text>
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<h2 class="hd hd-2 unit-title">22. Stacking blocks</h2>
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<p>
We end this section with the explanation of the stacking blocks that we started with in this section. </p><p>
The following three videos make up a continuous portion of a lecture. <br/></p>
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<h3 class="hd hd-2">Stacking blocks</h3>
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<h3 class="hd hd-2">Center of mass of the stacked blocks</h3>
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<h2 class="hd hd-2 unit-title">23. Summary</h2>
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<p><b class="bfseries">Notation</b></p><p>
A <b class="bf"><span style="color:#27408C">partial sum</span></b> [mathjaxinline]\, \displaystyle S_ N\,[/mathjaxinline] is the <b class="bf">finite</b> sum </p><table id="a0000001305" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001306"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle S_ N[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{N} a_ n.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.330)</td></tr></table><p>
A <b class="bf"><span style="color:#27408C">series</span></b> [mathjaxinline]\, S\,[/mathjaxinline] is the <b class="bf">infinite sum</b> [mathjaxinline]\, \displaystyle \sum _{n=0}^{\infty } a_ n[/mathjaxinline].<br/></p><p>
If the limit of the partial sum [mathjaxinline]\displaystyle \lim _{N\rightarrow \infty } \, S_ N\,[/mathjaxinline] exists, then </p><table id="a0000001307" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001308"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle S[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \sum _{n=0}^{\infty } a_ n[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \lim _{N\rightarrow \infty }S_ N[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.331)</td></tr></table><p>
and we say that the series [mathjaxinline]\, S\,[/mathjaxinline] <b class="bf"><span style="color:#27408C">converges</span></b>.<br/></p><p>
If the limit does not exist, we say the series [mathjaxinline]\, S\,[/mathjaxinline] <b class="bf"><span style="color:#27408C">diverges</span></b>. </p><p>
Note that a divergent series does not have to tend to [mathjaxinline]\, \infty ,\,[/mathjaxinline] and we have already seen the different divergent behaviors of the geometric series.<br/></p><p><b class="bf">Short hands for the summation notation:</b> We also sometimes use the following abbreviated notation: </p><ul class="itemize"><li><p>
[mathjaxinline]\, \displaystyle \sum _{0}^{\infty } a_ n\,[/mathjaxinline] for [mathjaxinline]\, \sum _{n=0}^{\infty } a_ n\,[/mathjaxinline] when it is clear which index to sum over,<br/></p></li><li><p>
[mathjaxinline]\, \displaystyle \sum ^{\infty } a_ n\,[/mathjaxinline] for [mathjaxinline]\, \sum _{n=N}^{\infty } a_ n\,[/mathjaxinline] when we are concerned with only the tail of the series. </p></li></ul><p><b class="bfseries">The geometric series</b></p><p>
A <b class="bf"><span style="color:#27408C">geometric series</span></b> is defined as </p><table id="a0000001309" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001310"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{n=0}^{\infty } a^ n[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \lim _{N\rightarrow \infty } \sum _{n=0}^{N} a^ n \qquad (a\, \text { is any number}).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.332)</td></tr></table><p>
Notice that each term (except the first) is [mathjaxinline]a[/mathjaxinline] times the previous term. In other words, [mathjaxinline]\, a\,[/mathjaxinline] is the ratio of consecutive terms.<br/></p><p>
Here is a formula for the partial sum: </p><table id="a0000001311" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001312"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{n=0}^{N} a^ n[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{1-a^{N+1}}{1-a} \qquad (a\, \text { is any number}).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.333)</td></tr></table><p>
When [mathjaxinline]\, \displaystyle |a|<1,\,[/mathjaxinline] the geometric series is convergent and converges to </p><table id="a0000001313" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001314"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \sum _{n=0}^{\infty } a^ n[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{1}{1-a} \qquad (|a|<1).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.334)</td></tr></table><p>
When [mathjaxinline]\, \displaystyle |a|\geq 1,\,[/mathjaxinline] the geometric series is divergent. </p><p><b class="bfseries">Divergence test</b></p><p>
One of the first and simplest tests on a series is the <b class="bf"><span style="color:#27408C">divergence test</span></b>:<br/></p><p>
If the sequence of numbers [mathjaxinline]\, a_1, a_2, a_3,\ldots \,[/mathjaxinline] does not tend to [mathjaxinline]\, 0,\,[/mathjaxinline] that is, if [mathjaxinline]\, \displaystyle \lim _{n\rightarrow \infty } a_ n\neq 0,\,[/mathjaxinline] then the series [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } a_ n[/mathjaxinline] diverges. </p><p>
This is very intuitive. For an infinite sum to approach a finite number, the terms being added had better approach [mathjaxinline]\, 0[/mathjaxinline]. </p><p><b class="bfseries">Mathematical induction</b></p><p>
Sometimes we can guess a formula for the partial sum [mathjaxinline]\, S_ N,\,[/mathjaxinline] but how do we know that our guess is correct for all [mathjaxinline]\, N\,[/mathjaxinline]? </p><p>
One way to show that the formula indeed works for all [mathjaxinline]\, N\,[/mathjaxinline] is by <b class="bf"><span style="color:#27408C">mathematical induction</span></b>. <br/><b class="bf"><span style="color:#27408C">Mathematical induction </span></b> consists of two steps:<br/></p><dl class="description"><dt><b class="bf"><span style="color:#27408C">Base case</span></b>: </dt><dd><p>
Show the formula is true for the [mathjaxinline]\, N=1\,[/mathjaxinline], </p></dd><dt><b class="bf"><span style="color:#27408C">Induction step</span></b>:</dt><dd><p>
Show that <b class="bf">if </b> the formula is true for [mathjaxinline]S_ N\,[/mathjaxinline], then formula would also be true for [mathjaxinline]\displaystyle \, S_{N+1}[/mathjaxinline]. </p></dd></dl><p>
If both statements are true, then the formula works for all [mathjaxinline]S_ N[/mathjaxinline]. <br/></p><p>
Mathematical induction works like the domino. <br/></p><div id="a0000001315" class="figure"><center><img src="/assets/courseware/v1/baadd4af5239f10c3b58c9f05efb4f32/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/images_series_bluedomino.svg" width="450px" alt="See caption" style="margin: 10px 25px 25px 25px"/><div class="caption"><b>Figure 36</b>: <span>All tiles in a domino falls with a push on the first tile if each tile is placed close enough to the one before.</span></div></center></div><p>
Showing a formula is true for [mathjaxinline]\, N\,[/mathjaxinline] is analogous to having the [mathjaxinline]N^{\text {th}}\,[/mathjaxinline] tile fall. The base case is analogous to the push on the first tile. The induction step is analogous to making sure that if one tile falls, it pushes the next one down. </p><p><b class="bfseries">Integral comparison test</b></p><p>
If [mathjaxinline]\, f(x)>0\,[/mathjaxinline] and is decreasing, then [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } f(n)\,[/mathjaxinline] and [mathjaxinline]\, \displaystyle \int _1^{\infty } f(x)\, dx\,[/mathjaxinline] either <b class="bf">both converge</b> or <b class="bf">both diverge</b>.<br/>Moreoever, we have the following inequality </p><table id="a0000001316" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001317"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \left| \sum _{n=1}^{\infty } f(n)- \int _1^{\infty } f(x)\, dx\right|[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle <[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle f(1).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.335)</td></tr></table><p>
This inequality is equivalent to the pair of inequalities shown in the figure below.<br/></p><table class="tabular" cellspacing="0" style="table-layout:auto"><tr><td style="text-align:center; border:none"><img src="/assets/courseware/v1/840e019aa437eaee8fc567003407af97/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/images_series_integraltest.svg" width="380px" alt="" style="margin: 10px 25px 25px 25px"/></td><td style="text-align:center; border:none"> </td><td style="text-align:center; border:none"><img src="/assets/courseware/v1/fdcc0a0d1f26a343c7ce7b45785b4858/asset-v1:MITx+18.01.3x+1T2020+type@asset+block/images_series_integraltest2.svg" width="380px" alt="" style="margin: 10px 25px 25px 25px"/></td></tr><tr><td style="text-align:center; border:none">
[mathjaxinline]\, \displaystyle \int _1^\infty f(x) \, dx > \sum _{n=1}^{\infty } f(n)- f(1)[/mathjaxinline]</td><td style="text-align:center; border:none"> </td><td style="text-align:center; border:none">
[mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } f(n)> \int _1^\infty f(x) \, dx[/mathjaxinline]</td></tr></table><p><b class="bfseries">Direct comparison test</b></p><p>
Let [mathjaxinline]\, \displaystyle 0\leq a_ n \leq b_ n\,[/mathjaxinline] for all [mathjaxinline]\, n\geq N[/mathjaxinline].<br/></p><p>
Then </p><ul class="itemize"><li><p>
[mathjaxinline]\, \displaystyle \sum _{n=N}^{\infty } b_ n\,[/mathjaxinline] converges implies [mathjaxinline]\, \displaystyle \sum _{n=N}^{\infty } a_ n\,[/mathjaxinline] converges; </p></li><li><p>
[mathjaxinline]\, \displaystyle \sum _{n=N}^{\infty } a_ n\,[/mathjaxinline] diverges implies [mathjaxinline]\, \displaystyle \sum _{n=N}^{\infty } b_ n[/mathjaxinline] diverges. </p></li></ul><p><b class="bfseries">Limit comparison</b></p><p>
If </p><ol class="enumerate"><li value="1"><p>
[mathjaxinline]\, \displaystyle \frac{f(n)}{g(n)}\rightarrow c\,[/mathjaxinline] where [mathjaxinline]\, \, c\neq 0\,[/mathjaxinline] is finite, </p></li><li value="2"><p>
[mathjaxinline]\, \displaystyle g(n)>0\,[/mathjaxinline] for all [mathjaxinline]\, n>N\,[/mathjaxinline] for some [mathjaxinline]\, N>0[/mathjaxinline] </p></li></ol><p>
then [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } f(n)\,[/mathjaxinline] and [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } g(n)\,[/mathjaxinline] either <b class="bf">both converge</b> or <b class="bf">both diverge</b>.<br/></p><p>
In other words, if [mathjaxinline]\, f(n)\,[/mathjaxinline] and [mathjaxinline]\, g(n)\,[/mathjaxinline] decay at the same rate as [mathjaxinline]\, n\,[/mathjaxinline] tends to [mathjaxinline]\, \infty ,\,[/mathjaxinline] then the series [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } f(n)\,[/mathjaxinline] and [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } g(n)\,[/mathjaxinline] converge or diverge together.<br/></p><p>
This is analogous to limit comparison for improper integrals.<br/></p><p><b class="bf">Note:</b> The condition [mathjaxinline]\, \displaystyle \frac{f(n)}{g(n)}\rightarrow c\neq 0\,[/mathjaxinline] is equivalent to </p><table id="a0000001318" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001319"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle f(n)\sim c g(n),\ \text {that is,}\, \displaystyle \frac{f(n)}{c g(n)}\rightarrow 1\, \, \text {as} \, n\, \rightarrow \, \infty .[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.336)</td></tr></table><p><b class="bfseries">More on limit comparison</b></p><p>
We can also use limit comparison in the following way. <br/></p><p>
Suppose [mathjaxinline]\, f(n), g(n)> 0[/mathjaxinline] for all [mathjaxinline]n\geq N[/mathjaxinline] for some large [mathjaxinline]\, N[/mathjaxinline].<br/></p><p>
If [mathjaxinline]\, \displaystyle \frac{f(n)}{g(n)}\rightarrow 0\,[/mathjaxinline] as [mathjaxinline]\, \displaystyle n\rightarrow \infty ,\,[/mathjaxinline] that is, [mathjaxinline]\, f(n)\,[/mathjaxinline] decays faster than [mathjaxinline]\, g(n),\,[/mathjaxinline] then </p><ul class="itemize"><li><p>
[mathjaxinline]\displaystyle \, \sum _{n=N}^{\infty } g(n)\,[/mathjaxinline] converges implies [mathjaxinline]\displaystyle \, \sum _{n=N}^{\infty } f(n)\,[/mathjaxinline] converges, </p></li><li><p>
[mathjaxinline]\displaystyle \, \sum _{n=N}^{\infty } f(n)\,[/mathjaxinline] diverges implies [mathjaxinline]\displaystyle \sum _{N}^{\infty } g(n)\,[/mathjaxinline] diverges. </p></li></ul><p><b class="bfseries">Absolute convergence versus conditional convergence</b></p><p>
So far, we have focused on series whose terms are positive (with the exception of the geometric series and the divergence test). For general series, including series with both positive and negative terms, there are two notions of convergence.<br/></p><p>
Consider the series </p><table id="a0000001320" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001321"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle S = \sum _{n=1}^{\infty } a_ n \qquad (a_ n \, \text {can be positive or negative}).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.337)</td></tr></table><p>
The series [mathjaxinline]S[/mathjaxinline] is <b class="bf"><span style="color:#27408C">absolutely convergent</span></b> if [mathjaxinline]\, \sum _{n=1}^{\infty } |a_ n |[/mathjaxinline] converges.<br/></p><p>
The series [mathjaxinline]S[/mathjaxinline] is <b class="bf"><span style="color:#27408C">conditionally convergent</span></b> if it converges but is <b class="bf">not absolutely convergent</b>.<br/></p><p>
For series with only positive terms, the two notions are the same. <br/></p><p>
In general, absolute convergence of a series implies convergence. <br/></p><p>
Because absolute convergence concerns the convergence of [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } \left|a_ n\right|,\,[/mathjaxinline] we can apply all of the techniques we have learned to determine absolute convergence. </p><p><b class="bfseries">Ratio test</b></p><p>
The <b class="bf"><span style="color:#27408C">ratio test</span></b> is another way to determine convergence of a series. </p><p>
Consider [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } a_ n\,[/mathjaxinline]. </p><p>
Define </p><table id="a0000001322" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001323"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle L=\lim _{n\rightarrow \infty }\left| \frac{a_{n+1}}{a_ n}\right|.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.338)</td></tr></table><p>
There are three cases: </p><ol class="enumerate"><li value="1"><p>
If [mathjaxinline]L<1,\,[/mathjaxinline] then the series <b class="bf">absolutely converges</b>; </p></li><li value="2"><p>
If [mathjaxinline]L>1,\,[/mathjaxinline] then the series <b class="bf">diverges</b>; </p></li><li value="3"><p>
If [mathjaxinline]L=1,\,[/mathjaxinline] then there is <b class="bf">no conclusion</b>. </p></li></ol><p>
When we talk about Taylor series, we will use the ratio test to find what is known as the radius of convergence.<br/></p><p><b class="bfseries">Root test</b></p><p>
The <b class="bf"><span style="color:#27408C">root test</span></b> is yet another test to determine the convergence of a series. </p><p>
Consider [mathjaxinline]\, \displaystyle \sum _{n=1}^{\infty } a_ n\,[/mathjaxinline]. </p><p>
Define </p><table id="a0000001324" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001325"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle L=\lim _{n\rightarrow \infty } \sqrt [n]{\left|a_ n\right|} =\lim _{n\rightarrow \infty } \left|a_ n\right|^{\frac{1}{n}}.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.339)</td></tr></table><p>
The conclusions are the same as for the ratio test, that is, </p><ol class="enumerate"><li value="1"><p>
If [mathjaxinline]L<1,\,[/mathjaxinline] then the series <b class="bf">absolutely converges</b>; </p></li><li value="2"><p>
If [mathjaxinline]L>1,\,[/mathjaxinline] then the series <b class="bf">diverges</b>; </p></li><li value="3"><p>
If [mathjaxinline]L=1,\,[/mathjaxinline] then there is <b class="bf">no conclusion</b>. </p></li></ol><p>
When using the root test, we often need to evaluate the [mathjaxinline]n[/mathjaxinline]-th root of expression. Here are some examples. <br/></p><ul class="itemize"><li><p>
For any constant [mathjaxinline]\, b>0,\,[/mathjaxinline] [mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } b^{\frac{1}{n}}\, =\, 1[/mathjaxinline] </p></li><li><p>
For any power [mathjaxinline]\, p>0,\,[/mathjaxinline] [mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } \left(n^ p\right)^{\frac{1}{n}}\, =\, 1[/mathjaxinline] </p></li><li><p>
[mathjaxinline]\displaystyle \lim _{n\rightarrow \infty } \left(n!\right)^{\frac{1}{n}}\, =\, \infty[/mathjaxinline] </p></li></ul><p>
To evaluate these limits, we use </p><table id="a0000001326" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001327"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \ln \left(\lim _{n\rightarrow \infty } b(n)\right)= \lim _{n\rightarrow \infty } \ln (b(n))[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.340)</td></tr></table><p>
provided [mathjaxinline]\, \displaystyle \lim _{n\rightarrow \infty } b(n)\,[/mathjaxinline] is positive or [mathjaxinline]\, +\infty[/mathjaxinline]. </p><p><b class="bfseries">Alternating series</b></p><p>
An <b class="bf"><span style="color:#27408C">alternating series</span></b> is a series whose terms alternate in signs. That is, an alternating series takes the form </p><table id="a0000001328" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001329"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle \pm \sum _{n=1}^{\infty } (-1)^ n c_ n \qquad \text {where}\, \, \, c_ n\geq 0.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.341)</td></tr></table><p><b class="bfseries">The alternating series test</b></p><p>
There is a simple test for convergence of an alternating series.<br/></p><p>
If for all [mathjaxinline]\, n\,[/mathjaxinline] large enough, [mathjaxinline]\quad \begin{cases} \displaystyle \lim _{n\rightarrow \infty } c_ n =0,& \\ c_ n\, \text { decreases as}\, \, n\, \, \text {increases},\\ \end{cases}[/mathjaxinline]<br/>then [mathjaxinline]\, \displaystyle \pm \sum _{n=1}^{\infty } (-1)^ n c_ n,\,[/mathjaxinline] where [mathjaxinline]c_ n\geq 0,\,[/mathjaxinline] converges.<br/></p><p><b class="bfseries">Examples of series</b></p><p>
In all the series below, the subscripts and superscripts of the summation notation is suppressed. That is, [mathjaxinline]\, \displaystyle \sum \,[/mathjaxinline] is the abbreviation of [mathjaxinline]\, \displaystyle \sum _{n=N}^{\infty }\,[/mathjaxinline] for some [mathjaxinline]N[/mathjaxinline].<br/></p><table class="tabular" cellspacing="0" style="table-layout:auto"><tr><td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \sum x^ n[/mathjaxinline]</td><td style="text-align:left; border:none">
(Geometric series)</td><td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \begin{cases} \text {converges absolutely}& \text {if } \, |x|<1\\ \text {diverges}& \text {if }\, |x|\geq 1\\ \end{cases}[/mathjaxinline]</td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \sum \frac1{n^ p}[/mathjaxinline]</td><td style="text-align:left; border:none">
([mathjaxinline]p[/mathjaxinline]-series)</td><td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \begin{cases} \text {converges}& \text {if } \, p>1\\ \text {diverges}& \text {if }\, p\leq 1\\ \end{cases}[/mathjaxinline]</td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \sum \frac1{n\left(\ln (n)\right)^ p}[/mathjaxinline]</td><td style="text-align:left; border:none"> </td><td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \begin{cases} \text {converges}& \text {if } \, p>1\\ \text {diverges}& \text {if }\, p\leq 1\\ \end{cases}[/mathjaxinline]</td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \sum \frac{\ln (n)}{n^ p}[/mathjaxinline]</td><td style="text-align:left; border:none"> </td><td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \begin{cases} \text {converges}& \text {if } \, p>1\\ \text {diverges}& \text {if }\, p\leq 1\\ \end{cases}[/mathjaxinline]</td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \sum \frac{x^ n}{n!}[/mathjaxinline]</td><td style="text-align:left; border:none"> </td><td style="text-align:left; border:none">
[mathjaxinline]\text {converges absolutely for all }\, x[/mathjaxinline]</td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \sum \frac{x^ n}{n^ p}[/mathjaxinline]</td><td style="text-align:left; border:none"> </td><td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \begin{cases} \text {converges absolutely for all } \, p & \text {if } \, |x|<1\\ \text {diverges for all } \, p& \text {if } \, |x|>1\\ \end{cases}[/mathjaxinline]</td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]\displaystyle \sum \frac{n!}{n^ n}[/mathjaxinline]</td><td style="text-align:left; border:none"> </td><td style="text-align:left; border:none">
[mathjaxinline]\text {converges}[/mathjaxinline]</td></tr></table>
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