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2016-04-19T01:21:01Z

Created page with "<span class="exam"> Find the interval of convergence of the following series. ::::::<math>\sum_{n=0}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math> {| class="mw-collapsible mw-c..."

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<span class="exam"> Find the interval of convergence of the following series.

::::::<math>\sum_{n=0}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
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|Recall:
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::'''1. Ratio Test''' Let <math style="vertical-align: -7px">\sum a_n</math> be a series and <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math> Then,
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:::If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.
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:::If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
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:::If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
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::'''2.''' After you find the radius of convergence, you need to check the endpoints of your interval
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:::for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math>
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'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|We proceed using the ratio test to find the interval of convergence. So, we have
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::<math>\begin{array}{rcl}
\displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(-1)^{n+1}(x+2)^{n+1}}{(n+1)^2}}\frac{n^2}{(-1)^n(x+2)^n}\bigg|\\
&&\\
& = & \displaystyle{|x+2|\lim_{n \rightarrow \infty}\frac{n^2}{(n+1)^2}}\\
&&\\
& = & \displaystyle{|x+2|\lim_{n \rightarrow \infty}\bigg(\frac{n}{n+1}\bigg)^2}\\
&&\\
& = & \displaystyle{|x+2|\bigg(\lim_{n \rightarrow \infty}\frac{n}{n+1}\bigg)^2}\\
&&\\
& = & \displaystyle{|x+2|(1)^2}\\
&&\\
& = & \displaystyle{|x+2|.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|So, we have <math style="vertical-align: -6px">|x+2|<1.</math> Hence, our interval is <math style="vertical-align: -3px">(-3,-1).</math> But, we still need to check the endpoints of this interval
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|to see if they are included in the interval of convergence.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
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|First, we let <math style="vertical-align: -1px">x=-1.</math> Then, our series becomes
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::<math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}.</math>
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|Since <math style="vertical-align: -5px">n^2<(n+1)^2,</math> we have <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}.</math> Thus, <math>\frac{1}{n^2}</math> is decreasing.
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|So, <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}</math> converges by the Alternating Series Test.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
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|Now, we let <math style="vertical-align: -1px">x=-3.</math> Then, our series becomes
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::<math>\begin{array}{rcl}
\displaystyle{\sum_{n=0}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=0}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\
&&\\
& = & \displaystyle{\sum_{n=0}^{\infty} \frac{1}{n^2}.}\\
\end{array}</math>
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|This is a convergent series by the p-test.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 5:  
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|Thus, the interval of convergence for this series is <math>[-3,-1].</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|   <math>[-3,-1]</math>
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[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

009C Sample Final 1, Problem 4 - Revision history

MathAdmin: Created page with " Find the interval of convergence of the following series. ::::::\sum_{n=0}^{\infty} (-1)^n \frac{(x+2)^n}{n^2} {| class="mw-collapsible mw-c..."

MathAdmin: Created page with " Find the interval of convergence of the following series. :::::: $\sum_{n=0}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}$ {| class="mw-collapsible mw-c..."