009C Sample Midterm 2, Problem 4 - Revision history

MathAdmin: Replaced content with " Find the radius of convergence and interval of convergence of the series. (a) $\sum_{n=1}^\infty n^nx^n$
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MathAdmin at 17:31, 14 April 2017

2017-04-14T17:31:35Z

MathAdmin: Created page with " Find the radius of convergence and interval of convergence of the series. (a) $\sum_{n=1}^\infty n^nx^n$
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Created page with "<span class="exam"> Find the radius of convergence and interval of convergence of the series. <span class="exam">(a) <math>\sum_{n=1}^\infty n^nx^n</math> <span class=..."

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

<span class="exam">(a)  <math>\sum_{n=1}^\infty n^nx^n</math>

<span class="exam">(b)  <math>\sum_{n=1}^\infty \frac{(x+1)^n}{\sqrt{n}}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' '''Root Test'''
|-
|        Let  <math>\{a_n\}</math>  be a positive sequence and let  <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} |a_n|^{\frac{1}{n}}=L.</math>
|-
|        Then,
|-
|        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|-
|'''2.''' '''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,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by applying the Root Test.
|-
|We have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \sqrt[n]{|a_n|}} & = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt[n]{|n^nx^n|}}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} |n^nx^n|^{\frac{1}{n}}}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} |nx|}\\
&&\\
& = & \displaystyle{n|x|}\\
&&\\
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} n}\\
&&\\
& = & \displaystyle{\infty.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|This means that as long as  <math style="vertical-align: -6px">x\ne 0,</math>  this series diverges.
|-
|Hence, the radius of convergence is  <math style="vertical-align: -1px">R=0</math>  and
|-
|the interval of convergence is  <math style="vertical-align: -5px">\{0\}.</math>
|-
|
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We first use the Ratio Test to determine the radius of convergence.
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(x+1)^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{(x+1)^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| (x+1)\frac{\sqrt{n}}{\sqrt{n+1}}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} |x+1|\frac{\sqrt{n}}{\sqrt{n+1}}}\\
&&\\
& = & \displaystyle{|x+1|\lim_{n\rightarrow \infty} \sqrt{\frac{n}{n+1}}}\\
&&\\
& = & \displaystyle{|x+1|\sqrt{\lim_{n\rightarrow \infty} \frac{n}{n+1}}}\\
&&\\
& = & \displaystyle{|x+1|\sqrt{1}}\\
&&\\
&=& \displaystyle{|x+1|.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|The Ratio Test tells us this series is absolutely convergent if  <math style="vertical-align: -4px">|x+1|<1.</math>
|-
|Hence, the Radius of Convergence of this series is  <math style="vertical-align: -1px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we need to determine the interval of convergence.
|-
|First, note that  <math style="vertical-align: -4px">|x+1|<1</math>  corresponds to the interval  <math style="vertical-align: -4px">(-2,0).</math>
|-
|To obtain the interval of convergence, we need to test the endpoints of this interval
|-
|for convergence since the Ratio Test is inconclusive when  <math style="vertical-align: -1px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|First, let  <math style="vertical-align: -1px">x=0.</math>
|-
|Then, the series becomes  <math>\sum_{n=1}^\infty \frac{1}{\sqrt{n}}.</math>
|-
|We note that this is a  <math style="vertical-align: -3px">p</math>-series with  <math style="vertical-align: -12px">p=\frac{1}{2}.</math>
|-
|Since  <math>p<1,</math>  the series diverges.
|-
|Hence, we do not include  <math style="vertical-align: -1px">x=0</math>  in the interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 5:  
|-
|Now, let  <math style="vertical-align: -1px">x=-2.</math>
|-
|Then, the series becomes  <math>\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
|-
|This series is alternating.
|-
|Let  <math style="vertical-align: -20px">b_n=\frac{1}{\sqrt{n}}.</math>
|-
|First, we have
|-
|       <math>\frac{1}{\sqrt{n}}\ge 0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|The sequence  <math>\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{\sqrt{n}}=0.</math>
|-
|Therefore, the series converges by the Alternating Series Test.
|-
|Hence, we include  <math style="vertical-align: -1px">x=-2</math>  in our interval of convergence.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 6:  
|-
|The interval of convergence is  <math style="vertical-align: -4px">[-2,0).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     The radius of convergence is  <math style="vertical-align: -1px">R=0</math>  and the interval of convergence is  <math>\{0\}.</math>
|-
|    '''(b)'''     The radius of convergence is  <math style="vertical-align: -1px">R=1</math>  and the interval of convergence is  <math style="vertical-align: -4px">[-2,0).</math>
|}
[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]