009C Sample Midterm 1, Problem 5 - Revision history

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MathAdmin: Created page with " Find the radius of convergence and interval of convergence of the series. (a) $\sum_{n=0}^\infty \sqrt{n}x^n$
2017-04-09T19:27:57Z

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=0}^\infty \sqrt{n}x^n</math> <span c..."

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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=0}^\infty \sqrt{n}x^n</math>

<span class="exam">(b)  <math>\sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''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 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{\sqrt{n+1}x^{n+1}}{\sqrt{n}x^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}x\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}|x|}\\
&&\\
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\
&&\\
& = & \displaystyle{|x|\sqrt{\lim_{n\rightarrow \infty} \frac{n+1}{n}}}\\
&&\\
& = & \displaystyle{|x|\sqrt{1}}\\
&&\\
&=& \displaystyle{|x|.}
\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: -5px">|x|<1.</math>
|-
|Hence, the Radius of Convergence of this series is  <math style="vertical-align: -2px">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: -5px">|x|<1</math>  corresponds to the interval  <math style="vertical-align: -4px">(-1,1).</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: -2px">L=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|First, let  <math style="vertical-align: -1px">x=1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty \sqrt{n}.</math>
|-
|We note that
|-
|        <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty.</math>
|-
|Therefore, the series diverges by the  <math style="vertical-align: 0px">n</math>th term test.
|-
|Hence, we do not include  <math style="vertical-align: -1px">x=1</math>  in the interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 5:  
|-
|Now, let  <math style="vertical-align: -1px">x=-1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty (-1)^n \sqrt{n}.</math>
|-
|Since  <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty,</math>
|-
|we have
|-
|        <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=\text{DNE}.</math>
|-
|Therefore, the series diverges by the  <math style="vertical-align: 0px">n</math>th term test.
|-
|Hence, we do not include  <math style="vertical-align: -1px">x=-1 </math>  in the interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 6:  
|-
|The interval of convergence is  <math style="vertical-align: -4px">(-1,1).</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{(-1)^{n+1}(x-3)^{n+1}}{2(n+1)+1}\frac{2n+1}{(-1)^n(x-3)^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x-3)\frac{2n+1}{2n+3}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} |x-3|\frac{2n+1}{2n+3}}\\
&&\\
& = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\
&&\\
& = & \displaystyle{|x-3|.}
\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: -5px">|x-3|<1.</math>
|-
|Hence, the Radius of Convergence of this series is  <math style="vertical-align: -2px">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: -5px">|x-3|<1</math>  corresponds to the interval  <math style="vertical-align: -4px">(2,4).</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: -2px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|First, let  <math style="vertical-align: -1px">x=4.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty (-1)^n \frac{1}{2n+1}.</math>
|-
|This is an alternating series.
|-
|Let  <math style="vertical-align: -16px">b_n=\frac{1}{2n+1}.</math>.
|-
|First, we have
|-
|       <math>\frac{1}{2n+1}\ge 0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|The sequence  <math>\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{2(n+1)+1}<\frac{1}{2n+1}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{2n+1}=0.</math>
|-
|Therefore, this series converges by the Alternating Series Test
|-
|and we include  <math style="vertical-align: -1px">x=4</math>  in our 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=0}^\infty \frac{1}{2n+1}.</math>
|-
|First, we note that  <math>\frac{1}{2n+1}>0</math>  for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|Thus, we can use the Limit Comparison Test.
|-
|We compare this series with the series  <math>\sum_{n=1}^\infty \frac{1}{n},</math>
|-
|which is the harmonic series and divergent.
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \frac{\frac{1}{2n+1}}{\frac{1}{n}}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{2n+1}}\\
&&\\
& = & \displaystyle{\frac{1}{2}.}
\end{array}</math>
|-
|Since this limit is a finite number greater than zero,
|-
|       <math>\sum_{n=0}^\infty \frac{1}{2n+1}</math> 
|-
|diverges by the Limit Comparison Test.
|-
|Therefore, we do not include  <math style="vertical-align: -1px">x=2</math>  in our interval.
|}

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

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     The radius of convergence is  <math style="vertical-align: -2px">R=1</math>  and the interval of convergence is  <math style="vertical-align: -4px">(-1,1).</math>
|-
|    '''(b)'''     The radius of convergence is  <math style="vertical-align: -2px">R=1</math>  and the interval of convergence is  <math style="vertical-align: -4px">(2,4].</math>
|}
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]