MathAdmin: Created page with " Consider the power series :: $\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}$ (a) Find the radius of convergence of the above..."

2017-03-19T18:52:54Z

Created page with " Consider the power series ::<math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}</math> (a) Find the radius of convergence of the above..."

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Consider the power series

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

(a) Find the radius of convergence of the above power series.

(b) Find the interval of convergence of the above power series.

(c) Find the closed formula for the function  <math style="vertical-align: -5px">f(x)</math>  to which the power series converges.

(d) Does the series

::<math>\sum_{n=0}^\infty \frac{1}{(n+1)3^{n+1}}</math>

converge?

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''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,
|-
|
        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.''' '''Direct Comparison Test'''
|-
|        Let  <math>\{a_n\}</math>  and  <math>\{b_n\}</math>  be positive sequences where  <math style="vertical-align: -3px">a_n\le b_n</math>
|-
|        for all  <math style="vertical-align: -3px">n\ge N</math>  for some  <math style="vertical-align: -3px">N\ge 1.</math>
|-
|        '''1.''' If  <math>\sum_{n=1}^\infty b_n</math>  converges, then  <math>\sum_{n=1}^\infty a_n</math>  converges.
|-
|        '''2.''' If  <math>\sum_{n=1}^\infty a_n</math>  diverges, then  <math>\sum_{n=1}^\infty b_n</math>  diverges.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We 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)^{n+2}}{(n+2)}\frac{n+1}{(-1)^n(x)^{n+1}}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x)\frac{n+1}{n+2}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} |x|\frac{n+1}{n+2}}\\
&&\\
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \frac{n+1}{n+2}}\\
&&\\
& = & \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: -1px">R=1.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|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: -1px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|First, let  <math style="vertical-align: -1px">x=1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty (-1)^n \frac{1}{n+1}.</math>
|-
|This is an alternating series.
|-
|Let  <math style="vertical-align: -15px">b_n=\frac{1}{n+1}.</math>.
|-
|First, we have
|-
|       <math>\frac{1}{n+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}{n+2}<\frac{1}{n+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}{n+1}=0.</math>
|-
|Therefore, this series converges by the Alternating Series Test
|-
|and we include  <math style="vertical-align: -1px">x=1</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, let  <math style="vertical-align: -1px">x=-1.</math>
|-
|Then, the series becomes
|-
|        <math>\begin{array}{rcl}
\displaystyle{\sum_{n=0}^\infty \frac{(-1)^{2n+1}}{n+1}} & = & \displaystyle{\sum_{n=1}^\infty \frac{-1}{n+1}}\\
&&\\
& = & \displaystyle{(-1)\sum_{n=1}^\infty \frac{1}{n+1}.}
\end{array}</math>
|-
|Now, we note that
|-
|        <math>\frac{1}{n+1}>0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|This means that we can use the limit comparison test on this series.
|-
|Let  <math style="vertical-align: -16px">a_n=\frac{1}{n+1}.</math>
|-
|Let  <math style="vertical-align: -14px">b_n=\frac{1}{n}.</math>
|-
|Then,  <math style="vertical-align: -20px">\sum_{n=1}^\infty b_n</math>  diverges since it is the harmonic series.
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \frac{a_n}{b_n}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(\frac{1}{n+1})}{(\frac{1}{n})}}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{n+1}}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} 1.}
\end{array}</math>
|-
|Therefore, the series
|-
|        <math>\sum_{n=1}^{\infty} \frac{1}{n+1}</math>
|-
|diverges by the Limit Comparison Test.
|-
|Therefore, we do not include  <math style="vertical-align: -1px">x=-1</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|The interval of convergence is  <math style="vertical-align: -4px">(-1,1].</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let
|-
|       <math>f(x)=\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.</math>
|-
|Then,
|-
|        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\frac{d}{dx}\bigg(\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}\bigg)}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{d}{dx}\bigg( (-1)^n \frac{x^{n+1}}{n+1}\bigg)}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty (-1)^n x^n}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty (-x)^n}\\
&&\\
& = & \displaystyle{\frac{1}{1-(-x)}}\\
&&\\
& = & \displaystyle{\frac{1}{1+x}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Thus,
|-
|        <math>\begin{array}{rcl}
\displaystyle{f(x)} & = & \displaystyle{\int \frac{1}{1+x}~dx}\\
&&\\
& = & \displaystyle{\ln(1+x)+C.}
\end{array}</math>
|-
|Since there is no constant term in the series <math style="vertical-align: -20px">\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1},</math>
|-
|       <math>C=0.</math>
|-
|Hence,
|-
|        <math>f(x)=\ln(1+x).</math>
|}

'''(d)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we note that
|-
|        <math>\frac{1}{(n+1)3^{n+1}}>0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|This means that we can use a comparison test on this series.
|-
|Let  <math style="vertical-align: -19px">a_n=\frac{1}{(n+1)3^{n+1}}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Let  <math style="vertical-align: -14px">b_n=\frac{1}{3^{n+1}}.</math>
|-
|We want to compare the series in this problem with
|-
|        <math>\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \frac{1}{3}\bigg(\frac{1}{3}\bigg)^n.</math>
|-
|This is a geometric series with  <math style="vertical-align: -13px">r=\frac{1}{3}.</math>
|-
|Since   <math>|r|<1,</math>  the series  <math style="vertical-align: -20px">\sum_{n=1}^\infty b_n</math>  converges.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Also, we have  <math style="vertical-align: -4px">a_n<b_n</math>  since
|-
|        <math>\frac{1}{(n+1)3^{n+1}}<\frac{1}{3^{n+1}}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|Therefore, the series  <math>\sum_{n=1}^\infty a_n</math>  converges
|-
|by the Direct Comparison Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     The radius of convergence is  <math style="vertical-align: -1px">R=1.</math>
|-
|    '''(b)'''     <math>(-1,1]</math>
|-
|    '''(c)'''     <math>f(x)=\ln(1+x)</math>
|-
|    '''(d)'''     converges (by the Direct Comparison Test)
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3, Problem 6 - Revision history

MathAdmin: Created page with " Consider the power series ::\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} (a) Find the radius of convergence of the above..."

MathAdmin: Created page with " Consider the power series :: $\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}$ (a) Find the radius of convergence of the above..."