009C Sample Final 2, Problem 2 - Revision history

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Created page with "<span class="exam"> For each of the following series, find the sum if it converges. If it diverges, explain why. <span class="exam">(a) <math style="vertical-align: -14..."

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<span class="exam"> For each of the following series, find the sum if it converges. If it diverges, explain why.

<span class="exam">(a)  <math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math>

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

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' The sum of a convergent geometric series is   <math>\frac{a}{1-r}</math>
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|        where  <math style="vertical-align: 0px">r</math>  is the ratio of the geometric series
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|        and  <math style="vertical-align: 0px">a</math>  is the first term of the series.
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|'''2.''' The  <math style="vertical-align: 0px">n</math>th partial sum,  <math style="vertical-align: -3px">s_n</math>  for a series  <math>\sum_{n=1}^\infty a_n </math>  is defined as
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|
        <math>s_n=\sum_{i=1}^n a_i.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -3px">a_n</math>  be the  <math style="vertical-align: 0px">n</math>th term of this sum.
|-
|We notice that
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|        <math>\frac{a_2}{a_1}=\frac{-2}{4},~\frac{a_3}{a_2}=\frac{1}{-2},</math>  and  <math>\frac{a_4}{a_2}=\frac{-1}{2}.</math>
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|So, this is a geometric series with  <math style="vertical-align: -14px">r=-\frac{1}{2}.</math>
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|Since  <math style="vertical-align: -5px">|r|<1,</math>  this series converges.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Hence, the sum of this geometric series is
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\frac{a_1}{1-r}} & = & \displaystyle{\frac{4}{1-(-\frac{1}{2})}}\\
&&\\
& = & \displaystyle{\frac{4}{\frac{3}{2}}}\\
&&\\
& = & \displaystyle{\frac{8}{3}.}
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by using partial fraction decomposition. Let
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|       <math>\frac{1}{(2x-1)(2x+1)}=\frac{A}{2x-1}+\frac{B}{2x+1}.</math>
|-
|If we multiply this equation by  <math style="vertical-align: -5px">(2x-1)(2x+1),</math>  we get
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|       <math>1=A(2x+1)+B(2x-1).</math>
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|If we let  <math style="vertical-align: -14px">x=\frac{1}{2},</math>  we get  <math style="vertical-align: -14px">A=\frac{1}{2}.</math>
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|If we let  <math style="vertical-align: -14px">x=-\frac{1}{2},</math>  we get  <math style="vertical-align: -14px">B=-\frac{1}{2}.</math>
|-
|So, we have
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|        <math>\begin{array}{rcl}
\displaystyle{\sum_{n=1}^\infty \frac{1}{(2n-1)(2n+1)}} & = & \displaystyle{\sum_{n=1}^\infty \frac{\frac{1}{2}}{2n-1}+\frac{-\frac{1}{2}}{2n+1}}\\
&&\\
& = & \displaystyle{\frac{1}{2} \sum_{n=1}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we look at the partial sums,  <math>s_n</math>  of this series.
|-
|First, we have
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|        <math>s_1=\frac{1}{2}\bigg(1-\frac{1}{3}\bigg).</math>
|-
|Also, we have
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|        <math>\begin{array}{rcl}
\displaystyle{s_2} & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{5}\bigg)}
\end{array}</math>
|-
|and
|-
|        <math>\begin{array}{rcl}
\displaystyle{s_3} & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{7}\bigg).}
\end{array}</math>
|-
|If we compare  <math>s_1,s_2,s_3,</math>  we notice a pattern.
|-
|We have
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|       <math>s_n=\frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, to calculate the sum of this series we need to calculate
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|       <math>\lim_{n\rightarrow \infty} s_n.</math>
|-
|We have
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|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} s_n} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{2}.}
\end{array}</math>
|-
|Since the partial sums converge, the series converges and the sum of the series is  <math style="vertical-align: -15px">\frac{1}{2}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>\frac{8}{3}</math>
|-
|   '''(b)'''    <math>\frac{1}{2}</math>
|}
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]