MathAdmin: Created page with " Find the sum of the following series: ::a) $\sum_{n=0}^{\infty} (-2)^ne^{-n}$ ::b) $\sum_{n=1}^{\i..."$

2016-04-19T01:14:29Z

Created page with " Find the sum of the following series: ::a) <math>\sum_{n=0}^{\infty} (-2)^ne^{-n}</math> ::b) <math>\sum_{n=1}^{\i..."

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Find the sum of the following series:

::a) <math>\sum_{n=0}^{\infty} (-2)^ne^{-n}</math>

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

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
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|
::'''1.''' For a geometric series <math>\sum_{n=0}^{\infty} ar^n</math> with <math>|r|<1,</math>
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:::<math>\sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}.</math>
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|
::'''2.''' For a telescoping series, we find the sum by first looking at the partial sum <math style="vertical-align: -3px">s_k</math>
|-
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:::and then calculate <math style="vertical-align: -14px">\lim_{k\rightarrow\infty} s_k.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
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|
::<math>\begin{array}{rcl}
\displaystyle{\sum_{n=0}^{\infty} (-2)^n e^{-n}} & = & \displaystyle{\sum_{n=0}^{\infty} \frac{(-2)^n}{e^n}}\\
&&\\
& = & \displaystyle{\sum_{n=0}^{\infty} \bigg(\frac{-2}{e}\bigg)^n.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since <math style="vertical-align: -16px">2<e,~\bigg|-\frac{2}{e}\bigg|<1.</math> So,
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|
::<math>\begin{array}{rcl}
\displaystyle{\sum_{n=0}^{\infty} (-2)^ne^{-n}} & = & \displaystyle{\frac{1}{1+\frac{2}{e}}}\\
&&\\
& = & \displaystyle{\frac{1}{\frac{e+2}{e}}}\\
&&\\
& = & \displaystyle{\frac{e}{e+2}.}\\
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|This is a telescoping series. First, we find the partial sum of this series.
|-
|Let <math style="vertical-align: -20px">s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
|-
|Then,
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::<math style="vertical-align: -14px">s_k=\frac{1}{2}-\frac{1}{2^{k+1}}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Thus,
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|
::<math>\begin{array}{rcl}
\displaystyle{\sum_{n=1}^{\infty}\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)} & = & \displaystyle{\lim_{k\rightarrow \infty} s_k}\\
&&\\
& = & \displaystyle{\lim_{k\rightarrow \infty}\frac{1}{2}-\frac{1}{2^{k+1}}}\\
&&\\
& = & \displaystyle{\frac{1}{2}.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|   '''(a)''' <math>\frac{e}{e+2}</math>
|-
|   '''(b)''' <math>\frac{1}{2}</math>
|}
[[009C_Sample_Final_1|'''Return to Sample Exam''']]

009C Sample Final 1, Problem 2 - Revision history

MathAdmin: Created page with " Find the sum of the following series: ::a) \sum_{n=0}^{\infty} (-2)^ne^{-n} ::b) \sum_{n=1}^{\i..."

MathAdmin: Created page with " Find the sum of the following series: ::a) $\sum_{n=0}^{\infty} (-2)^ne^{-n}$ ::b) $\sum_{n=1}^{\i..."$