Difference between revisions of "009C Sample Final 1, Problem 2"

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<span class="exam"> Find the sum of the following series:  
 
<span class="exam"> Find the sum of the following series:  
  
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<span class="exam">a) <math>\sum_{n=0}^{\infty} (-2)^ne^{-n}</math>
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::<span class="exam">a) <math>\sum_{n=0}^{\infty} (-2)^ne^{-n}</math>
  
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<span class="exam">b) <math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)</math>
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::<span class="exam">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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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|Recall:
 
|Recall:
 
|-
 
|-
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|'''1.''' For a geometric series <math>\sum_{n=0}^{\infty} ar^n</math> with <math>|r|<1,</math>  
+
|
 +
::'''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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:::<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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|
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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>
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:::and then calculate <math style="vertical-align: -14px">\lim_{k\rightarrow\infty} s_k.</math>
 
|}
 
|}
  
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|-
 
|-
 
|
 
|
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::<math>\sum_{n=0}^{\infty} (-2)^ne^{-n}=\frac{1}{1+\frac{2}{e}}=\frac{1}{\frac{e+2}{e}}=\frac{e}{e+2}.</math>
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::<math>\begin{array}{rcl}
 +
\displaystyle{\sum_{n=0}^{\infty} (-2)^ne^{-n}} & = & \displaystyle{\frac{1}{1+\frac{2}{e}}}\\
 +
&&\\
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& = & \displaystyle{\frac{1}{\frac{e+2}{e}}}\\
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&&\\
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& = & \displaystyle{\frac{e}{e+2}.}\\
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\end{array}</math>
 
|}
 
|}
  
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|Let <math style="vertical-align: -20px">s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
 
|Let <math style="vertical-align: -20px">s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
 
|-
 
|-
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|Then, <math style="vertical-align: -14px">s_k=\frac{1}{2}-\frac{1}{2^{k+1}}.</math>
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|Then,  
 +
|-
 +
|
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::<math style="vertical-align: -14px">s_k=\frac{1}{2}-\frac{1}{2^{k+1}}.</math>
 
|}
 
|}
  
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|-
 
|-
 
|
 
|
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::<math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)=\lim_{k\rightarrow \infty} s_k=\lim_{k\rightarrow \infty}\frac{1}{2}-\frac{1}{2^{k+1}}=\frac{1}{2}.</math>
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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}}}\\
 +
&&\\
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& = & \displaystyle{\frac{1}{2}.}\\
 +
\end{array}</math>
 
|}
 
|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
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|'''(a)''' <math>\frac{e}{e+2}</math>
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|&nbsp;&nbsp; '''(a)''' <math>\frac{e}{e+2}</math>
 
|-
 
|-
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|'''(b)''' <math>\frac{1}{2}</math>
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|&nbsp;&nbsp; '''(b)''' <math>\frac{1}{2}</math>
 
|}
 
|}
 
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 17:26, 18 April 2016

Find the sum of the following series:

a)
b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)}
Foundations:  
Recall:
1. For a geometric series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^{\infty} ar^n} with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |r|<1,}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}.}
2. For a telescoping series, we find the sum by first looking at the partial sum Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_k}
and then calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{k\rightarrow\infty} s_k.}

Solution:

(a)

Step 1:  
First, we write
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}
Step 2:  
Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2<e,~\bigg|-\frac{2}{e}\bigg|<1.} So,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

(b)

Step 1:  
This is a telescoping series. First, we find the partial sum of this series.
Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).}
Then,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_k=\frac{1}{2}-\frac{1}{2^{k+1}}.}
Step 2:  
Thus,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}
Final Answer:  
   (a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{e}{e+2}}
   (b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}}

Return to Sample Exam

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