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) &nbsp; <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) &nbsp; <math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)</math>
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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<hr>
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!Foundations: &nbsp;
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[[009C Sample Final 1, Problem 2 Solution|'''<u>Solution</u>''']]
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|Review geometric series.
 
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'''Solution:'''
 
  
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'''(a)'''
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[[009C Sample Final 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|First, we write
 
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::<math>\begin{array}{rcl}
 
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\displaystyle{\sum_{n=0}^{\infty} (-2)^n e^{-n}} & = & \displaystyle{\sum_{n=0}^{\infty} \frac{(-2)^n}{e^n}}\\
 
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&&\\
 
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& = & \displaystyle{\sum_{n=0}^{\infty} \bigg(\frac{-2}{e}\bigg)^n}\\
 
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\end{array}</math>
 
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!Step 2: &nbsp;
 
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|Since <math>2<e</math>, <math>\bigg|-\frac{2}{e}\bigg|<1</math>. So,
 
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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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'''(b)'''
 
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!Step 1: &nbsp;
 
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!Step 2: &nbsp;
 
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!Step 3: &nbsp;
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Final Answer: &nbsp;
 
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|'''(a)''' <math>\frac{e}{e+2}</math>
 
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|'''(b)'''
 
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[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 17:57, 2 December 2017

Find the sum of the following series:

(a)  

(b)  

Solution


Detailed Solution


Return to Sample Exam

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