Difference between revisions of "009C Sample Midterm 1, Problem 4"

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!Foundations:    
 
!Foundations:    
 
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|'''Direct Comparison Test'''
 
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'''Solution:'''
 
'''Solution:'''
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!Step 1:    
 
!Step 1:    
 
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|First, we note that
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n^23^n}>0</math>
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|for all <math>n\ge 1.</math>
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|This means that we can use a comparison test on this series.
 
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|Let <math>a_n=\frac{1}{n^23^n}.</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Let <math>b_n=\frac{1}{n^2}.</math>
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|We want to compare the series in this problem with
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \frac{1}{n^2}.</math>
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|This is a <math>p</math>-series with <math>p=2.</math>
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|Hence, <math>\sum_{n=1}^\infty b_n</math> converges.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 3: &nbsp;
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|-
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|Also, we have <math>a_n<b_n</math> since
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n^23^n}<\frac{1}{n^2}</math>
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|-
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|for all <math>n\ge 1.</math>
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|-
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|Therefore, the series <math>\sum_{n=1}^\infty a_n</math> converges
 
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|by the Direct Comparison Test.
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|&nbsp; &nbsp; &nbsp; &nbsp; converges
 
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[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 16:22, 12 February 2017

Determine the convergence or divergence of the following series.

Be sure to justify your answers!


Foundations:  
Direct Comparison Test


Solution:

Step 1:  
First, we note that
       
for all
This means that we can use a comparison test on this series.
Let
Step 2:  
Let
We want to compare the series in this problem with
       
This is a -series with
Hence, converges.
Step 3:  
Also, we have since
       
for all
Therefore, the series converges
by the Direct Comparison Test.


Final Answer:  
        converges

Return to Sample Exam