Difference between revisions of "009C Sample Midterm 2, Problem 2"

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!Foundations:    
 
!Foundations:    
 
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| Direct Comparison Test
 
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!Step 1:    
 
!Step 1:    
 
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|-
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|First, we note that
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{3^n}{n}>0</math>
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|-
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|for all <math>n\ge 1.</math>
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|-
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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{3^n}{n}.</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|Let <math>b_n=\frac{1}{n}.</math>
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|-
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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 \frac{1}{n}.</math>
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|-
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|This is the harmonic series (or <math>p</math>-series with <math>p=1.</math>)
 
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|Hence, <math>\sum_{n=1}^\infty b_n</math> diverges.
 
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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>b_n<a_n</math> since
 +
|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n}<\frac{3^n}{n}</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> diverges
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|-
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|by the Direct Comparison Test.
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|}
  
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|&nbsp; &nbsp; &nbsp; &nbsp; diverges
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[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 11:06, 13 February 2017

Determine convergence or divergence:


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 the harmonic series (or -series with )
Hence, diverges.
Step 3:  
Also, we have since
       
for all
Therefore, the series diverges
by the Direct Comparison Test.


Final Answer:  
        diverges

Return to Sample Exam