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

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(Created page with "<span class="exam">Determine convergence or divergence: ::<math>\sum_{n=1}^\infty \frac{3^n}{n}</math> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !F...")
 
 
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::<math>\sum_{n=1}^\infty \frac{3^n}{n}</math>
 
::<math>\sum_{n=1}^\infty \frac{3^n}{n}</math>
  
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[[009C Sample Midterm 2, Problem 2 Solution|'''<u>Solution</u>''']]
  
  
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[[009C Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
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!Foundations: &nbsp;
 
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|'''Direct Comparison Test'''
 
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|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math>\{a_n\}</math>&nbsp; and &nbsp;<math>\{b_n\}</math>&nbsp; be positive sequences where &nbsp;<math style="vertical-align: -3px">a_n\le b_n</math>
 
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|&nbsp; &nbsp; &nbsp; &nbsp; for all &nbsp;<math style="vertical-align: -3px">n\ge N</math>&nbsp; for some &nbsp;<math style="vertical-align: -3px">N\ge 1.</math>
 
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|&nbsp; &nbsp; &nbsp; &nbsp; '''1.''' If &nbsp;<math>\sum_{n=1}^\infty b_n</math>&nbsp; converges, then &nbsp;<math>\sum_{n=1}^\infty a_n</math>&nbsp; converges.
 
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|&nbsp; &nbsp; &nbsp; &nbsp; '''2.''' If &nbsp;<math>\sum_{n=1}^\infty a_n</math>&nbsp; diverges, then &nbsp;<math>\sum_{n=1}^\infty b_n</math>&nbsp; diverges.
 
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'''Solution:'''
 
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!Step 1: &nbsp;
 
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|First, we note that
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{3^n}{n}>0</math>
 
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|for all &nbsp;<math style="vertical-align: -3px">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 &nbsp;<math style="vertical-align: -13px">a_n=\frac{3^n}{n}.</math>
 
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!Step 2: &nbsp;
 
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|Let &nbsp;<math style="vertical-align: -14px">b_n=\frac{1}{n}.</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 \frac{1}{n}.</math>
 
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|This is the harmonic series (or &nbsp;<math style="vertical-align: -4px">p</math>-series with &nbsp;<math style="vertical-align: -4px">p=1.</math>&nbsp;)
 
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|Hence, &nbsp;<math>\sum_{n=1}^\infty b_n</math>&nbsp; diverges.
 
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!Step 3: &nbsp;
 
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|Also, we have &nbsp;<math style="vertical-align: -4px">b_n<a_n</math>&nbsp; since
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n}<\frac{3^n}{n}</math>
 
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| for all &nbsp;<math style="vertical-align: -3px">n\ge 1.</math>
 
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|Therefore, the series &nbsp;<math>\sum_{n=1}^\infty a_n</math>&nbsp; diverges
 
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|by the Direct Comparison Test.
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp; &nbsp; &nbsp; diverges (by the Direct Comparison Test)
 
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[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 11:37, 12 November 2017

Determine convergence or divergence:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{n=1}^{\infty }{\frac {3^{n}}{n}}}

Solution


Detailed Solution


Return to Sample Exam

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