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

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(Created page with "<span class="exam">Determine the convergence or divergence of the following series. <span class="exam"> Be sure to justify your answers! ::<math>\sum_{n=1}^\infty \frac{1}{n...")
 
 
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!Foundations: &nbsp;
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[[009C Sample Midterm 1, Problem 4 Solution|'''<u>Solution</u>''']]
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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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[[009C Sample Midterm 1, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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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{1}{n^23^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: -14px">a_n=\frac{1}{n^23^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^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 &nbsp;<math style="vertical-align: -4px">p</math>-series with &nbsp;<math style="vertical-align: -4px">p=2.</math>
 
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|Hence, &nbsp;<math>\sum_{n=1}^\infty b_n</math>&nbsp; converges.
 
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!Step 3: &nbsp;
 
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|Also, we have &nbsp;<math style="vertical-align: -4px">a_n<b_n</math>&nbsp; since
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n^23^n}<\frac{1}{n^2}</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; converges
 
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|by the Direct Comparison Test.
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp; &nbsp; &nbsp; converges (by the Direct Comparison Test)
 
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[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 22:58, 11 November 2017

Determine the convergence or divergence of the following series.

Be sure to justify your answers!

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 \frac{1}{n^23^n}}


Solution


Detailed Solution


Return to Sample Exam

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