MathAdmin at 19:37, 12 November 2017

2017-11-12T19:37:40Z

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2017-04-09T19:30:12Z

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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<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;"
!Foundations:  
|-
|'''Direct Comparison Test'''
|-
|        Let  <math>\{a_n\}</math>  and  <math>\{b_n\}</math>  be positive sequences where  <math style="vertical-align: -3px">a_n\le b_n</math>
|-
|        for all  <math style="vertical-align: -3px">n\ge N</math>  for some  <math style="vertical-align: -3px">N\ge 1.</math>
|-
|        '''1.''' If  <math>\sum_{n=1}^\infty b_n</math>  converges, then  <math>\sum_{n=1}^\infty a_n</math>  converges.
|-
|        '''2.''' If  <math>\sum_{n=1}^\infty a_n</math>  diverges, then  <math>\sum_{n=1}^\infty b_n</math>  diverges.
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we note that
|-
|        <math>\frac{3^n}{n}>0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|This means that we can use a comparison test on this series.
|-
|Let  <math style="vertical-align: -13px">a_n=\frac{3^n}{n}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Let  <math style="vertical-align: -14px">b_n=\frac{1}{n}.</math>
|-
|We want to compare the series in this problem with
|-
|        <math>\sum_{n=1}^\infty \frac{1}{n}.</math>
|-
|This is the harmonic series (or  <math style="vertical-align: -4px">p</math>-series with  <math style="vertical-align: -4px">p=1.</math> )
|-
|Hence,  <math>\sum_{n=1}^\infty b_n</math>  diverges.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Also, we have  <math style="vertical-align: -4px">b_n<a_n</math>  since
|-
|        <math>\frac{1}{n}<\frac{3^n}{n}</math>
|-
| for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|Therefore, the series  <math>\sum_{n=1}^\infty a_n</math>  diverges
|-
|by the Direct Comparison Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|        diverges (by the Direct Comparison Test)
|}
[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

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 ::<math>\sum_{n=1}^\infty \frac{3^n}{n}</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009C Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

009C Sample Midterm 2, Problem 2 - Revision history

MathAdmin at 19:37, 12 November 2017

MathAdmin: Created page with "Determine convergence or divergence: ::\sum_{n=1}^\infty \frac{3^n}{n} {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !F..."

MathAdmin: Created page with "Determine convergence or divergence: :: $\sum_{n=1}^\infty \frac{3^n}{n}$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !F..."