MathAdmin: Created page with " Consider the series :: $\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}}.$ (a) Test if the series converges absolutely. Give reason..."

2017-03-19T18:40:48Z

Created page with " Consider the series ::<math>\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}}.</math> (a) Test if the series converges absolutely. Give reason..."

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Consider the series

::<math>\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}}.</math>

(a) Test if the series converges absolutely. Give reasons for your answer.

(b) Test if the series converges conditionally. Give reasons for your answer.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' A series  <math>\sum a_n</math>  is '''absolutely convergent''' if
|-
|        the series  <math>\sum |a_n|</math>  converges.
|-
|'''2.''' A series  <math>\sum a_n</math>  is '''conditionally convergent''' if
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|        the series  <math>\sum |a_n|</math>  diverges and the series  <math>\sum a_n</math>  converges.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we take the absolute value of the terms in the original series.
|-
|Let  <math style="vertical-align: -20px">a_n=\frac{(-1)^n}{\sqrt{n}}.</math>
|-
|Therefore,
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|        <math>\begin{array}{rcl}
\displaystyle{\sum_{n=1}^\infty |a_n|} & = & \displaystyle{\sum_{n=1}^\infty \bigg|\frac{(-1)^n}{\sqrt{n}}\bigg|}\\
&&\\
& = & \displaystyle{\sum_{n=1}^\infty \frac{1}{\sqrt{n}}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|This series is a  <math style="vertical-align: -5px">p</math>-series with  <math style="vertical-align: -4px">p=1/2.</math> 
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|Therefore, it diverges.
|-
|Hence, the series
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|        <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math>
|-
|is not absolutely convergent.
|-
|
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|For
|-
|        <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}},</math>
|-
|we notice that this series is alternating.
|-
|Let  <math style="vertical-align: -20px"> b_n=\frac{1}{\sqrt{n}}.</math>
|-
|First, we have
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|       <math>\frac{1}{\sqrt{n}}\ge 0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|The sequence  <math style="vertical-align: -4px">\{b_n\}</math>  is decreasing since
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|        <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|Also,
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|        <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}=0.</math>
|-
|Therefore, the series  <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math>   converges
|-
|by the Alternating Series Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since the series  <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math>   is not absolutely convergent but convergent,
|-
|this series is conditionally convergent.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    not absolutely convergent (by the  <math style="vertical-align: -5px">p</math>-series test)
|-
|   '''(b)'''    conditionally convergent (by the Alternating Series Test)
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3, Problem 2 - Revision history

MathAdmin: Created page with " Consider the series ::\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}}. (a) Test if the series converges absolutely. Give reason..."

MathAdmin: Created page with " Consider the series :: $\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}}.$ (a) Test if the series converges absolutely. Give reason..."