009C Sample Midterm 1, Problem 3 - Revision history

MathAdmin at 05:56, 12 November 2017

2017-11-12T05:56:16Z

MathAdmin at 17:28, 14 April 2017

2017-04-14T17:28:13Z

MathAdmin: Created page with " Determine whether the following series converges absolutely, conditionally or whether it diverges. Be sure to ju..."

2017-04-09T19:27:05Z

Created page with " Determine whether the following series converges absolutely, conditionally or whether it diverges. Be sure to ju..."

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Determine whether the following series converges absolutely,

 conditionally or whether it diverges.

 Be sure to justify your answers!

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

{| 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
|-
|        the series  <math>\sum |a_n|</math>  diverges and the series  <math>\sum a_n</math>  converges.
|}

'''Solution:'''

{| 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: -14px">a_n=\frac{(-1)^n}{n}.</math>
|-
|Therefore,
|-
|        <math>\begin{array}{rcl}
\displaystyle{\sum_{n=1}^\infty |a_n|} & = & \displaystyle{\sum_{n=1}^\infty \bigg|\frac{(-1)^n}{n}\bigg|}\\
&&\\
& = & \displaystyle{\sum_{n=1}^\infty \frac{1}{n}.}
\end{array}</math>
|}

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

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we need to look back at the original series to see
|-
|if it conditionally converges.
|-
|For
|-
|        <math>\sum_{n=1}^\infty \frac{(-1)^n}{n},</math>
|-
|we notice that this series is alternating.
|-
|Let  <math style="vertical-align: -14px"> b_n=\frac{1}{n}.</math>
|-
|First, we have
|-
|       <math>\frac{1}{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
|-
|        <math>\frac{1}{n+1}<\frac{1}{n}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n}=0.</math>
|-
|Therefore, the series  <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math>   converges
|-
|by the Alternating Series Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|Since the series  <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math>   is not absolutely convergent but convergent,
|-
|this series is conditionally convergent.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|        conditionally convergent (by the p-test and the Alternating Series Test)
|-
|
|}
[[009C_Sample_Midterm_1|'''Return to Sample Exam''']]

@@ Line 8: / Line 8: @@
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Midterm 1, Problem 3 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Midterm 1, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 90: / Line 90: @@
 !Step 4: &nbsp;
 |-
-|Since the series &nbsp;<math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> &nbsp; is not absolutely convergent but convergent,
+|Since the series
 |-
-|this series is conditionally convergent.
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> &nbsp;
 |}