Difference between revisions of "009C Sample Final 3, Problem 2"

Latest revision as of 18:41, 2 December 2017

Consider the series

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=2}^\infty \frac{(-1)^n}{\sqrt{n}}.}

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

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

Solution

Detailed Solution

Return to Sample Exam

@@ Line 7: / Line 7: @@
 <span class="exam">(b) Test if the series converges conditionally. Give reasons for your answer.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 3, Problem 2 Solution|'''<u>Solution</u>''']]
-|-
-|'''1.''' A series &nbsp;<math>\sum a_n</math>&nbsp; is '''absolutely convergent''' if
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; the series &nbsp;<math>\sum |a_n|</math>&nbsp; converges.
-|-
-|'''2.''' A series &nbsp;<math>\sum a_n</math>&nbsp; is '''conditionally convergent''' if
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; the series &nbsp;<math>\sum |a_n|</math>&nbsp; diverges and the series &nbsp;<math>\sum a_n</math>&nbsp; converges.
-|}
-'''Solution:'''
+[[009C Sample Final 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, we take the absolute value of the terms in the original series.
-|-
-|Let &nbsp;<math style="vertical-align: -20px">a_n=\frac{(-1)^n}{\sqrt{n}}.</math>
-|-
-|Therefore,
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;
-|-
-|This series is a &nbsp;<math style="vertical-align: -5px">p</math>-series with &nbsp;<math style="vertical-align: -14px">p=\frac{1}{2}.</math>&nbsp;
-|-
-|Therefore, it diverges.
-|-
-|Hence, the series
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;
-|-
-|For
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}},</math>
-|-
-|we notice that this series is alternating.
-|-
-|Let &nbsp;<math style="vertical-align: -20px"> b_n=\frac{1}{\sqrt{n}}.</math>
-|-
-|First, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{\sqrt{n}}\ge 0</math>
-|-
-|for all &nbsp;<math style="vertical-align: -3px">n\ge 1.</math>
-|-
-|The sequence &nbsp;<math style="vertical-align: -4px">\{b_n\}</math>&nbsp; is decreasing since
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
-|-
-|for all &nbsp;<math style="vertical-align: -3px">n\ge 1.</math>
-|-
-|Also,
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}=0.</math>
-|-
-|Therefore, the series &nbsp;<math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math> &nbsp; converges
-|-
-|by the Alternating Series Test.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Since the series &nbsp;<math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math> &nbsp; is not absolutely convergent but convergent,
-|-
-|this series is conditionally convergent.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; not absolutely convergent (by the p test)
-|-
-|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; conditionally convergent (by the Alternating Series Test)
-|}
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009C Sample Final 3, Problem 2"

Latest revision as of 18:41, 2 December 2017

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