Difference between revisions of "009C Sample Midterm 1, Problem 3"

Latest revision as of 22:56, 11 November 2017

Determine whether the following series converges absolutely,

conditionally or whether it diverges.

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}{n}}

Solution

Detailed Solution

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>''']]
-|-
-|'''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 Midterm 1, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
-{| 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: -14px">a_n=\frac{(-1)^n}{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}{n}\bigg|}\\
-&&\\
-& = & \displaystyle{\sum_{n=1}^\infty \frac{1}{n}.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|This series is the harmonic series (or &nbsp;<math style="vertical-align: -5px">p</math>-series with &nbsp;<math style="vertical-align: -5px">p=1</math>&nbsp;).
-|-
-|Thus, it diverges. Hence, the series
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;
-|-
-|Now, we need to look back at the original series to see
-|-
-|if it conditionally converges.
-|-
-|For
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^\infty \frac{(-1)^n}{n},</math>
-|-
-|we notice that this series is alternating.
-|-
-|Let &nbsp;<math style="vertical-align: -14px"> b_n=\frac{1}{n}.</math>
-|-
-|First, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{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}{n+1}<\frac{1}{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}{n}=0.</math>
-|-
-|Therefore, the series &nbsp;<math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> &nbsp; converges
-|-
-|by the Alternating Series Test.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 4: &nbsp;
-|-
-|Since the series
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> &nbsp;
-|-
-|converges but does not converge absolutely,
-|-
-|the series converges conditionally.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; conditionally convergent (by the p-test and the Alternating Series Test)
-|-
-|
-|}
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009C Sample Midterm 1, Problem 3"

Latest revision as of 22:56, 11 November 2017

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