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

Latest revision as of 22:54, 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 1: / Line 1: @@
-<span class="exam">Determine whether the following series converges absolutely, conditionally or whether it diverges.
+<span class="exam"> Determine whether the following series converges absolutely,
+<span class="exam"> conditionally or whether it diverges.
 <span class="exam"> Be sure to justify your answers!
-::::::<math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math>
+::<math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations: &nbsp;
-|-
-|'''1.''' A series <math>\sum a_n</math> is '''absolutely convergent''' if
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; the series <math>\sum |a_n|</math> converges.
-|-
-|'''2.''' A series <math>\sum a_n</math> is '''conditionally convergent''' if
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; 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: &nbsp;
-|-
-|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,
-|-
-|&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;"
+<hr>
-!Step 2: &nbsp;
+[[009C Sample Midterm 1, Problem 3 Solution|'''<u>Solution</u>''']]
-|-
-|This series is the harmonic series (or <math style="vertical-align: -5px">p</math>-series with <math style="vertical-align: -5px">p=1</math>).
-|-
-|So, 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 <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>
-|-
-|The sequence <math style="vertical-align: -4px">\{b_n\}</math> is decreasing since
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n+1}<\frac{1}{n}</math>
-|-
-|for all <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 <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;"
+[[009C Sample Midterm 1, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
-!Step 4: &nbsp;
-|-
-|Since the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{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; &nbsp; &nbsp; Conditionally convergent
-|-
-|
-|}
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

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

Latest revision as of 22:54, 11 November 2017

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