Difference between revisions of "009C Sample Midterm 2, Problem 4"

Latest revision as of 12:41, 12 November 2017

Find the radius of convergence and interval of convergence of the series.

(a) 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 n^nx^n}

(b) 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{(x+1)^n}{\sqrt{n}}}

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Detailed Solution

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@@ Line 4: / Line 4: @@
 <span class="exam">(b) &nbsp;<math>\sum_{n=1}^\infty \frac{(x+1)^n}{\sqrt{n}}</math>
+<hr>
+[[009C Sample Midterm 2, Problem 4 Solution|'''<u>Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009C Sample Midterm 2, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']]
-!Foundations: &nbsp;
-|-
-|'''1.''' '''Root Test'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math>\{a_n\}</math>&nbsp; be a positive sequence and let &nbsp;<math style="vertical-align: -12px">\lim_{n\rightarrow \infty} |a_n|^{\frac{1}{n}}=L.</math>
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then,
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L<1,</math>&nbsp; the series is absolutely convergent.
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.
-|-
-|'''2.''' '''Ratio Test'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -7px">\sum a_n</math>&nbsp; be a series and &nbsp;<math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then,
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L<1,</math>&nbsp; the series is absolutely convergent.
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.
-|}
-'''Solution:'''
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We begin by applying the Root Test.
-|-
-|We have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{n\rightarrow \infty} \sqrt[n]{|a_n|}} & = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt[n]{|n^nx^n|}}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} |n^nx^n|^{\frac{1}{n}}}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} |nx|}\\
-&&\\
-& = & \displaystyle{n|x|}\\
-&&\\
-& = & \displaystyle{|x|\lim_{n\rightarrow \infty} n}\\
-&&\\
-& = & \displaystyle{\infty.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|This means that as long as &nbsp;<math style="vertical-align: -6px">x\ne 0,</math>&nbsp; this series diverges.
-|-
-|Hence, the radius of convergence is &nbsp;<math style="vertical-align: -1px">R=0</math>&nbsp; and
-|-
-|the interval of convergence is &nbsp;<math style="vertical-align: -5px">\{0\}.</math>
-|-
-|
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We first use the Ratio Test to determine the radius of convergence.
-|-
-|We have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(x+1)^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{(x+1)^n}\bigg|}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| (x+1)\frac{\sqrt{n}}{\sqrt{n+1}}\bigg|}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} |x+1|\frac{\sqrt{n}}{\sqrt{n+1}}}\\
-&&\\
-& = & \displaystyle{|x+1|\lim_{n\rightarrow \infty} \sqrt{\frac{n}{n+1}}}\\
-&&\\
-& = & \displaystyle{|x+1|\sqrt{\lim_{n\rightarrow \infty} \frac{n}{n+1}}}\\
-&&\\
-& = & \displaystyle{|x+1|\sqrt{1}}\\
-&&\\
-&=& \displaystyle{|x+1|.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|The Ratio Test tells us this series is absolutely convergent if &nbsp;<math style="vertical-align: -4px">|x+1|<1.</math>
-|-
-|Hence, the Radius of Convergence of this series is &nbsp;<math style="vertical-align: -1px">R=1.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 3: &nbsp;
-|-
-|Now, we need to determine the interval of convergence.
-|-
-|First, note that &nbsp;<math style="vertical-align: -4px">|x+1|<1</math>&nbsp; corresponds to the interval &nbsp;<math style="vertical-align: -4px">(-2,0).</math>
-|-
-|To obtain the interval of convergence, we need to test the endpoints of this interval
-|-
-|for convergence since the Ratio Test is inconclusive when &nbsp;<math style="vertical-align: -1px">R=1.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 4: &nbsp;
-|-
-|First, let &nbsp;<math style="vertical-align: -1px">x=0.</math>
-|-
-|Then, the series becomes &nbsp;<math>\sum_{n=1}^\infty \frac{1}{\sqrt{n}}.</math>
-|-
-|We note that this is a &nbsp;<math style="vertical-align: -3px">p</math>-series with &nbsp;<math style="vertical-align: -12px">p=\frac{1}{2}.</math>
-|-
-|Since &nbsp;<math>p<1,</math>&nbsp; the series diverges.
-|-
-|Hence, we do not include &nbsp;<math style="vertical-align: -1px">x=0</math>&nbsp; in the interval.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 5: &nbsp;
-|-
-|Now, let &nbsp;<math style="vertical-align: -1px">x=-2.</math>
-|-
-|Then, the series becomes &nbsp;<math>\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
-|-
-|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>\{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 converges by the Alternating Series Test.
-|-
-|Hence, we include &nbsp;<math style="vertical-align: -1px">x=-2</math>&nbsp; in our interval of convergence.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 6: &nbsp;
-|-
-|The interval of convergence is &nbsp;<math style="vertical-align: -4px">[-2,0).</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is &nbsp;<math style="vertical-align: -1px">R=0</math>&nbsp; and the interval of convergence is &nbsp;<math>\{0\}.</math>
-|-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; The radius of convergence is &nbsp;<math style="vertical-align: -1px">R=1</math>&nbsp; and the interval of convergence is &nbsp;<math style="vertical-align: -4px">[-2,0).</math>
-|}
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009C Sample Midterm 2, Problem 4"

Latest revision as of 12:41, 12 November 2017

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