Difference between revisions of "009C Sample Midterm 1, Problem 5"

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(Created page with "<span class="exam"> Find the radius of convergence and interval of convergence of the series. <span class="exam">(a)  <math>\sum_{n=0}^\infty \sqrt{n}x^n</math> <span c...")
 
(Replaced content with "<span class="exam"> Find the radius of convergence and interval of convergence of the series. <span class="exam">(a)  <math>\sum_{n=0}^\infty \sqrt{n}x^n</math> <sp...")
 
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<span class="exam">(b) &nbsp;<math>\sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}</math>
 
<span class="exam">(b) &nbsp;<math>\sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}</math>
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+
<hr>
!Foundations: &nbsp;
+
[[009C Sample Midterm 1, Problem 5 Solution|'''<u>Solution</u>''']]
|-
 
|'''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:'''
+
[[009C Sample Midterm 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
'''(a)'''
 
{| 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{\sqrt{n+1}x^{n+1}}{\sqrt{n}x^n}\bigg|}\\
 
&&\\
 
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}x\bigg|}\\
 
&&\\
 
& = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}|x|}\\
 
&&\\
 
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\
 
&&\\
 
& = & \displaystyle{|x|\sqrt{\lim_{n\rightarrow \infty} \frac{n+1}{n}}}\\
 
&&\\
 
& = & \displaystyle{|x|\sqrt{1}}\\
 
&&\\
 
&=& \displaystyle{|x|.}
 
\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: -5px">|x|<1.</math>
 
|-
 
|Hence, the Radius of Convergence of this series is &nbsp;<math style="vertical-align: -2px">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: -5px">|x|<1</math>&nbsp; corresponds to the interval &nbsp;<math style="vertical-align: -4px">(-1,1).</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: -2px">L=1.</math>
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 4: &nbsp;
 
|-
 
|First, let &nbsp;<math style="vertical-align: -1px">x=1.</math>
 
|-
 
|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty \sqrt{n}.</math>
 
|-
 
|We note that
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty.</math>
 
|-
 
|Therefore, the series diverges by the &nbsp;<math style="vertical-align: 0px">n</math>th term test.
 
|-
 
|Hence, we do not include &nbsp;<math style="vertical-align: -1px">x=1</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=-1.</math>
 
|-
 
|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty (-1)^n \sqrt{n}.</math>
 
|-
 
|Since &nbsp;<math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty,</math>
 
|-
 
|we have
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=\text{DNE}.</math>
 
|-
 
|Therefore, the series diverges by the &nbsp;<math style="vertical-align: 0px">n</math>th term test.
 
|-
 
|Hence, we do not include &nbsp;<math style="vertical-align: -1px">x=-1 </math>&nbsp; in the interval.
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 6: &nbsp;
 
|-
 
|The interval of convergence is &nbsp;<math style="vertical-align: -4px">(-1,1).</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{(-1)^{n+1}(x-3)^{n+1}}{2(n+1)+1}\frac{2n+1}{(-1)^n(x-3)^n}\bigg|}\\
 
&&\\
 
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x-3)\frac{2n+1}{2n+3}\bigg|}\\
 
&&\\
 
& = & \displaystyle{\lim_{n\rightarrow \infty} |x-3|\frac{2n+1}{2n+3}}\\
 
&&\\
 
& = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\
 
&&\\
 
& = & \displaystyle{|x-3|.}
 
\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: -5px">|x-3|<1.</math>
 
|-
 
|Hence, the Radius of Convergence of this series is &nbsp;<math style="vertical-align: -2px">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: -5px">|x-3|<1</math>&nbsp; corresponds to the interval &nbsp;<math style="vertical-align: -4px">(2,4).</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: -2px">R=1.</math>
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 4: &nbsp;
 
|-
 
|First, let &nbsp;<math style="vertical-align: -1px">x=4.</math> 
 
|-
 
|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty (-1)^n \frac{1}{2n+1}.</math>
 
|-
 
|This is an alternating series.
 
|-
 
|Let &nbsp;<math style="vertical-align: -16px">b_n=\frac{1}{2n+1}.</math>.
 
|-
 
|First, we have
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{2n+1}\ge 0</math>
 
|-
 
|for all &nbsp;<math style="vertical-align: -3px">n\ge 0.</math>
 
|-
 
|The sequence &nbsp;<math>\{b_n\}</math>&nbsp; is decreasing since
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{2(n+1)+1}<\frac{1}{2n+1}</math>
 
|-
 
|for all &nbsp;<math style="vertical-align: -3px">n\ge 0.</math>
 
|-
 
|Also,
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{2n+1}=0.</math>
 
|-
 
|Therefore, this series converges by the Alternating Series Test
 
|-
 
|and we include &nbsp;<math style="vertical-align: -1px">x=4</math>&nbsp; in our 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=0}^\infty \frac{1}{2n+1}.</math>
 
|-
 
|First, we note that &nbsp;<math>\frac{1}{2n+1}>0</math>&nbsp; for all &nbsp;<math style="vertical-align: -3px">n\ge 0.</math>
 
|-
 
|Thus, we can use the Limit Comparison Test.
 
|-
 
|We compare this series with the series &nbsp;<math>\sum_{n=1}^\infty \frac{1}{n},</math>
 
|-
 
|which is the harmonic series and divergent.
 
|-
 
|Now, we have
 
|-
 
|
 
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\lim_{n\rightarrow \infty} \frac{\frac{1}{2n+1}}{\frac{1}{n}}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{2n+1}}\\
 
&&\\
 
& = & \displaystyle{\frac{1}{2}.}
 
\end{array}</math>
 
|-
 
|Since this limit is a finite number greater than zero,
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^\infty \frac{1}{2n+1}</math>&nbsp; 
 
|-
 
|diverges by the Limit Comparison Test. 
 
|-
 
|Therefore, we do not include &nbsp;<math style="vertical-align: -1px">x=2</math>&nbsp; in our interval.
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 6: &nbsp;
 
|-
 
|The interval of convergence is &nbsp;<math style="vertical-align: -4px">(2,4].</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: -2px">R=1</math>&nbsp; and the interval of convergence is &nbsp;<math style="vertical-align: -4px">(-1,1).</math>
 
|-
 
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; The radius of convergence is &nbsp;<math style="vertical-align: -2px">R=1</math>&nbsp; and the interval of convergence is &nbsp;<math style="vertical-align: -4px">(2,4].</math>
 
|}
 
 
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 23:00, 11 November 2017

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

(a)  

(b)  


Solution


Detailed Solution


Return to Sample Exam