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

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<span class="exam"> Find the radius of convergence and interval of convergence of the series.
 
<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 class="exam">(a) &nbsp;<math>\sum_{n=0}^\infty \sqrt{n}x^n</math>
::<span class="exam">b) <math>\sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}</math>
 
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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<span class="exam">(b) &nbsp;<math>\sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}</math>
!Foundations: &nbsp;  
 
|-
 
|Ratio Test
 
|-
 
|
 
::
 
|-
 
|
 
::
 
|}
 
  
'''Solution:'''
+
<hr>
 +
[[009C Sample Midterm 1, Problem 5 Solution|'''<u>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}|\frac{a_{n+1}}{a_n}|} & = & \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;"
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[[009C Sample Midterm 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
!Step 2: &nbsp;
 
|-
 
|The Ratio Test tells us this series is absolutely convergent if <math>|x|<1.</math>
 
|-
 
|Hence, the Radius of Convergence of this series is <math>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 <math>|x|<1</math> corresponds to the interval <math>(-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 <math>R=1.</math>
 
|}
 
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 4: &nbsp;
 
|-
 
|First, let <math>x=1.</math>
 
|-
 
|Then, the series becomes <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 <math>n</math>th term test.
 
|-
 
|Hence, we do not include <math>x=1</math> in the interval.
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 5: &nbsp;
 
|-
 
|Now, let <math>x=-1.</math>
 
|-
 
|Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \sqrt{n}.</math>
 
|-
 
|Since <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty,</math>
 
|-
 
|we have
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=</math>DNE.
 
|-
 
|Therefore, the series diverges by the <math>n</math>th term test.
 
|-
 
|Hence, we do not include <math>x=-1 </math> in the interval.
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 6: &nbsp;
 
|-
 
|The interval of convergence is <math>(-1,1).</math>
 
|}
 
 
'''(b)'''
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;
 
|-
 
|
 
|-
 
|
 
|-
 
|
 
|-
 
|
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
|-
 
|
 
|-
 
|
 
|-
 
|
 
|-
 
|
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;
 
|-
 
|'''(a)'''
 
|-
 
|'''(b)'''
 
|}
 
 
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 22:56, 11 November 2017

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

(a)  

(b)  


Solution


Detailed Solution


Return to Sample Exam