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

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
|  
+
|Ratio Test
 
|-
 
|-
 
|
 
|
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'''(a)'''
 
'''(a)'''
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:    
+
!Step 1:  
 
|-
 
|-
|
+
|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>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!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>
 
|}
 
|}
  

Revision as of 09:17, 13 February 2017

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

a)
b)
Foundations:  
Ratio Test

Solution:

(a)

Step 1:  
We first use the Ratio Test to determine the radius of convergence.
We have
       
Step 2:  
The Ratio Test tells us this series is absolutely convergent if
Hence, the Radius of Convergence of this series is
Step 3:  
Now, we need to determine the interval of convergence.
First, note that corresponds to the interval
To obtain the interval of convergence, we need to test the endpoints of this interval
for convergence since the Ratio Test is inconclusive when
Step 4:  
First, let
Then, the series becomes
We note that
       
Therefore, the series diverges by the th term test.
Hence, we do not include in the interval.
Step 5:  
Now, let
Then, the series becomes
Since
we have
        DNE.
Therefore, the series diverges by the th term test.
Hence, we do not include in the interval.
Step 6:  
The interval of convergence is

(b)

Step 1:  
Step 2:  
Final Answer:  
(a)
(b)

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