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

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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 n^nx^n</math>
+
<span class="exam">(a) &nbsp;<math>\sum_{n=0}^\infty n^nx^n</math>
  
<span class="exam">(b) <math>\sum_{n=0}^\infty \frac{(x+1)^n}{\sqrt{n}}</math>
+
<span class="exam">(b) &nbsp;<math>\sum_{n=0}^\infty \frac{(x+1)^n}{\sqrt{n}}</math>
  
  
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|'''1.''' '''Root Test'''
 
|'''1.''' '''Root Test'''
 
|-
 
|-
|&nbsp; &nbsp; &nbsp; &nbsp; Let <math>\{a_n\}</math> be a positive sequence and let <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} |a_n|^{\frac{1}{n}}=L.</math>
+
|&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; Then,
 
|-
 
|-
|&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent.
+
|&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 <math style="vertical-align: -4px">L>1,</math> the series is divergent.
+
&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
 
|-
 
|-
 
|
 
|
&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.  
+
&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.  
 
|-
 
|-
 
|'''2.''' '''Ratio Test'''  
 
|'''2.''' '''Ratio Test'''  
 
|-
 
|-
|&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -7px">\sum a_n</math> be a series and <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>  
+
|&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; Then,
 
|-
 
|-
 
|
 
|
&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent.  
+
&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 <math style="vertical-align: -4px">L>1,</math> the series is divergent.
+
&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
 
|-
 
|-
 
|
 
|
&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
+
&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|This means that as long as <math style="vertical-align: -6px">x\ne 0,</math> this series diverges.
+
|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 <math style="vertical-align: -1px">R=0</math> and  
+
|Hence, the radius of convergence is &nbsp;<math style="vertical-align: -1px">R=0</math>&nbsp; and  
 
|-
 
|-
|the interval of convergence is <math style="vertical-align: -5px">\{0\}.</math>
+
|the interval of convergence is &nbsp;<math style="vertical-align: -5px">\{0\}.</math>
 
|-
 
|-
 
|
 
|
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|The Ratio Test tells us this series is absolutely convergent if <math style="vertical-align: -4px">|x+1|<1.</math>
+
|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 <math style="vertical-align: -1px">R=1.</math>
+
|Hence, the Radius of Convergence of this series is &nbsp;<math style="vertical-align: -1px">R=1.</math>
 
|}
 
|}
  
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|Now, we need to determine the interval of convergence.  
 
|Now, we need to determine the interval of convergence.  
 
|-
 
|-
|First, note that <math style="vertical-align: -4px">|x+1|<1</math> corresponds to the interval <math style="vertical-align: -4px">(-2,0).</math>
+
|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
 
|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 style="vertical-align: -1px">R=1.</math>
+
|for convergence since the Ratio Test is inconclusive when &nbsp;<math style="vertical-align: -1px">R=1.</math>
 
|}
 
|}
  
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!Step 4: &nbsp;
 
!Step 4: &nbsp;
 
|-
 
|-
|First, let <math style="vertical-align: -1px">x=0.</math>  
+
|First, let &nbsp;<math style="vertical-align: -1px">x=0.</math>  
 
|-
 
|-
|Then, the series becomes <math>\sum_{n=0}^\infty \frac{1}{\sqrt{n}}.</math>
+
|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty \frac{1}{\sqrt{n}}.</math>
 
|-
 
|-
|We note that this is a <math style="vertical-align: -3px">p</math>-series with <math style="vertical-align: -12px">p=\frac{1}{2}.</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 <math>p<1,</math> the series diverges.
+
|Since &nbsp;<math>p<1,</math>&nbsp; the series diverges.
 
|-
 
|-
|Hence, we do not include <math style="vertical-align: -1px">x=0</math> in the interval.
+
|Hence, we do not include &nbsp;<math style="vertical-align: -1px">x=0</math>&nbsp; in the interval.
 
|}
 
|}
  
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!Step 5: &nbsp;
 
!Step 5: &nbsp;
 
|-
 
|-
|Now, let <math style="vertical-align: -1px">x=-2.</math>
+
|Now, let &nbsp;<math style="vertical-align: -1px">x=-2.</math>
 
|-
 
|-
|Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
+
|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
 
|-
 
|-
 
|This series is alternating.  
 
|This series is alternating.  
 
|-
 
|-
|Let <math style="vertical-align: -20px">b_n=\frac{1}{\sqrt{n}}.</math>
+
|Let &nbsp;<math style="vertical-align: -20px">b_n=\frac{1}{\sqrt{n}}.</math>
 
|-
 
|-
|The sequence <math>\{b_n\}</math> is decreasing since
+
|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>
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
 
|-
 
|-
|for all <math style="vertical-align: -3px">n\ge 1.</math>
+
|for all &nbsp;<math style="vertical-align: -3px">n\ge 1.</math>
 
|-
 
|-
 
|Also,
 
|Also,
Line 160: Line 160:
 
|Therefore, the series converges by the Alternating Series Test.
 
|Therefore, the series converges by the Alternating Series Test.
 
|-
 
|-
|Hence, we include <math style="vertical-align: -1px">x=-2</math> in our interval of convergence.
+
|Hence, we include &nbsp;<math style="vertical-align: -1px">x=-2</math>&nbsp; in our interval of convergence.
 
|}
 
|}
  
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!Step 6: &nbsp;
 
!Step 6: &nbsp;
 
|-
 
|-
|The interval of convergence is <math style="vertical-align: -4px">[-2,0).</math>
+
|The interval of convergence is &nbsp;<math style="vertical-align: -4px">[-2,0).</math>
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is <math style="vertical-align: -1px">R=0</math> and the interval of convergence is <math>\{0\}.</math>
+
|&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 <math style="vertical-align: -1px">R=1</math> and the interval fo convergence is <math style="vertical-align: -4px">[-2,0).</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>''']]
 
[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 19:12, 26 February 2017

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

(a)  

(b)  


Foundations:  
1. Root Test
        Let    be a positive sequence and let  
        Then,
        If    the series is absolutely convergent.

        If    the series is divergent.

        If    the test is inconclusive.

2. Ratio Test
        Let    be a series and  
        Then,

        If    the series is absolutely convergent.

        If    the series is divergent.

        If    the test is inconclusive.


Solution:

(a)

Step 1:  
We begin by applying the Root Test.
We have

       

Step 2:  
This means that as long as    this series diverges.
Hence, the radius of convergence is    and
the interval of convergence is  

(b)

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  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 R=1.}
Step 3:  
Now, we need to determine the interval of convergence.
First, note that  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 |x+1|<1}   corresponds to the interval  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 (-2,0).}
To obtain the interval of convergence, we need to test the endpoints of this interval
for convergence since the Ratio Test is inconclusive when  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 R=1.}
Step 4:  
First, let  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 x=0.}
Then, the series becomes  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=0}^\infty \frac{1}{\sqrt{n}}.}
We note that this is 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 p} -series with  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 p=\frac{1}{2}.}
Since  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 p<1,}   the series diverges.
Hence, we do not include  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 x=0}   in the interval.
Step 5:  
Now, let  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 x=-2.}
Then, the series becomes  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=0}^\infty (-1)^n \frac{1}{\sqrt{n}}.}
This series is alternating.
Let  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 b_n=\frac{1}{\sqrt{n}}.}
The sequence  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 \{b_n\}}   is decreasing since
        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 \frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}}
for all  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 n\ge 1.}
Also,
        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 \lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{\sqrt{n}}=0.}
Therefore, the series converges by the Alternating Series Test.
Hence, we include  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 x=-2}   in our interval of convergence.
Step 6:  
The interval of convergence is  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 [-2,0).}


Final Answer:  
    (a)     The radius of convergence is  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 R=0}   and the interval of convergence is  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 \{0\}.}
    (b)     The radius of convergence is  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 R=1}   and the interval of convergence is  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 [-2,0).}

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