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

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
−
| Root Test
+
|'''1.''' '''Root Test'''
 
|-
 
|-
−
| Ratio 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; Then,
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L<1,</math> the series converges.
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L>1,</math> the series is divergent.
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
 +
|-
 +
|'''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; Then,
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L<1,</math> 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 <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
 
|}
 
|}
  
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|
 
|
 
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
−
\displaystyle{\lim_{n\rightarrow \infty} \sqrt{|a_n|}} & = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt{|n^nx^n|}}\\
+
\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} |n^nx^n|^{\frac{1}{n}}}\\
Line 38: Line 61:
 
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} n}\\
 
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} n}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\infty}
+
& = & \displaystyle{\infty.}
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
Line 45: Line 68:
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|This means that as long as <math>x\ne 0,</math> this series diverges.
+
|This means that as long as <math style="vertical-align: -6px">x\ne 0,</math> this series diverges.
 
|-
 
|-
−
|Hence, the radius of convergence is <math>R=0</math> and  
+
|Hence, the radius of convergence is <math style="vertical-align: -1px">R=0</math> and  
 
|-
 
|-
−
|the interval of convergence is <math>\{0\}.</math>
+
|the interval of convergence is <math style="vertical-align: -5px">\{0\}.</math>
 
|-
 
|-
 
|
 
|
Line 63: Line 86:
 
|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
|&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{(x+1)^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{(x+1)^n}\bigg|}\\
+
\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} \bigg| (x+1)\frac{\sqrt{n}}{\sqrt{n+1}}\bigg|}\\
Line 82: Line 105:
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|The Ratio Test tells us this series is absolutely convergent if <math>|x+1|<1.</math>
+
|The Ratio Test tells us this series is absolutely convergent if <math style="vertical-align: -4px">|x+1|<1.</math>
 
|-
 
|-
−
|Hence, the Radius of Convergence of this series is <math>R=1.</math>
+
|Hence, the Radius of Convergence of this series is <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>|x+1|<1</math> corresponds to the interval <math>(-2,0).</math>
+
|First, note that <math style="vertical-align: -4px">|x+1|<1</math> corresponds to the interval <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>R=1.</math>
+
|for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">R=1.</math>
 
|}
 
|}
  
Line 102: Line 125:
 
!Step 4: &nbsp;
 
!Step 4: &nbsp;
 
|-
 
|-
−
|First, let <math>x=0.</math>  
+
|First, let <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 <math>\sum_{n=0}^\infty \frac{1}{\sqrt{n}}.</math>
 
|-
 
|-
−
|We note that this is a <math>p</math>-series with <math>p=\frac{1}{2}.</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>
 
|-
 
|-
 
|Since <math>p<1,</math> the series diverges.
 
|Since <math>p<1,</math> the series diverges.
 
|-
 
|-
−
|Hence, we do not include <math>x=0</math> in the interval.
+
|Hence, we do not include <math style="vertical-align: -1px">x=0</math> in the interval.
 
|}
 
|}
  
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!Step 5: &nbsp;
 
!Step 5: &nbsp;
 
|-
 
|-
−
|Now, let <math>x=-2.</math>
+
|Now, let <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 <math>\sum_{n=0}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
Line 122: Line 145:
 
|This series is alternating.  
 
|This series is alternating.  
 
|-
 
|-
−
|Let <math>b_n=\frac{1}{\sqrt{n}}.</math>
+
|Let <math style="vertical-align: -20px">b_n=\frac{1}{\sqrt{n}}.</math>
 
|-
 
|-
 
|The sequence <math>\{b_n\}</math> is decreasing since
 
|The sequence <math>\{b_n\}</math> is decreasing since
Line 128: Line 151:
 
|&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>n\ge 1.</math>
+
|for all <math style="vertical-align: -3px">n\ge 1.</math>
 
|-
 
|-
 
|Also,
 
|Also,
Line 136: Line 159:
 
|Therefore, the series converges by the Alternating Series Test.
 
|Therefore, the series converges by the Alternating Series Test.
 
|-
 
|-
−
|Hence, we include <math>x=-2</math> in our interval of convergence.
+
|Hence, we include <math style="vertical-align: -1px">x=-2</math> in our interval of convergence.
 
|}
 
|}
  
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!Step 6: &nbsp;
 
!Step 6: &nbsp;
 
|-
 
|-
−
|The interval of convergence is <math>[-2,0).</math>
+
|The interval of convergence is <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>R=0</math> and the interval of convergence is <math>\{0\}.</math>
+
|&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; '''(b)''' &nbsp; &nbsp; The radius of convergence is <math>R=1</math> and the interval fo convergence is <math>[-2,0).</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>
 
|}
 
|}
 
[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 16:39, 15 February 2017

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

a)
b)


Foundations:  
1. Root Test
        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 \{a_n\}} be a positive sequence and 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 \lim_{n\rightarrow \infty} (a_n)^{\frac{1}{n}}=L.}
        Then,
        If 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 L<1,} the series converges.

        If 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 L>1,} the series is divergent.

        If 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 L=1,} the test is inconclusive.

2. Ratio Test
        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 \sum a_n} be a series and 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 L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.}
        Then,

        If 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 L<1,} the series is absolutely convergent.

        If 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 L>1,} the series is divergent.

        If 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 L=1,} the test is inconclusive.


Solution:

(a)

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

        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 \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}}

Step 2:  
This means that as long as 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\ne 0,} this series diverges.
Hence, 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)

Step 1:  
We first use the Ratio Test to determine the radius of convergence.
We have
        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 \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}}
Step 2:  
The Ratio Test tells us this series is absolutely convergent if 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.}
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 fo 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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