Difference between revisions of "009C Sample Final 3, Problem 6"

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
−
|
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|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>
 
|-
 
|-
−
|
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|To obtain the interval of convergence, we need to test the endpoints of this interval
 
|-
 
|-
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|
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|for convergence since the Ratio Test is inconclusive when &nbsp;<math style="vertical-align: -1px">R=1.</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|
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|First, let &nbsp;<math style="vertical-align: -1px">x=1.</math> 
 +
|-
 +
|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty (-1)^n \frac{1}{n+1}.</math>
 +
|-
 +
|This is an alternating series.
 +
|-
 +
|Let &nbsp;<math style="vertical-align: -15px">b_n=\frac{1}{n+1}.</math>.
 +
|-
 +
|The sequence &nbsp;<math>\{b_n\}</math>&nbsp; is decreasing since
 +
|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n+2}<\frac{1}{n+1}</math>
 +
|-
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|for all &nbsp;<math style="vertical-align: -3px">n\ge 0.</math>
 +
|-
 +
|Also,
 +
|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{n+1}=0.</math>
 +
|-
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|Therefore, this series converges by the Alternating Series Test
 +
|-
 +
|and we include &nbsp;<math style="vertical-align: -1px">x=1</math>&nbsp; in our interval.
 +
|}
 +
 
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 3: &nbsp;
 +
|-
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|Now, let &nbsp;<math style="vertical-align: -1px">x=-1.</math>
 +
|-
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|Then, the series becomes
 +
|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\sum_{n=0}^\infty \frac{(-1)^{2n+1}}{n+1}} & = & \displaystyle{\sum_{n=1}^\infty \frac{-1}{n+1}}\\
 +
&&\\
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& = & \displaystyle{(-1)\sum_{n=1}^\infty \frac{1}{n+1}.}
 +
\end{array}</math>
 +
|-
 +
|Now, we note that
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n+1}>0</math>
 +
|-
 +
|for all &nbsp;<math style="vertical-align: -3px">n\ge 0.</math>
 +
|-
 +
|This means that we can use the limit comparison test on this series.
 +
|-
 +
|Let &nbsp;<math style="vertical-align: -19px">a_n=\frac{1}{n+1}.</math>
 +
|-
 +
|Let &nbsp;<math style="vertical-align: -14px">b_n=\frac{1}{n}.</math>
 +
|-
 +
|Then, &nbsp;<math>\sum_{n=1}^\infty b_n</math>&nbsp; diverges since it is the harmonic series.
 +
|-
 +
|We have
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\lim_{n\rightarrow \infty} \frac{a_n}{b_n}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(\frac{1}{n+1})}{(\frac{1}{n})}}\\
 +
&&\\
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& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{n+1}}\\
 +
&&\\
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& = & \displaystyle{\lim_{n\rightarrow \infty} 1.}
 +
\end{array}</math>
 +
|-
 +
|Therefore, the series
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^{\infty} \frac{1}{n+1}</math>
 +
|-
 +
|diverges by the Limit Comparison Test.
 +
|-
 +
|Therefore, we do not include &nbsp;<math style="vertical-align: -1px">x=-1</math>&nbsp; in our interval.
 +
|}
 +
 
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 4: &nbsp;
 
|-
 
|-
−
|
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|The interval of convergence is &nbsp;<math style="vertical-align: -4px">(-1,1].</math>
 
|}
 
|}
  
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is &nbsp;<math style="vertical-align: -1px">R=1.</math>
 
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is &nbsp;<math style="vertical-align: -1px">R=1.</math>
 
|-
 
|-
−
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>(-1,1]</math>
 
|-
 
|-
 
|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;
 
|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;

Revision as of 15:45, 5 March 2017

Consider the power series

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{x^{n+1}}{n+1}}

(a) Find the radius of convergence of the above power series.

(b) Find the interval of convergence of the above power series.

(c) Find the closed formula for the function  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 f(x)}   to which the power series converges.

(d) Does the series

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}{(n+1)3^{n+1}}}

converge? If so, find its sum.

Foundations:  
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 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{(-1)^{n+1}(x)^{n+2}}{(n+2)}\frac{n+1}{(-1)^n(x)^{n+1}}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x)\frac{n+1}{n+2}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} |x|\frac{n+1}{n+2}}\\ &&\\ & = & \displaystyle{|x|\lim_{n\rightarrow \infty} \frac{n+1}{n+2}}\\ &&\\ & = & \displaystyle{|x|.} \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.}
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.}

(b)

Step 1:  
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}   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 (-1,1).}
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 2:  
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=1.}
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}{n+1}.}
This is an alternating series.
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}{n+1}.} .
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}{n+2}<\frac{1}{n+1}}
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 0.}
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}{n+1}=0.}
Therefore, this series converges by the Alternating Series Test
and 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=1}   in our interval.
Step 3:  
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=-1.}
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 \begin{array}{rcl} \displaystyle{\sum_{n=0}^\infty \frac{(-1)^{2n+1}}{n+1}} & = & \displaystyle{\sum_{n=1}^\infty \frac{-1}{n+1}}\\ &&\\ & = & \displaystyle{(-1)\sum_{n=1}^\infty \frac{1}{n+1}.} \end{array}}
Now, we 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 \frac{1}{n+1}>0}
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 0.}
This means that we can use the limit comparison test on this series.
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=\frac{1}{n+1}.}
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}{n}.}
Then,  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=1}^\infty b_n}   diverges since it is the harmonic series.
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} \frac{a_n}{b_n}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(\frac{1}{n+1})}{(\frac{1}{n})}}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{n+1}}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} 1.} \end{array}}
Therefore, the series
        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=1}^{\infty} \frac{1}{n+1}}
diverges by the Limit Comparison Test.
Therefore, 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=-1}   in our interval.
Step 4:  
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 (-1,1].}

(c)

Step 1:  
Step 2:  

(d)

Step 1:  
Step 2:  


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=1.}
    (b)     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 (-1,1]}
    (c)    
    (d)    

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