Difference between revisions of "009C Sample Final 1, Problem 4"

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::If <math style="vertical-align: -1px">L<1</math>, the series is absolutely convergent.  
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::If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.  
 
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::If <math style="vertical-align: -1px">L>1</math>, the series is divergent.
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::If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
 
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::If <math style="vertical-align: -1px">L=1</math>, the test is inconclusive.
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::If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
 
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|'''2.''' After you find the radius of convergence, you need to check the endpoints of your interval  
 
|'''2.''' After you find the radius of convergence, you need to check the endpoints of your interval  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|So, we have <math style="vertical-align: -6px">|x+2|<1</math>. Hence, our interval is <math style="vertical-align: -3px">(-3,-1)</math>. But, we still need to check the endpoints of this interval  
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|So, we have <math style="vertical-align: -6px">|x+2|<1.</math> Hence, our interval is <math style="vertical-align: -3px">(-3,-1).</math> But, we still need to check the endpoints of this interval  
 
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|to see if they are included in the interval of convergence.
 
|to see if they are included in the interval of convergence.
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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
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|First, we let <math style="vertical-align: -1px">x=-1</math>. Then, our series becomes <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}.</math>
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|First, we let <math style="vertical-align: -1px">x=-1.</math> Then, our series becomes <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}.</math>
 
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|Since <math style="vertical-align: -5px">n^2<(n+1)^2</math>, we have <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}.</math> Thus, <math>\frac{1}{n^2}</math> is decreasing.
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|Since <math style="vertical-align: -5px">n^2<(n+1)^2,</math> we have <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}.</math> Thus, <math>\frac{1}{n^2}</math> is decreasing.
 
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|So, <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}</math> converges by the Alternating Series Test.
 
|So, <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}</math> converges by the Alternating Series Test.
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!Step 4: &nbsp;
 
!Step 4: &nbsp;
 
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|Now, we let <math style="vertical-align: -1px">x=-3</math>. Then, our series becomes  
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|Now, we let <math style="vertical-align: -1px">x=-3.</math> Then, our series becomes  
 
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Revision as of 10:33, 1 March 2016

Find the interval of convergence of the following series.

Foundations:  
Recall:
1. 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.
2. After you find the radius of convergence, you need to check the endpoints of your 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 L=1.}

Solution:

Step 1:  
We proceed using the ratio test to find the interval of convergence. So, 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+2)^{n+1}}{(n+1)^2}}\frac{n^2}{(-1)^n(x+2)^n}\bigg|\\ &&\\ & = & \displaystyle{|x+2|\lim_{n \rightarrow \infty}\frac{n^2}{(n+1)^2}}\\ &&\\ & = & \displaystyle{|x+2|\lim_{n \rightarrow \infty}\bigg(\frac{n}{n+1}\bigg)^2}\\ &&\\ & = & \displaystyle{|x+2|\bigg(\lim_{n \rightarrow \infty}\frac{n}{n+1}\bigg)^2}\\ &&\\ & = & \displaystyle{|x+2|(1)^2}\\ &&\\ & = & \displaystyle{|x+2|.}\\ \end{array}}
Step 2:  
So, 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 |x+2|<1.} Hence, our interval 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 (-3,-1).} But, we still need to check the endpoints of this interval
to see if they are included in the interval of convergence.
Step 3:  
First, we 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, our 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^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 n^2<(n+1)^2,} 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 \frac{1}{(n+1)^2}<\frac{1}{n^2}.} Thus, 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}} is decreasing.
So, 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^2}} converges by the Alternating Series Test.
Step 4:  
Now, we 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=-3.} Then, our 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} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=0}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\ &&\\ & = & \displaystyle{\sum_{n=0}^{\infty} \frac{1}{n^2}.}\\ \end{array}}
This is a convergent series by the p-test.
Step 5:  
Thus, the interval of convergence for 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 [-3,-1].}
Final Answer:  
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 [-3,-1]}

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