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

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<span class="exam"> Find the interval of convergence of the following series.
 
<span class="exam"> Find the interval of convergence of the following series.
  
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::<math>\sum_{n=0}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math>
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::<math>\sum_{n=1}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math>
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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|
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}.</math>
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^2}.</math>
 
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|-
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|Since &nbsp;<math style="vertical-align: -5px">n^2<(n+1)^2,</math> &nbsp; we have
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|We notice that this series is alternating.
 
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}.</math>  
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|Let &nbsp;<math style="vertical-align: -14px"> b_n=\frac{1}{n^2}.</math>
 
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|-
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|Thus, &nbsp; <math>\frac{1}{n^2}</math> &nbsp; is decreasing.
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|First, we have
 
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|-
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|So, &nbsp;<math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}</math> &nbsp; converges by the Alternating Series Test.
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{n^2}\ge 0</math>
 +
|-
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|for all &nbsp;<math style="vertical-align: -3px">n\ge 1.</math>
 +
|-
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|The sequence &nbsp;<math style="vertical-align: -4px">\{b_n\}</math>&nbsp; is decreasing since
 +
|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}</math>
 +
|-
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|for all &nbsp;<math style="vertical-align: -3px">n\ge 1.</math>
 +
|-
 +
|Also,
 +
|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n^2}=0.</math>
 +
|-
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|So, &nbsp;<math>\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^2}</math> &nbsp; converges by the Alternating Series Test.
 
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|}
  
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
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\displaystyle{\sum_{n=0}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=0}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\
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\displaystyle{\sum_{n=1}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=1}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\
 
&&\\
 
&&\\
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& = & \displaystyle{\sum_{n=0}^{\infty} \frac{1}{n^2}.}\\
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& = & \displaystyle{\sum_{n=1}^{\infty} \frac{1}{n^2}.}\\
 
\end{array}</math>
 
\end{array}</math>
 
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|-

Revision as of 10:12, 17 March 2017

Find the interval of convergence of the following 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} (-1)^n \frac{(x+2)^n}{n^2}}
Foundations:  
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=1}^{\infty} (-1)^n \frac{1}{n^2}.}

We notice that 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}{n^2}.}
First, 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^2}\ge 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 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+1)^2}<\frac{1}{n^2}}
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}{n^2}=0.}
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=1}^{\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=1}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=1}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\ &&\\ & = & \displaystyle{\sum_{n=1}^{\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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