Difference between revisions of "009C Sample Final 1, Problem 4"
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|'''1. Ratio Test''' | |'''1. Ratio Test''' | ||
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| − | | 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> Then, | + | | 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> Then, |
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| − | If <math style="vertical-align: - | + | If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent. |
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| − | If <math style="vertical-align: - | + | If <math style="vertical-align: -4px">L>1,</math> the series is divergent. |
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| − | If <math style="vertical-align: - | + | If <math style="vertical-align: -4px">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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| − | for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math> | + | for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math> |
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!Step 2: | !Step 2: | ||
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| − | |So, we have <math style="vertical-align: -6px">|x+2|<1.</math> Hence, our interval is <math style="vertical-align: - | + | |So, we have <math style="vertical-align: -6px">|x+2|<1.</math> Hence, our interval is <math style="vertical-align: -5px">(-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: | !Step 3: | ||
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| − | |First, we let <math style="vertical-align: -1px">x=-1.</math> Then, our series becomes | + | |First, we let <math style="vertical-align: -1px">x=-1.</math> Then, our series becomes |
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<math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}.</math> | <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 | + | |Since <math style="vertical-align: -5px">n^2<(n+1)^2,</math> we have |
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| − | |So, <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}</math> converges by the Alternating Series Test. | + | | <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}.</math> |
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| + | |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. | ||
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!Step 4: | !Step 4: | ||
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| − | |Now, we let <math style="vertical-align: -1px">x=-3.</math> Then, our series becomes | + | |Now, we let <math style="vertical-align: -1px">x=-3.</math> Then, our series becomes |
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!Step 5: | !Step 5: | ||
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| − | |Thus, the interval of convergence for this series is <math>[-3,-1].</math> | + | |Thus, the interval of convergence for this series is <math>[-3,-1].</math> |
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Revision as of 16:10, 26 February 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=0}^{\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, |
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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. |
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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. |
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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: |
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| 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: |
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| 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: |
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| 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 |
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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: |
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| 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]} |