Difference between revisions of "009C Sample Final 3, Problem 2"
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|For | |For | ||
|- | |- | ||
| − | | <math>\sum_{n=1}^\infty \frac{(-1)^n}{n},</math> | + | | <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}},</math> |
|- | |- | ||
|we notice that this series is alternating. | |we notice that this series is alternating. | ||
|- | |- | ||
| − | |Let <math style="vertical-align: - | + | |Let <math style="vertical-align: -20px"> b_n=\frac{1}{\sqrt{n}}.</math> |
|- | |- | ||
|The sequence <math style="vertical-align: -4px">\{b_n\}</math> is decreasing since | |The sequence <math style="vertical-align: -4px">\{b_n\}</math> is decreasing since | ||
|- | |- | ||
| − | | <math>\frac{1}{n+1}<\frac{1}{n}</math> | + | | <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math> |
|- | |- | ||
|for all <math style="vertical-align: -3px">n\ge 1.</math> | |for all <math style="vertical-align: -3px">n\ge 1.</math> | ||
| Line 77: | Line 77: | ||
|Also, | |Also, | ||
|- | |- | ||
| − | | <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n}=0.</math> | + | | <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}=0.</math> |
|- | |- | ||
| − | |Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> converges | + | |Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math> converges |
|- | |- | ||
|by the Alternating Series Test. | |by the Alternating Series Test. | ||
Revision as of 12:58, 5 March 2017
Consider 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=2}^\infty \frac{(-1)^n}{\sqrt{n}}.}
(a) Test if the series converges absolutely. Give reasons for your answer.
(b) Test if the series converges conditionally. Give reasons for your answer.
| Foundations: |
|---|
| 1. A 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 a_n} is absolutely convergent if |
| 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 |a_n|} converges. |
| 2. A 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 a_n} is conditionally convergent if |
| 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 |a_n|} diverges and 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 a_n} converges. |
Solution:
(a)
| Step 1: |
|---|
| First, we take the absolute value of the terms in the original 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}{\sqrt{n}}.} |
| Therefore, |
| 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 |a_n|} & = & \displaystyle{\sum_{n=1}^\infty \bigg|\frac{(-1)^n}{\sqrt{n}}\bigg|}\\ &&\\ & = & \displaystyle{\sum_{n=1}^\infty \frac{1}{\sqrt{n}}.} \end{array}} |
| Step 2: |
|---|
| This series 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}.} |
| Therefore, it diverges. |
| Hence, 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}{\sqrt{n}}} |
| is not absolutely convergent. |
(b)
| Step 1: |
|---|
| For |
| 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}{\sqrt{n}},} |
| 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}{\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 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}{\sqrt{n}}} converges |
| by the Alternating Series Test. |
| Step 2: |
|---|
| Since 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}{\sqrt{n}}} is not absolutely convergent but convergent, |
| this series is conditionally convergent. |
| Final Answer: |
|---|
| (a) not absolutely convergent |
| (b) conditionally convergent |