Difference between revisions of "009C Sample Midterm 1, Problem 3"
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|This series is the harmonic series (or <math style="vertical-align: -5px">p</math>-series with <math style="vertical-align: -5px">p=1</math>). | |This series is the harmonic series (or <math style="vertical-align: -5px">p</math>-series with <math style="vertical-align: -5px">p=1</math>). | ||
|- | |- | ||
| − | | | + | |Thus, it diverges. Hence, the series |
|- | |- | ||
| <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> | | <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> | ||
| Line 58: | Line 58: | ||
|if it conditionally converges. | |if it conditionally converges. | ||
|- | |- | ||
| − | |For <math>\sum_{n=1}^\infty \frac{(-1)^n}{n},</math> | + | |For |
| + | |- | ||
| + | | <math>\sum_{n=1}^\infty \frac{(-1)^n}{n},</math> | ||
|- | |- | ||
|we notice that this series is alternating. | |we notice that this series is alternating. | ||
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| <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}{n}=0.</math> | ||
|- | |- | ||
| − | |Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> converges by the Alternating Series Test. | + | |Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> converges |
| + | |- | ||
| + | |by the Alternating Series Test. | ||
|} | |} | ||
Revision as of 08:47, 15 February 2017
Determine whether the following series converges absolutely,
conditionally or whether it diverges.
Be sure to justify your answers!
- 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}{n}}
| 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 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:
| 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}{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}{n}\bigg|}\\ &&\\ & = & \displaystyle{\sum_{n=1}^\infty \frac{1}{n}.} \end{array}} |
| Step 2: |
|---|
| This series is the harmonic series (or 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=1} ). |
| Thus, 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}{n}} |
| is not absolutely convergent. |
| Step 3: |
|---|
| Now, we need to look back at the original series to see |
| if it conditionally converges. |
| 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}{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}{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}{n+1}<\frac{1}{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}{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}{n}} converges |
| by the Alternating Series Test. |
| Step 4: |
|---|
| 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}{n}} is not absolutely convergent but convergent, |
| this series is conditionally convergent. |
| Final Answer: |
|---|
| Conditionally convergent |