Difference between revisions of "009C Sample Midterm 1, Problem 3"
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| − | | | + | | conditionally convergent (by the p-test and the Alternating Series Test) |
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[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 10:16, 18 March 2017
Determine whether the following series converges absolutely,
conditionally or whether it diverges.
Be sure to justify your answers!
- Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}}
| Foundations: |
|---|
| 1. A series is absolutely convergent if |
| the series converges. |
| 2. A series is conditionally convergent if |
| the series diverges and the series converges. |
Solution:
| Step 1: |
|---|
| First, we take the absolute value of the terms in the original series. |
| Let |
| Therefore, |
| Step 2: |
|---|
| This series is the harmonic series (or -series with ). |
| Thus, it diverges. Hence, the series |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 |
| we notice that this series is alternating. |
| Let |
| 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}\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}<\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 (by the p-test and the Alternating Series Test) |