009C Sample Final 3, Problem 2 Detailed Solution
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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.
| Background Information: |
|---|
| 1. A series is absolutely convergent if |
| the series Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum |a_{n}|} converges. |
| 2. A series is conditionally convergent if |
| the series Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum |a_{n}|} diverges and the series converges. |
Solution:
(a)
| Step 1: |
|---|
| First, we take the absolute value of the terms in the original series. |
| Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle a_{n}={\frac {(-1)^{n}}{\sqrt {n}}}.} |
| Therefore, |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 -series with |
| Therefore, 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}}{\sqrt {n}}}} |
| is not absolutely convergent. |
(b)
| Step 1: |
|---|
| For |
| we notice that this series is alternating. |
| Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle b_{n}={\frac {1}{\sqrt {n}}}.} |
| First, we have |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{\sqrt {n}}}\geq 0} |
| for all |
| The sequence is decreasing since |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{\sqrt {n+1}}}<{\frac {1}{\sqrt {n}}}} |
| for all |
| Also, |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{\sqrt {n}}}=0.} |
| Therefore, 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}}{\sqrt {n}}}} converges |
| by the Alternating Series Test. |
| Step 2: |
|---|
| Since 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}}{\sqrt {n}}}} is not absolutely convergent but convergent, |
| this series is conditionally convergent. |
| Final Answer: |
|---|
| (a) not absolutely convergent (by the -series test) |
| (b) conditionally convergent (by the Alternating Series Test) |