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:
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1. A series is absolutely convergent if
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the series converges.
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2. A series is conditionally convergent if
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the series diverges and the series converges.
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Solution:
(a)
| Step 1:
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| First, we take the absolute value of the terms in the original series.
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Let
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| Therefore,
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| Step 2:
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This series is a -series with Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle p=1/2.}
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| Therefore, it diverges.
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| Hence, the series
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| 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}}}}
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| is not absolutely convergent.
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(b)
| Step 1:
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| For
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| we notice that this series is alternating.
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| 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}}}.}
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| First, we have
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| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{\sqrt {n}}}\geq 0}
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for all
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The sequence is decreasing since
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| 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}}}}
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for all
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| Also,
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| 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.}
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Therefore, the series converges
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| by the Alternating Series Test.
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| Step 2:
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Since the series is not absolutely convergent but convergent,
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| this series is conditionally convergent.
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| Final Answer:
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(a) not absolutely convergent (by the -series test)
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| (b) conditionally convergent (by the Alternating Series Test)
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