Use the Alternating Series Test, if applicable, to determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{(-1)^n}{n^5}

chillywilly12a 2021-03-02 Answered
Use the Alternating Series Test, if applicable, to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty\frac{(-1)^n}{n^5}\)

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Expert Answer

Margot Mill
Answered 2021-03-03 Author has 19175 answers

The given series is \(\sum_{n=1}^\infty\frac{(-1)^n}{n^5}\)
To check its convergence or divergence using Alternating series test.
Solution:
The alternating series says that if we have series of form, \(\sum_{n=1}^\infty(-1)^nb_n\), then if,
1)\(\lim_{n\rightarrow\infty}b_n=0\) and
2)\(b_n\) is a decreasing sequence, then the series \(\sum_{n=1}^\infty(-1)^nb_n\) is said to be convergent.
Since we have series \(\sum_{n=1}^\infty\frac{(-1)^n}{n^5}\) we have sequence \(b_n\) as \(b_n=\frac{1}{n^5}\),
Now to check both the condition using alternating series test for convergence,
1) \(\lim_{n\rightarrow\infty}b_n=\lim_{n\rightarrow\infty}\frac{1}{n^5}=0\)
2) \(n^5<(n+1)^5\)
\(\frac{1}{n^5}>\frac{1}{(n+1)^5}\)
\(b_n>b_{n+1}\)
So \(b_n\) is the decreasing sequence as well,
Since , both the condition are satisficed so the given series is convergent.
Hence, the given series \(\sum_{n=1}^\infty\frac{(-1)^n}{n^5}\) is convergent.

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Answered 2021-12-26 Author has 11052 answers

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