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

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

2021-03-03
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) $$n^5<(n+1)^5$$</span>
$$\frac{1}{n^5}>\frac{1}{(n+1)^5}$$
$$b_n>b_{n+1}$$
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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