chillywilly12a

2021-03-02

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

### Answer & Explanation

Margot Mill

The given series is $\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{{n}^{5}}$
To check its convergence or divergence using Alternating series test.
Solution:
The alternating series says that if we have series of form, $\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n}{b}_{n}$, then if,
1)$\underset{n\to \mathrm{\infty }}{lim}{b}_{n}=0$ and
2)${b}_{n}$ is a decreasing sequence, then the series $\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n}{b}_{n}$ is said to be convergent.
Since we have series $\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{{n}^{5}}$ 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) $\underset{n\to \mathrm{\infty }}{lim}{b}_{n}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{5}}=0$
2) ${n}^{5}<\left(n+1{\right)}^{5}$
$\frac{1}{{n}^{5}}>\frac{1}{\left(n+1{\right)}^{5}}$
${b}_{n}>{b}_{n+1}$
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 $\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{{n}^{5}}$ is convergent.

Jeffrey Jordon