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## Answered question

2020-10-25

Use the Alternating Series Test, if applicable, to determine the convergence or divergence of the series.
$\sum _{n=2}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}n}{{n}^{2}-3}$

### Answer & Explanation

rogreenhoxa8

Skilled2020-10-26Added 109 answers

We have given series
$\sum _{n=2}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}n}{{n}^{2}-3}$
Let, $\sum _{n=2}^{\mathrm{\infty }}\left(-1{\right)}^{n}{a}_{n}=\sum _{n=2}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}n}{{n}^{2}-3}$
Then, ${a}_{n}=\frac{n}{{n}^{2}-3}$
${a}_{n+1}=\frac{n+1}{\left(n+1{\right)}^{2}-3}$
$\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=\underset{n\to \mathrm{\infty }}{lim}\frac{n}{{n}^{2}-3}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{n}{{n}^{2}\left(1-\frac{3}{{n}^{2}}\right)}$
$=0$
${n}^{2}-3<\left(n+1{\right)}^{2}-3$
$\frac{1}{{n}^{2}-3}>\frac{1}{\left(n+1{\right)}^{2}-3}$
$\frac{n}{{n}^{2}-3}>\frac{n}{\left(n+1{\right)}^{2}-3}$
${a}_{n}>{a}_{n+1}$
Therefore, sequence $\left\{{a}_{n}\right\}$ is decreaing.
So, by alternating series test given series is convergent.

Jeffrey Jordon

Expert2021-12-16Added 2607 answers

Answer is given below (on video)

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