Expert Answer to "Use the Alternating Series Test, if applicable, to..."

sibuzwaW

Answered question

2020-10-25

Use the Alternating Series Test, if applicable, to determine the convergence or divergence of the series.

\sum_{n = 2}^{\infty} \frac{(- 1)^{n} n}{n^{2} - 3}

Answer & Explanation

rogreenhoxa8

Skilled2020-10-26Added 109 answers

We have given series
$\sum_{n = 2}^{\infty} \frac{(- 1)^{n} n}{n^{2} - 3}$
Let, $\sum_{n = 2}^{\infty} (- 1)^{n} a_{n} = \sum_{n = 2}^{\infty} \frac{(- 1)^{n} n}{n^{2} - 3}$
Then, $a_{n} = \frac{n}{n^{2} - 3}$
$a_{n + 1} = \frac{n + 1}{(n + 1)^{2} - 3}$
$lim_{n \to \infty} a_{n} = lim_{n \to \infty} \frac{n}{n^{2} - 3}$
$= lim_{n \to \infty} \frac{n}{n^{2} (1 - \frac{3}{n^{2}})}$
$= 0$
$n^{2} - 3 < (n + 1)^{2} - 3$
$\frac{1}{n^{2} - 3} > \frac{1}{(n + 1)^{2} - 3}$
$\frac{n}{n^{2} - 3} > \frac{n}{(n + 1)^{2} - 3}$
$a_{n} > a_{n + 1}$
Therefore, sequence ${a_{n}}$ is decreaing.
So, by alternating series test given series is convergent.

Jeffrey Jordon

Expert2021-12-16Added 2605 answers

Answer is given below (on video)

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