Answer
series is convergent and the sum is $\frac{1}{4}$.
Work Step by Step
$\Sigma^{\infty}_{n=2} \frac{1}{n^{3}-n}$
$a_{n}=\frac{1}{n^{3}-n}$
$=\frac{1}{n(n+1)(n-1)}$
$\frac{1}{n(n+1)(n-1)}= \frac{A}{n}+\frac{B}{n-1}+\frac{C}{n+1}$
$A=-1$, $B=\frac{1}{2}$, $C=\frac{1}{2}$
$a_{n}=\frac{-1}{n}+\frac{\frac{1}{2}}{n-1}+\frac{\frac{1}{2}}{n+1}$
$a_{n}=\frac{-\frac{1}{2}}{n}+\frac{\frac{1}{2}}{n-1}+\frac{\frac{1}{2}}{n+1}-\frac{\frac{1}{2}}{n}$
$=-\frac{1}{2}(\frac{1}{n}-\frac{1}{n-1})+\frac{1}{2}(\frac{1}{n+1}-\frac{1}{n})$
$\Sigma^{\infty}_{i=2}a_{n} = -\frac{1}{2}[(\frac{1}{2}-1)+(\frac{1}{3}-\frac{1}{2})+(\frac{1}{4}-\frac{1}{3})+...+(\frac{1}{n}-\frac{1}{n-1})] +\frac{1}{2}[(\frac{1}{3}-\frac{1}{2})+(\frac{1}{4}-\frac{1}{3})+(\frac{1}{5}-\frac{1}{4})+...+(\frac{1}{n+1}-\frac{1}{n})]$
$=\frac{1}{2} -\frac{1}{2n}-\frac{1}{4}+\frac{1}{2(n+1)}$
$=\frac{1}{4}-\frac{1}{2n}+\frac{1}{2(n+1)}$
$\lim\limits_{n \to \infty}S_{n} = \lim\limits_{n \to \infty}[\frac{1}{4}-\frac{1}{2n}+\frac{1}{2(n+1)}]$
$=\frac{1}{4}$
Therefore the series is convergent and the sum is $\frac{1}{4}$.
