Answer
The series is convergent and the sum is $\frac{1}{4}$.
Work Step by Step
Define the $n-$th partial sum: $s_n=\sum_{i=2}^n\frac{1}{i^3-i}$.
Simplify $s_n$:
$s_n=\sum_{i=2}^n\frac{1}{i^3-i}$
$=\sum_{i=2}^n\frac{1}{(i-1)i(i+1)}$
$=\sum_{i=2}^n(\frac{1/2}{i-1}-\frac{1/2}{i}+\frac{1/2}{i+1}-\frac{1/2}{i})$
$=\sum_{i=2}^n(\frac{1/2}{i-1}-\frac{1/2}{i})+\sum_{i=2}^n(\frac{1/2}{i+1}-\frac{1/2}{i})$
$=\frac{1/2}{2-1}-\frac{1/2}{n}+\frac{1/2}{n+1}-\frac{1/2}{2}$
$=\frac{1}{2}-\frac{1}{2n}+\frac{1}{2n+2}-\frac{1}{4}$
$=\frac{1}{4}-\frac{1}{2n}+\frac{1}{2n+2}$
Find the sum of the series:
$\sum_{n=1}^\infty \frac{1}{n^3-n}=\lim\limits_{n \to \infty}s_n=\lim\limits_{n \to \infty}(\frac{1}{4}-\frac{1}{2n}+\frac{1}{2n+2})=\frac{1}{4}-0+0=\frac{1}{4}$
So, the series is convergent and the sum is $\frac{1}{4}$.
