Find $ \sum_{n=2}^{+\infty}$ $\frac{1}{n^{3}-n}$ Find:
$$ \sum_{n=2}^{+\infty}\frac{1}{n^{3}-n}$$
I tried to resolve into partial fractions to see if there are some cancellations ,but that did not helped me .How do i do this ?Thanks

We know that the series at hand is convergent absolutely, so we can safely play with the sum.
$$\sum_{i=2}^\infty\frac{1}{n^3 - n} = -\sum_{i=2}^\infty\frac{1}{n} + 0.5\sum_{i=2}^\infty\frac{1}{n+1} + 0.5\sum_{i=2}^\infty\frac{1}{n-1} = -\sum_{i=2}^\infty\frac{1}{n} + 0.5\sum_{i=3}^\infty\frac{1}{n} + 0.5\sum_{i=1}^\infty\frac{1}{n} = -\sum_{i=1}^\infty\frac{1}{n} + 1 + 0.5\sum_{i=1}^\infty\frac{1}{n} - 0.5 - 0.25 + 0.5\sum_{i=1}^\infty\frac{1}{n} = 1 - 0.75 = 0.25 = \frac{1}{4}$$.
As $n \rightarrow \infty$, the limit is $\frac14$.
