Computing $\sum\limits_{n=2}^{\infty }\frac{1}{n^3-n}$ I don't understand why I can't get the telescopic sum after the partial fraction decomposition:
$$\frac{1}{n^3-n}= \frac{1}{n(n-1)(n+1)}=\frac{-1}{n}+\frac{1}{2(n-1)}+\frac{1}{2(n+1)}=\frac{-1}{n}+\frac{n}{(n-1)(n+1)}.$$
I have $\sum\limits_{n=2}^{\infty }\frac{1}{n^3-n}=\phi(n+1)-\phi(0).$
The answer should be $1/4$.
 A: Hint:  We have
$$\frac{1}{n(n+1)(n-1)}=\frac{1}{2}\left(\frac{1}{n(n-1)}-\frac{1}{n(n+1)}\right)$$
and the sum telescopes.
A: If you had stopped at the ${-1\over n}+{1\over2(n-1)}+{1\over2(n+1)}$ and rewritten it as
$${1\over n^3-n}={1\over2}\left({1\over n-1}-{1\over n} \right)-{1\over2}\left({1\over n}-{1\over n+1} \right)$$
then the series does telescope.  Note, however, you need the sum to start at $n=2$ rather than $n=0$, since $1/(n^3-n)$ is undefined for $n=0$ and $n=1$. (An edit to the OP corrected the starting point while I was posting this answer.)   The only term that remains after the telescopic cancellations is
$${1\over2}\left({1\over2-1}-{1\over2}\right)={1\over4}$$
A: $$\sum _{ n=2 }^{ \infty  } \frac { 1 }{ n^{ 3 }-n } =\sum _{ n=2 }^{ \infty  } \frac { 1 }{ \left( n-1 \right) n\left( n+1 \right)  } =\frac { 1 }{ 2 } \sum _{ n=2 }^{ \infty  } \left[ \frac { 1 }{ \left( n-1 \right)  } +\frac { 1 }{ \left( n+1 \right)  } -\frac { 2 }{ n }  \right] =\\ =\frac { 1 }{ 2 } \sum _{ n=2 }^{ \infty  } \left[ \frac { 1 }{ \left( n-1 \right)  } -\frac { 1 }{ n } +\frac { 1 }{ \left( n+1 \right)  } -\frac { 1 }{ n }  \right] =\color{red}{\frac { 1 }{ 4 }}  $$
