<s>[INST] 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 $1/4$.
 [/INST] $$\sum _{ n=2 }^{ \infty  } \frac { 1 }{ n^{ 3 }-n } =\sum _{ n=2 }^{ \infty  } \frac { 1 }{ \left( n-1 \right) n\left( n+1 \right)  } =\frac { 1 }{ 2 } \sum _{ n=2 }^{ \infty  } \left[ \frac { 1 }{ \left( n-1 \right)  } +\frac { 1 }{ \left( n+1 \right)  } -\frac { 2 }{ n }  \right] =\\ =\frac { 1 }{ 2 } \sum _{ n=2 }^{ \infty  } \left[ \frac { 1 }{ \left( n-1 \right)  } -\frac { 1 }{ n } +\frac { 1 }{ \left( n+1 \right)  } -\frac { 1 }{ n }  \right] =\color{red}{\frac { 1 }{ 4 }}  $$
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