How can I calculate the sum of this series :
$$\sum_{n=2}^{\infty}\frac{3n-5}{n(n^2-1)}=?$$
I've tried to divide in factors $\frac{3n-5}{n(n^2-1)}$ and obtained $\frac{-5}{n(n-1)}+\frac{8}{(n-1)(n+1)}$. But when I try to expand the series I cannot make any simplifications. Can you please help me ? I've tried to divide in factors in different ways, but also got nothing. Thanks!
Answer
You can split it one step further as follows:
$$
\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n},\qquad \frac{1}{(n-1)(n+1)}=\frac{1}{2}\left[\frac{1}{n-1}-\frac{1}{n+1}\right]
$$
Now both series telescope.
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