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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!

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  • $\begingroup$ Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. $\endgroup$ – AlexR Jan 19 '15 at 8:07
  • $\begingroup$ Ok, I understand $\endgroup$ – Ivan Gandacov Jan 19 '15 at 8:08
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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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Using partial fractions, we get $$ \begin{align} \frac{3n-5}{(n-1)n(n+1)} &=\frac{-1}{n-1}+\frac5n+\frac{-4}{n+1}\\ &=-\left(\frac1{n-1}-\frac1n\right)+4\left(\frac1n-\frac1{n+1}\right) \end{align} $$ Now sum two Telescoping Series.

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Hint:

$$\frac{-5}{n(n-1)}+\frac{8}{(n-1)(n+1)}=-5(\frac{1}{n-1}-\frac{1}{n})+4(\frac{1}{(n-1)}-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1})$$

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