The sum of the infinite series

Question

$\sum\limits_{n=3}^{\infty} \frac{n+5}{(n+2)(n^2-1)} = \sum\limits_{n=3}^{\infty} (\frac{1}{n+2} + \frac{1}{n-1} + \frac{-2}{n+1}) $ So I do not know how I should write partial sum.

Answer 1

$$\dfrac1{n+2}+\dfrac1{n-1}-\dfrac2{n+1}$$

$$=\dfrac1{n+2}-\dfrac1{n+1}+\dfrac1{n-1}-\dfrac1{n+1}$$

$$=\underbrace{f(n+2)-f(n+1)}-\underbrace{f(n+1)-f(n-1)}$$

where $f(n)=\dfrac1n$

Set a few values of $n$ to recognize the telescoping series.

Answer 2

Hint. You have telescoping sums by writing $$ \begin{align} &\sum\limits_{n=3}^{N} \left(\frac{1}{n+2} + \frac{1}{n-1} + \frac{-2}{n+1}\right) \\\\&=\sum\limits_{n=3}^{N} \left(\frac{1}{n+2} - \frac{1}{n+1}\right)-\sum\limits_{n=3}^{N} \left(\frac{1}{n+1} - \frac{1}{n}\right)-\sum\limits_{n=3}^{N} \left(\frac{1}{n} - \frac{1}{n-1}\right) \end{align} $$hope you can take it from here.