Q:

Computing the value of $\frac{1}{3^2+1} + \frac{1}{4^2+2} + \frac{1}{5^2+3}...\infty$=? 

I have tried converting this series into a telescopic sum whose terms could cancel out but haven't succeeded in that effort.  How should I proceed further?  

A:

You can use:
$$\dfrac{1}{n^2+(n-2)}=\dfrac{1}{(n-1)(n+2)}=\dfrac{1}{3}\left(\dfrac{1}{n-1}-\dfrac{1}{n+2}\right)$$

A:

The hint:
Use the telescopic sum and $$(n+2)^2+n=(n+4)(n+1).$$
Now, $$\sum_{n=1}^{+\infty}\frac{1}{(n+2)^2+n}=\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=\frac{13}{36}.$$

