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

Hint: $$\frac{1}{(n+1)(n+4)}=\frac{1}{3} \Bigg[\frac{1}{n+1}-\frac{1}{n+4}\Bigg]$$
By Cancelling terms we have remaining $$ \frac{1}{3} \Bigg[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\Bigg]= \frac{1}{3} \cdot\frac{13}{12}=\frac{13}{36}$$ 
