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