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? 

Note that $\frac{1}{n^2+(n-2)}=\frac{1}{(n-1)(n+2)}=\frac{1}{3}(\frac1{n-1}-\frac1{n+2})$
Therefore the expression becomes $$\frac{1}{3}(\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}...)=\frac13(\frac{1}{2}+\frac{1}{3}+\frac{1}{4})=\frac{36}{13}$$ since everything else cancel out.
