Telescoping Sum

$\begin{aligned} S & = \sum_{n=1}^\infty \frac {2n+1}{n^2(n+1)^2} \\ & = \sum_{n=1}^\infty \frac {(n+1)^2 -n^2}{n^2(n+1)^2} \\ & = \sum_{n=1}^\infty \left(\frac 1{n^2} - \frac 1{(n+1)^2} \right) \\ & = \sum_{n=1}^\infty \frac 1{n^2} - \sum_{n=2}^\infty \frac 1{n^2} \\ & = \frac 1{1^2} = \boxed 1 \end{aligned}$

The title of the problem gives it's solution . For any number $a$ , we can write $\dfrac{1}{a^2}-\dfrac{1}{(1+a)^2}=\dfrac{2a+1}{a^2\times (a+1)^2}$ . Hence the given sum is $\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...=\boxed 1$ .

The desired sum is $\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}=\sum_{n=1}^{\infty} (\frac{1}{n^2}-\frac{1}{(n+1)^2})=\lim_{n\rightarrow \infty} (\frac{1}{1}-\frac{1}{(n+1)^2})=1$