$\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = ?$ $S_1=\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = ?$ 
Attempt: $S_1=\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = \dfrac {2n+2-1}{n^2(n+1)^2}$
$S_1=2  \sum_{n=1}^\infty [\dfrac {n+1}{n^2(n+1)^2} ] - \sum_{n=1}^\infty \dfrac {1}{n^2(n+1)^2}$
$=2 \sum_{n=1}^\infty [\dfrac {1}{n^2(n+1) } ] - \sum_{n=1}^\infty \dfrac {1}{n^2(n+1)^2}$
I am stuck on how to move ahead. Please guide me.
Thank you for your help.
 A: First note that $$\sum_{n=1}^{\infty}\frac{2n+1}{n^{2}(n+1)^{2}}=\sum_{n=1}^{\infty}\left[\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right]$$
By telescoping the series, we can see that
$$
\lim\limits_{z\to\infty}\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\dots +\frac{1}{z^2}-\frac{1}{(z+1)^2}\right]
$$
$$
= \lim\limits_{z\to\infty}\left[1-\frac{1}{(z+1)^2}\right]=1-0=1
$$
A: $$\sum_{n=1}^{\infty}\frac{2n+1}{n^{2}(n+1)^{2}}=\sum_{n=1}^{\infty}\left[\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right]$$
$$=\bigl[\frac{\pi^{2}}{6}-\bigl(\frac{\pi^{2}}{6}-1\bigr)\bigr]=1.$$
using the fact that $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$.
Otherwise:

$$\sum_{n=1}^{\infty}\frac{2n+1}{n^{2}(n+1)^{2}}=\sum_{n=1}^{\infty}\left[\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right]=\sum_{n=1}^{\infty}\frac{1}{n^{2}}-\sum_{n=1}^{\infty}\frac{1}{(n+1)^{2}}$$
$$=\sum_{n=1}^{\infty}\frac{1}{n^{2}}-\left[\sum_{n=1}^{\infty}\frac{1}{n^{2}}-1\right]=1.$$
As, the series $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ is convergent( by p-test) so it cancel out.

