Calculate the sum of the series $\sum_{n=1}^\infty \frac 1{n(n+1)^2}$ I have already proven that
$$\int_0^1 \ln(1-x)\ln (x)\mathrm d x=\sum_{n=1}^\infty \frac 1{n(n+1)^2}.$$
So now I want so calculate

$$S:=\sum_{n=1}^\infty \frac 1{n(n+1)^2}.$$

I think I should use the fact that 
$$\sum_{n=1}^\infty \frac 1{n^2}=\frac{\pi^2}6$$
and I know using Wolfram Alpha that $S=2-\frac{\pi^2}6$.
So finally I will get
$$\int_0^1 \ln(1-x)\ln (x)\mathrm d x=2-\frac{\pi^2}6.$$
But how to calculate $S$ ?

we have that 
$$\sum_{n=1}^{m }\frac{1}{n}- \frac{1}{n+1}=\frac{m}{m+1}=1-\frac{1}{m+1}$$ the sum is $1$when the $m\rightarrow \infty$
$$\sum_{n=1}^{\infty }\frac{1}{n(n+1)^2} = \sum_{n=1}^{\infty }(\frac{1}{n}- \frac{1}{n+1}) -\sum_{n=1}^{\infty } \frac{1}{(n+1)^2}=1 -\sum_{n=2}^{\infty } \frac{1}{n^2}$$
$$=1-(\frac{\pi^2}{6}-1)=2-\frac{\pi^2}{6}$$
