Prove that $\sum_{n=1}^\infty \frac{1}{n^2}$=$1+\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$ I'm really at loss with this problem. I should prove that 
$$\sum_{n=1}^\infty \frac{1}{n^2}=1+\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$$
Only thing I managed to do by working the left side was
$$\sum_{n=1}^\infty \frac{1}{n^2} = 1+\sum_{n=2}^\infty \frac{1}{n^2}=1+ \sum_{n=1}^\infty \frac{1}{(n+1)^2}$$
How am I supposed to get to the right form?

HINT
Note the following:
$$\frac{1}{n^2(n+1)}=\frac{1}{n^2}-\frac{1}{n(n+1)}$$ $$\sum_{n=1}^\infty \frac{1}{n(n+1)}=\sum_{n=1}^\infty (\frac{1}{n}-\frac{1}{n+1})=1$$
Can you continue from here?
