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?
So if I got it right it goes something like this $$\sum_{n=1}^\infty \frac{1}{n^2}=1+\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$$ $$=1+\sum_{n=1}^\infty (\frac{1}{n^2} - \frac{1}{n(n+1)})$$ $$=1+\sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty (\frac{1}{n(n+1)})$$ $$=1+\sum_{n=1}^\infty \frac{1}{n^2}-\sum_{n=1}^\infty (\frac{1}{n}-\frac{1}{n+1})$$ and because of telescoping property $$\sum_{n=1}^\infty (\frac{1}{n}-\frac{1}{n+1}) = 1 $$ so $$\sum_{n=1}^\infty \frac{1}{n^2}=1 + \sum_{n=1}^\infty \frac{1}{n^2} - 1 = \sum_{n=1}^\infty \frac{1}{n^2}.$$
