Calculate $\sum_{n=1}^\infty\frac{1}{n^2}$ and $\sum_{n=1}^\infty\frac{1}{n^4}$ I need to do that using 
$$\sum_{n \in \mathbb{Z}}\frac{1}{(z-n)^2}=\left(\frac{\pi}{\sin \pi z}\right)^2$$
I've already prove that this is true. The thing is that this function in meromorphic and it's poles are $\mathbb{Z}$. So the natural evauation in $0$ is not possible. I tried to think of a way to do it with that using that the singular part of this function in $n\in\mathbb{Z}$ is $\frac{1}{(z-n)^2}$, but could find a way. What I was looking was sort of substract the term where $n=0$ and divide by $2$. 
For the second sum, I didn't find a clear way. The natural evaluation in $n^2-n$ is not possible since $n$ varies in the sum.
Thanks

Write this as
$$\frac1{z^2}+\sum_{n=1}^\infty\left(\frac{1}{(z-n)^2}+\frac{1}{(z+n)^2}\right).$$
Then
$$\frac1{(z-n)^2}=\frac1{n^2}\frac1{(1-z/n)^2}
=\sum_{k=0}^\infty\frac{(k+1)z^k}{n^{k+2}}$$
and so
$$\frac1{(z-n)^2}+\frac1{(z+n)^2}
=2\sum_{r=0}^\infty\frac{(2r+1)z^{2r}}{n^{2r+2}}.$$
Then
$$\frac1{z^2}+\sum_{n=1}^\infty\left(\frac{1}{(z-n)^2}+\frac{1}{(z+n)^2}\right)=\frac1{z^2}+2\sum_{r=0}^\infty(2r+1)\zeta(2r+2)z^{2r}$$
which you can now compare to the expansion of $\pi^2/\sin^2\pi z$.
