$\sum_{n=1}^\infty\frac{z^2}{1+n^2z^2}$ converges to analytic function 
For which $z$ does $\sum_{n=1}^\infty\dfrac{z^2}{1+n^2z^2}$ converge to an analytic function? What are its poles?

I think the poles should be $\pm\dfrac{i}{n}$, since those are the values at which one of the denominators disappear. I'm not sure about the converging to analytic function part

This sum have closed form. As it was proved earlier
$$
f(w)=\sum\limits_{n=1}^{\infty}\frac{1}{n^2-w^2}=\frac{1}{2w}\left(\frac{1}{w}-\pi\cot(\pi z)\right)
$$
so
$$
\sum_{n=1}^\infty\dfrac{z^2}{1+n^2z^2}=f(i/z)=-\frac{1}{2} z \left(z-\pi  \coth \left(\frac{\pi }{z}\right)\right)
$$
So the question simplifies significantly
