$\sum_{n=1}^\infty\frac{z^2}{1+n^2z^2}$ converges to analytic function

Question

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

It might be easier to write $z = 1/w$ and consider the sum as a function of $w$. — Daniel Fischer, Dec 19 '13 at 20:22
This should be $\frac{iz}{2}\left( \pi\cot (i\pi/z)+iz\right)$. — Julien, Dec 19 '13 at 20:27

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

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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

edited Apr 13 '17 at 12:20

Community

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answered Dec 19 '13 at 20:27

Norbert

56,803

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(+1) I don't see right away why this is true. Could you give me a link or some explanation? – tylerc0816 Dec 19 '13 at 20:29
soon I'll post an explanation – Norbert Dec 19 '13 at 20:31

$\sum_{n=1}^\infty\frac{z^2}{1+n^2z^2}$ converges to analytic function

1 Answers1