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For what values of $z$ does the series $\sum_{n=0}^\infty \frac{1}{n^2 + z^2}$ converge?

I tried to solve this using the ratio test to get the radius of convergence, but the results were inconclusive. For this reason I tried to compare it to $\sum_{n=0}^\infty \frac{1}{n^2}$. However, even using the comparison test, I am not sure how I can use these results to determine the values of $z$ that enable the series to converge.

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  • $\begingroup$ $z$ is a complex number? $\endgroup$
    – sranthrop
    Sep 14, 2016 at 4:38
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    $\begingroup$ The radius of convergence is defined for power series, and your series is not. It converges for any $z\in\mathbb{R}^*$ by the p-test and its value is given by $$ \frac{1}{2} \left(\frac{1}{z^2}+\frac{\pi \coth(\pi z)}{z}\right) $$ by considering the logarithmic derivative of a Weierstrass product. $\endgroup$
    – Jack D'Aurizio
    Sep 14, 2016 at 4:38
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    $\begingroup$ $z^{2} \not= 0,-1,-4,\ldots\implies z \not= 0,\pm\,\mathrm{i},\pm 2\,\mathrm{i},\ldots$ $\endgroup$
    – Felix Marin
    Sep 14, 2016 at 5:01

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First we have to find the values for which $1/(z^2+n^2)$ is well defined for any $n$. We have to be sure that $z^2+n^2$ does not vanish for any $n\in\mathbb N$. Since the solutions of the equation $z^2+n^2=0$ are $in$ and $-in$, we have to investigate the convergence of $S:=\sum_{n=1}^{ +\infty}1/(z^2+n^2)$ for $z\notin \mathbb Z i$. For these $z$, we have $$\lim_{n\to +\infty} \frac{n^2}{\left|z^2+n^2\right|}=1 $$ (using the triangle inequality for the denominator) and we derive the absolute convergence of $S$.

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  • $\begingroup$ Why do we have $n^2$ in the numerator of the final limit, instead of $1$. $\endgroup$
    – user366704
    Sep 14, 2016 at 12:36
  • $\begingroup$ Otherwise it would be wrong. We compare $1/|z^2+n^2|$ with $1/n^2$. $\endgroup$
    – Davide Giraudo
    Sep 14, 2016 at 14:33

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