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Prove that the following series define a meromorphic function in $\Bbb C$ $$\sum_{n=1}^\infty\frac{\sinh (n\pi z)}{n!(z^2+n)}$$ and find poles and function poles's order.

My attempt is the following, since $$ \sinh (n\pi z) = \frac{e^{\pi n(x+iy)}-e^{-\pi n(x+iy)}}{2}$$ I think that $$ |\sinh (n\pi z)|\leq \bigg (\frac{e^{\pi |x|}}{2}\bigg)^n$$ so I need $$e^{\pi |x|} < 2 \iff |x| < \frac{\log 2}{\pi}$$ but if this is valid the region of convergence is $x \in \big(-\frac{\log 2}{\pi}, \frac{\log 2}{\pi}\big)$ and the function isn't meromorphic in $\Bbb C$. This led me to think that this argument is valid if $n!$ isn't in the denominator but I actually don't know how to prove it rigorously.

The part regarding function's pole we study the denominator and $$(z^2+n)\neq 0\\ z\neq \pm i\sqrt n$$ each one is a 1-pole and I can easily evaluate the residue. Another thing I think I could say is that $\infty$ is not an isolated singularity because poles accumulates to infinity as $n$ increases. Thank you for your help and time!

Turquoise Tilt
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  • Note that: for any $M>0$, the series $$\sum\frac{M^n}{n!}$$ is convergent. – Riemann Sep 09 '24 at 16:14
  • $z^2+n^2$ in the denominator would seem more natural. – Oscar Lanzi Sep 09 '24 at 16:38
  • $$|\sinh (n\pi z) |\leq\frac{|e^{\pi n(x+iy)}|+|e^{-\pi n(x+iy)}|}{2}\leq\frac{e^{n\pi x}+e^{-n\pi x}}{z}\leq e^{n\pi|x|}.$$ – Riemann Sep 09 '24 at 16:48
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    My proof is similar to the proof in this post due to @Sangchul Lee(https://math.stackexchange.com/questions/1433527/a-n-0-for-all-n-in-mathbb-n-if-sum-a-nk-0-for-all-integers-k-ge-1?noredirect=1), but here is a little hard than that. – Riemann Sep 10 '24 at 10:55

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We have $$|\sinh (n\pi z)|\leq e^{n\pi|x|},\quad \sinh(\pi z)=0\iff z=ki,k=0,\pm1,\pm2,\cdots.$$

Take $D=\mathbb C\setminus\{\pm i\sqrt n\mid n=1,2,\cdots\}$, then $D$ is a domain. Take any compact subset $K$ of $D$, then there exist $\delta>0,R>0$, such that $$|z^2+n|\geq\delta>0,\quad |\sinh (n\pi z)\leq R^n,\quad \forall z\in K.$$ So when $z\in K$, we have $$\left|\frac{\sinh (n\pi z)}{n!(z^2+n)}\right| \leq\frac{R^n}{n!\cdot\delta},$$ since the series $$\sum \frac{R^n}{n!\cdot\delta}$$ is convergent, so $$\sum_{n=1}^\infty\frac{\sinh (n\pi z)}{n!(z^2+n)}$$ converges umiformly on $K$, and Weierstrass's theorem implies that the sum function $$S(z)=\sum_{n=1}^\infty\frac{\sinh (n\pi z)}{n!(z^2+n)}$$ is holomorphic in the domain $D$.

On the other hand, when $k$ is not square, then $z_k=i\sqrt k$ are the simple poles of $S(z)$, because $$\lim_{z\to z_k}S(z) =\lim_{z\to z_k}\left(\frac{\sinh (k\pi z)}{k!(z^2+k)}+\sum_{n\neq k}^\infty\frac{\sinh (n\pi z)}{n!(z^2+n)}\right) =\infty+\sum_{n\neq k}^\infty\frac{\sinh (n\pi z_k)}{n!(z_k^2+n)} =\infty,$$ and $$\lim_{z\to z_k}(z-z_k)S(z)=\frac{\sinh (k\pi z_k)}{k!(2i\sqrt k)}.$$ In the same way, when $k$ is not a square, $-z_k$ are also the simple poles of $S(z)$.

Hence, $S(z)$ is a meromorphic function in $\mathbb C$, the poles are $\pm i\sqrt k$, where $k$ is not a square, and all the poles are simple poles.

Riemann
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  • Why would you go on discuss if $z_k$ is a square or not? In my mind it's either way a pure imaginary number and I don't think that this could lead to some difference relatively to $k$, the rest is very clear! – Turquoise Tilt Sep 10 '24 at 08:55
  • I think I got it, we have to check that numerator and denominator do not vanish on the same values in order to have a pole, but this do not seems to happen for any value of $k\neq 0,1$ am I right? Are you sure about the zero of the numerator, is $\sinh$ periodic? – Turquoise Tilt Sep 10 '24 at 09:02
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    @Turquoise Tilt For fixed $n$, the numerator $$\sinh(n\pi z)=0\iff e^{n\pi z}=e^{-n\pi z}\iff e^{2n\pi z}=1\iff 2n\pi z=2m\pi i \iff z=\frac{m}{n}i, m\in\mathbb Z.$$ There are infinity many zeros for $\sinh(n \pi z)$, when choose $m=nk$, then $z=k i$. The common zeros of the numerator and denominator are only those cases. – Riemann Sep 10 '24 at 10:43
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    $\sinh(z)$ is not periodic, but $e^z$ is periodic, i.e. $$e^{z+2m\pi i}=e^z,\quad \forall z\in\mathbb C,m\in\mathbb Z.$$ – Riemann Sep 10 '24 at 10:45