This topic is about the following exercise:
(i) Show that the series $$(1) \;\;\;\;\;\; f(z) := \sum_{n=1}^\infty \frac{z^2}{8+z^2n^2}$$ converges absolutely for every $z \in \mathbb R$ and that the map$f \colon z \mapsto f(z)$ is continuous on $\mathbb R$.
(ii) Find the largest open set $U \subseteq \mathbb C$, such that $f$ given by $(1)$ is well-defined and holomorphic on $U$.
Question: I am looking for an elementary solution of part (ii) of the exercise given above. One solution I thought of is written down below, it uses the partial fraction expansion of the cotangent. On the other hand, the solution given below does more than required in the exercise, i.e., it gives a closed form of $f$. That is why I hope that one can find a more elementary solution to (ii) (one, which is independent of the partial fraction expansion of the cotangent, or one, which gives the set $U$ without having to calculate a closed form of $f$).
One solution involving the partial fraction expansion of the cotangent:
I will focus on (ii), as (i) is clear.
It is obvious that $f$ is not defined in the zeroes $\pm \frac{\sqrt 8 i}{n}$ of the denominators of the $n$-th summand of $f$. Since these zeroes tend to $0$ as $n \to \infty$, we obtain that
$$U \subseteq \mathbb C \setminus \left( \left\{ \frac{\sqrt 8 i}{n} \; \bigg| \; n \in \mathbb Z \setminus \{0\} \right\} \cup \{ 0 \} \right).$$
We will show that indeed this is an equality.
The partial fraction expansion of the cotangent tells us that
$$\forall x \in \mathbb C \setminus \mathbb Z \colon \;\;\; \pi \cot(\pi x) = \frac1z + \sum_{n=1}^\infty \frac{2x}{x^2 - n^2}.$$
Substituting $w := \frac1x$, one obtains
$$\frac{\pi}{2w} \cot\left(\frac{\pi}{w}\right) = \frac12 + \sum_{n=1}^\infty \frac{1}{1 - w^2n^2}.$$
Now, substituting $z := iw\sqrt{8}$ gives
$$\frac{\pi\sqrt{2}}{z} \cot\left(\frac{\pi i \sqrt{8}}{z}\right) = \frac12 + 8\sum_{n=1}^\infty \frac{1}{8 + z^2n^2},$$
or, equivalently,
$$\frac{z\pi}{4\sqrt 2} \cot\left(\frac{\pi i \sqrt{8}}{z}\right) - \frac{z^2}{16} = \sum_{n=1}^\infty \frac{z^2}{8 + z^2n^2}.$$
This proves the other inclusion.