Showing a series of functions converges to a periodic function. Let $z = x+iy$ and let $k \in \mathbb{C}$ be a constant.
I am reading a paper from Perelman where he considers the function,
\begin{align*}
\mathrm{f}&:\mathbb{C}\backslash\ D \longrightarrow \mathbb{R}\\[3mm]
&:z \longmapsto \mathfrak{R}[\ln\left(1+\frac{4k^{3}}{(z + k)^{2} (z - 2k)}\right)]
\end{align*}
where $k \in \mathbb{C}$ is a constant and $D$ is a discrete subset of $\mathbb{C}$
Perelma says that that the series converges
$$
\sum_{a,b\ \in\ \mathbb{Z}}\mathrm{f}\left(z + a + \mathrm{i}b\right)
$$
converges to a function, $\mathrm{g}\left(z\right)$, which $\textbf{is}$ $1$-periodic. It is not clear to me whether I have to find this function $\mathrm{g}\left(z\right)$ explicitly in order to show this $1$-periodicity ?. In any case, I am quite at a loss as how to move forward to show the $1$-periodicity and convergence of this series.
By "$1$-periodic", I mean with respect to $x$ and $y$.
$\textbf{EDIT/UPDATE}$: I am now clear on the $1$-periodicity of the series, however it is still unclear to me why this series converges?
$\textbf{EDIT/UPDATE}$: I realised I copied Perelman's function incorrectly, I have since changed this in the text above, also you can see the relevant part of his paper here.
 A: Edit: With the edit on the post, I'm undeleting my answer (with corrections thanks to Lukas' comments)
For completeness I'll port my comment to an answer.
First, periodicity comes for free by definition:
$$
g(z+n+mi) = \sum_{a,b\in\mathbb Z}f(z+n+mi+a+bi) = \sum_{a,b\in\mathbb Z}f(z+(n+a)+(m+b)i) = \sum_{a',b'\in\mathbb Z}f(z+a'+b'i) = g(z)
$$
by reindexing with $a' := n+a$ and $b' := m+b$.
As for convergence, we can bound $0\leq\log(1+x)\leq x$ for all $x\geq0$ (indeed, this is equality at $x=0$ and otherwise check their derivatives) and on the other hand $0\geq\log(1+x)\geq2x$ when $-\frac12\leq x\leq0$. For complex $w$, we have $\Re\log (1+w)=\log|1+w|$, so if we assume $|w|\leq\frac12$, then by both triangle inequalities and monotonicity of $\log$ as a real function, we get
$$
-2|w|\leq\log(1-|w|)=\log|1-|w||\leq\log|1+w|\leq\log(1+|w|)\leq|w|
$$
and thus $|\Re\log(1+w)|\leq2|w|$ for all $|w|\leq\frac12$.
In particular, $|f(z)| = \left|\Re\log\left(1+\frac{4k^3}{(z+k)^2(z-2k)}\right)\right| \leq \frac{2|4k^3|}{|z+k|^2|z-2k|} \in O\left(\frac1{|z|^3}\right)$ as $|z|\to\infty$, from which we can see that the series defining $g(z)$ converges absolutely for almost every $z$.
