This is from Complex Analysis by Shakarchi and Stein in the chapter of Elliptic functions.
How do we show that ${\sum}_{w\in\wedge}\frac{1}{(z+w)^2}$ is not absolutely convergent, where $\wedge$ is a double lattice in complex plane? It seems difficult for me to show the absolute convergence of the complex infinite sums. Maybe I can try to bound this series and use the fact from real series, but I don't know how.
By a rotation we can assume the lattice is $m+n\tau, \tau=a+ib, b>0$ and wlog we can assume $a \ge 0$ as otherwise we use $n <0$ in what follows.
Fix $z=x+iy$, so $|z+m+na+inb|^2=(m+na+x)^2+(nb+y)^2$.
Then if $Nb>|y|$, we get $(nb+y)^2<4b^2n^2, n \ge N$
and similarly $M>0, M+Na >|x|$ implies $(m+na+x)^2<4(m+na)^2, m \ge M, n \ge N$
This means that $\frac{1}{|z+m+na|^2} \ge \frac{1}{4b^2n^2+4(m+na)^2}, m \ge M, n \ge N$
But now summing only those terms and calling that sum $S$ we get that:
$S \ge \sum_{m \ge M, n \ge N}\frac{1}{4b^2n^2+4(m+na)^2}$
Using that a double series of positive numbers can be interchanged at will (with the same result either finite or infinite) we immediately get (as summand is decreasing in $m$) that for fixed $n \ge N$:
$\sum_{m \ge M}\frac{1}{b^2n^2+(m+na)^2} \ge \int_{M+1}^{\infty}\frac{dt}{b^2n^2+(t+na)^2}=$
$=\frac{1}{bn} \tan^{-1}(\frac{t+na}{nb})|_{t=M+1}^{t=\infty}=\frac{1}{bn}(\pi/2-\tan^{-1}(\frac{M+1+na}{nb})) \ge \frac{1}{bn}(\pi/2-c) =A/n, n \ge N$
where $c=\tan^{-1}(\frac{M+1+Na}{Nb})$ as $\frac{M+1+na}{nb} \le \frac{M+1+Na}{Nb}, n \ge N$ and the arctangent is increasing
But this shows that $S \ge \sum_{n \ge N}\frac{A}{4n}=\infty$ so the double series of absolute values on a lattice subset is already infinite and we are done!