Divergence of sum over lattice. This is  follow up to my last question on summing over countably infinite sets. I understand the idea conceptually now but I am still stuck when dealing with a concrete example. Specifically, consider 
$$\sum_{\omega \in \Lambda^*}\frac{1}{\omega^2}$$ where $\Lambda^*$ is a lattice in the complex plane without the origin. This sum has come up in several books when the author is defining the $\wp$ elliptic function. Apparently, it diverges but I haven't been able to find the details of why it does. Intuitively, I would think it would converge since for large $|\omega|$ one would expect $|1/\omega^2|$ to decay very rapidly. 
How would I even begin to analyze such a sum? I think the issue is my intuition from the sum $\sum \frac{1}{n^2}$ which does converge is corrupting my understanding. 
Thanks.
 A: Roughly speaking the sum behaves (for large $m,n$) like $\sum_{m,n \ne 0}\frac{1}{m^2+n^2}$ and that is divergent since the sum say in m for fixed $n$ is about $\frac{1}{n}$, so the double sum behaves like the harmonic sum. 
(if $a \ge 1$, $\int_1^{\infty}\frac{dx}{x^2+a^2}<\sum_{m \ge 1}\frac{1}{m^2+a^2}<\int_0^{\infty}\frac{dx}{x^2+a^2}$ so $\frac{\pi}{2a}-O(\frac{1}{a^2})< \sum_{m \ge 1}\frac{1}{m^2+a^2} < \frac{\pi}{2a}$
A: In short, the number of lattice points of given bounded magnitude grows linearly, giving a contribution proportional to $n \cdot \frac{1}{n^2} = \frac{1}{n}$, making the series diverge.
I give a proof that $\sum_{\omega \in \Lambda^*} \omega^{-s}$ converges absolutely iff $s>2$:
Fix a lattice $\Lambda \in \Bbb C$ and set $\Omega_r = \{m\lambda_1 + n\lambda_2 | m,n \in \Bbb Z \,\,\text{and}\,\, \max(|m|, |n|) = r \}$.
Then $\Lambda^*$ is a disjoint union of the $\Lambda_r$, $r > 0$. Observe that $|\Lambda_r| = 8r$.
Let $D$ and $d$ be the greatest and least moduli of the elements of the parallelogram $\Pi_1$ containing $\Lambda_1$. Then we have $rD \geq |\omega| \geq rd$ for all $\omega \in \Lambda_r$.
Define $\sigma_{r, s} = \sum_{\omega \in \Lambda_r} |\omega|^{-s}$.
$\sigma_{r, s}$ lies between $8r(rD)^{-s}$ and $8r(rd)^{-s}$. Therefore $\sum_{r=1}^\infty \sigma_{r, s}$ converges iff $\sum r^{1-s}$ converges, i.e. iff $s > 2$. 
The claim follows.
This proof follows the one in Jones and Singerman's Complex functions pp.91.
