How to show the "naive" Weierstrass elliptic function does not converge absolutely Several resources (e.g., Stein and Shakarchi, Complex Analysis) begin a discussion of the Weierstrass $\wp$ function by saying that, in order to construct a doubly periodic meromorphic function with lattice $L$, a good first guess is the function
$$ f(z) = \sum_{\omega \in L}\frac{1}{(z-\omega)^2}$$
however, the series fails to converge absolutely, which is why the $\wp$ function is defined the way it is.  I cannot, however, find any resource that actually goes about showing why the series fails to converge.  Maybe it is a trivial calculation and I am just not seeing the answer, but could someone please rigorously show that this series fails to converge absolutely?  Part of the problem I am having in understanding this series is that it is indexed over a set that is not the positive integers, so I'm not sure what a partial sum would even look like exactly.
Any help is greatly appreciated, thanks!
 A: When I first studied elliptic functions I had the exact same question.
In what follows, for convenience, I will write ${\sum \limits_{\omega \in L}} '$ instead of $ \sum \limits_{\omega \in L \setminus\{0\}}$.
I convinced myself that the function $f(z)$ you give is divergent by looking at
$$
f(0)={\sum \limits_{\omega \in L}} '\frac{1}{\omega^2}
$$
and proving that it diverges for a particular choice of $L$.
Let $L := \left\{ m \alpha + i n \beta : m,n \in \mathbb{Z}\right\}$, for fixed $\alpha, \beta \in \mathbb{R}$. Then
\begin{align*}
{\sum \limits_{\omega \in L}} '\frac{1}{\omega^2} & =  {\sum \limits_{m,n \in \mathbb{Z}}} '\frac{1}{(m \alpha + i n \beta )^2} \\ 
 & = \left(   {\sum \limits_{m,n \in \mathbb{Z}}} '\frac{m^2\alpha^2-n^2\beta^2}{(m^2 \alpha^2 + n^2 \beta^2)^2 } \right) + i  \left(   {\sum \limits_{m,n \in \mathbb{Z}}} '\frac{2mn\alpha\beta}{(m^2 \alpha^2 + n^2 \beta^2)^2} \right)
\end{align*}
We now show that the imaginary part diverges. Well,
$$
  {\sum \limits_{m,n \in \mathbb{Z}}} '\frac{2mn\alpha\beta}{(m^2 \alpha^2 + n^2 \beta^2)^2} = 2\alpha\beta   {\sum \limits_{m\in \mathbb{Z}}} ' m   {\sum \limits_{n\in \mathbb{Z}}} \frac{n}{(m^2 \alpha^2 + n^2 \beta^2)^2}.
$$
Define
$$
S_m(\alpha, \beta) :=  \sum \limits_{n=1}^{\infty} \frac{n}{(m^2 \alpha^2 + n^2 \beta^2)^2} 
$$
and a function $f_m(\alpha, \beta)$ by
$$
f_m(\alpha, \beta) (x) :=  \frac{x}{(m^2 \alpha^2 + n^2 \beta^2)^2} 
$$
A single variable calculus argument shows that $f_m(\alpha, \beta)$ is decreasing function, so we can use the integral test for convergence of series. Indeed,
$$
\int\limits_{m}^{\infty} f_m(\alpha, \beta) (x)  dx = \int\limits_{m}^{\infty}  \frac{xdx}{(m^2 \alpha^2 + n^2 \beta^2)^2} = \frac{1}{2\beta^2(\alpha+\beta^2)m^2}
$$
and
$$
S_m(\alpha, \beta)  \geq \frac{1}{2\beta^2(\alpha+\beta^2)m^2},
$$
whence
$$
 {\sum \limits_{m,n \in \mathbb{Z}}} '\frac{1}{(m \alpha + i n \beta )^2}  \geq  2\alpha\beta   {\sum \limits_{m\in \mathbb{Z}}} ' m S_m(\alpha, \beta) \geq 2\alpha\beta   {\sum \limits_{m\in \mathbb{Z}}} '  \frac{1}{2\beta^2(\alpha+\beta^2)m^2} = \infty
$$
as wanted.
