Why does absolute convergence allow the order of summation to be changed? (Laurent expansion of the Weierstrass $\wp$-function) I'm working through the derivation of the Laurent expansion of the Weierstrass P function (Theorem 1.11) in Tom Apostol's Modular functions and Dirichlet series in Number Theory.
The proof uses the series $\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}=\sum_{n=1}^\infty\frac{n+1}{\omega^{n+2}}z^n$, which converges absolutely for $|z/\omega|<1$. Here, $\omega\in\Omega$ where $\Omega\subset\mathbb{C}$ is a lattice. Next, Apostol sums these terms for all $\omega$, giving
$$
p(z)=\frac{1}{z^2}+\sum_{\omega\neq 0}\sum_{n=1}^\infty\frac{n+1}{\omega^{n+2}}z^n=\frac{1}{z^2}+\sum_{n=1}^\infty (n+1)\sum_{\omega\neq 0}\frac{z^n}{\omega^{n+2}}.
$$
Apostol hints that absolute convergence is what allows the interchange of these sums. I think the fact that $\sum_{\omega\neq 0}\frac{1}{\omega^\alpha}$ is absolutely convergent for all $\alpha>2$ is also important. However, I'm struggling to fill in the details (it feel like I missed something important in my analysis education!). Any help is appreciated :)
 A: As a preliminary remark, note that the fact that we can rearrange the double series is already implicit in the notation. Namely, we have not specified any ordering of the lattice points.

You suspect that you might be missing some results from analysis. As reuns hints at in the comments, the result you are looking for might be phrased as follows:
Theorem (Fubini's theorem for infinite series). Let $g\colon \mathbb{N}\rightarrow \mathbb{N}\times \mathbb{N}$ be a bijection. Let $\{a_{m,n}\}_{(m,n)\in \mathbb{N}^2}$ be a doubly-indexed set of complex numbers. Assume that the series $\sum_{k=1}^{\infty}a_{g(k)}$ converges absolutely. Then, for all $m,n\in \mathbb{N}$, the series $$\sum_{j=1}^{\infty}a_{m,j}\ \ \ \textit{ and } \ \ \ \sum_{i=1}^{\infty}a_{i,n} $$ converge absolutely. Additionally, we have
$$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}a_{n,m}=\sum_{k=1}^{\infty}a_{\sigma(k)}=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} a_{n,m}.$$
Proof sketch. This might be using a sledgehammer to crack a nut …
Namely, consider the function $f\colon \mathbb{N}^2\rightarrow \mathbb{C}; (n,m)\mapsto a_{n,m}.$ Since by assumption the series $\sum_{k=1}^{\infty}a_{g(k)}$ converges absolutely, the function $f\circ g$ is integrable with respect to the counting measure $\mu$ on $\mathbb{N}$. Additionally, we have $\int_{\mathbb{N}}(f\circ g) \, d\mu=\sum_{k=1}^{\infty}f(g(k))$, where $\int_{\mathbb{N}}(f\circ g) \, d\mu$ denotes the Lebesgue-integral of $f\circ \sigma$ on $\mathbb{N}$ with respect to the counting measure $\mu$ on $\mathbb{N}$. A proof of these statements in the case of real numbers can be found here. We conclude that $f$ is Lebesgue-integrable on $\mathbb{N}^2$ with respect to the counting measure and that $\int_{\mathbb{N}^2}f \ d\mu=\sum_{k=1}^{\infty}f(g(k))$.
Next, for $m,n\in \mathbb{N}$, set
$$A_m\colon=\{m\}\times \mathbb{N}, \ B_n\colon =\mathbb{N}\times \{n\}.$$
Then we have
$$\sum_{k=1}^{\infty}f(g(k))=\int_{\mathbb{N}^2}f \, d\mu=\int_{\bigcup_{m\in \mathbb{N}}A_m}f \, d\mu=\sum_{m=1}^{\infty}\int_{A_m}f(m,n) \, d\mu=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}a_{m,n}.$$
In the last step, we have used that the Lebesgue-integral is a sigma-additive set function. Analogously, by using $B_n$, one shows $\sum_{k=1}^{\infty}f(g(k))=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}a_{m,n}.$

Let us turn back to your problem. By the above theorem, to prove that we are allowed to exchange the order of summation, it suffices to prove that the series $$\sum_{\omega\neq 0 }\sum_{n=1}^{\infty}\frac{(n+1)}{\vert \omega \vert ^{n+2}}\cdot \vert z\vert ^n$$ converges for $0<\vert z\vert< \operatorname{min}\{\vert \omega \vert \colon \omega \in \Omega \text{ and } \omega\neq 0\}.$ Again, we do not specify an ordering of the lattice points. By above theorem, if we prove absolute convergence for any ordering, then we are done.
Using the formula for the differentiated geometric series (and the fact that with our choice of $z$ we have $\vert\frac{z}{\omega}\vert<1$ for all $\omega \neq 0$), one shows $$\sum_{\omega\neq 0 }\sum_{n=1}^{\infty}\frac{(n+1)}{\vert \omega \vert ^{n+2}}\cdot \vert z\vert ^n=\sum_{\omega\neq 0}\Big(\frac{1}{(\vert \omega \vert-\vert z \vert)^2}-\frac{1}{\vert \omega \vert^2}\Big).$$ The right-hand side is a convergent series. This can be established by using the absolute convergence of the Eisenstein series $\sum_{\omega \neq 0}\frac{1}{\omega^3}$ as follows:
Let $\omega \in \Omega \setminus \{0\}$ and $0<\vert z \vert \leq \frac{\vert \omega \vert}{2}$.
First, note that $$\frac{1}{(\vert z \vert -\vert \omega \vert)^2}-\frac{1}{\vert \omega \vert ^2}=\frac{2\vert z \vert \vert \omega \vert - \vert z \vert ^2}{(\vert z \vert - \vert \omega \vert)^2\vert \omega \vert ^2}.$$
Additionally, we then have $$(\vert z \vert - \vert \omega \vert)^2\vert \omega \vert ^2= (\vert \omega \vert - \vert z \vert)^2\vert \omega \vert ^2\geq \frac{\vert \omega \vert ^4}{4}.$$  By the triangle inequality
$$2\vert z \vert \vert \omega \vert -\vert z\vert^2 \leq \big\vert 2\vert z \vert \vert \omega \vert -\vert z\vert^2\big\vert \leq 2\vert z \vert \vert \omega \vert +\vert z\vert^2\leq \frac{5}{2}\vert z \vert \vert \omega \vert.$$
Thus, we have
$$\sum_{\vert \omega \vert \geq 2\vert z \vert} \Big(\frac{1}{(\vert \omega \vert-\vert z \vert)^2}-\frac{1}{\vert \omega \vert^2}\Big)\leq 10\vert z \vert \sum_{\vert \omega \vert \geq 2\vert z \vert} \frac{1}{\vert \omega \vert^3}.$$
Since the Eisenstein series $\sum_{\omega \neq 0}\frac{1}{\omega^3}$ converges absolutely, this shows $$\sum_{\vert \omega \vert \geq 2\vert z \vert} \Big(\frac{1}{(\vert \omega \vert-\vert z \vert)^2}-\frac{1}{\vert \omega \vert^2}\Big)<\infty.$$
Finally, note that
$$\sum_{\omega \neq 0} \Big(\frac{1}{(\vert \omega \vert-\vert z \vert)^2}-\frac{1}{\vert \omega \vert^2}\Big)= \sum_{0<\vert \omega \vert < 2\vert z \vert} \Big(\frac{1}{(\vert \omega \vert-\vert z \vert)^2}-\frac{1}{\vert \omega \vert^2}\Big)+ \sum_{\vert \omega \vert \geq 2\vert z \vert} \Big(\frac{1}{(\vert \omega \vert-\vert z \vert)^2}-\frac{1}{\vert \omega \vert^2}\Big).$$ The first sum is finite, while the second converges, as we have shown.
