I have the definition of the Weierstrass elliptical function $\wp$ as: $$\wp(z)=\frac{1}{z^2}+\sum_{\omega\neq 0}\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}$$
Where $w=n\omega_1+m\omega_2$, for $\omega_1$ and $\omega_2$ the periods of $\wp$.
Also I have the definition for the anti-derivative of $\wp$, $\zeta$. We have $\zeta'(z)=-\wp(z)$ with $$\zeta(z)=\frac{1}{z}+\sum_{\omega\neq0}\frac{1}{z-\omega}+\frac{1}{\omega}+\frac{z}{\omega^2}$$
In order to find the laurent expansion of $\wp$ I am trying to find the laurent expansion of $\zeta$ and then differentiating termwise as $\zeta$ and $\wp$ converge uniformly. Beginning:
$$\frac{1}{z-\omega}=\frac{-1}{\omega(1-\frac{z}{\omega})}=\frac{-1}{\omega}\sum_{n=0}^{\infty}\frac{z^n}{w^n}$$ and so $$\frac{1}{z-\omega}+\frac{1}{\omega}+\frac{z}{\omega^2}=-\frac{z^2}{\omega^3}-\frac{z^3}{\omega^4}$$ And now I am stuck. In Ahlfors complex analysis text Chapter $7$, Section $3.3$ he observes that when we sum over all the periods the terms with odd powers of the periods are zero (equivalently the even powers of $z$). I can't follow this step. I know it must be true as $\zeta$ is an odd function. However why can we jump to this conclusion?
$$ \wp(z)=\frac{1}{z^2}+\sum_{\omega\neq 0}\Big[\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\Big]=\frac{1}{z^2}+\sum_{\omega\neq 0}\Big[\frac{1}{\omega^2}\frac{1}{(1-\frac{z}{\omega})^2}-\frac{1}{\omega^2}\Big] $$
$$=\frac{1}{z^2}+\sum_{\omega\neq 0}\frac{1}{\omega^2}\sum_{m=1}^{\infty}(m+1)\Big(\frac{z}{\omega}\Big)^{m}=\frac{1}{z^2}+\sum_{m=1}^{\infty}\Big[\sum_{\omega\neq 0}\frac{m+1}{\omega^{m+2}}\Big]z^m$$
using the fact that $$ -1+\frac{1}{(1-x)^2}=\sum_{m=1}^{\infty}(m+1)x^m $$ for $|x|<1$, and using the absolute convergence of the $\wp$ function to interchange the sums.
Finally, if $m$ is odd then $\omega^{m+2}=-(-\omega)^{m+2}$, hence $$ \sum_{\omega\neq 0}\frac{m+1}{\omega^{m+2}}=0 $$ because for every non-zero element $\omega$ there is a corresponding term $-\omega$ in the sum. This shows that all the odd powers vanish in the Laurent expansion.