TITLE: Weierstrass $\wp$ function doubly periodic
QUESTION [9 upvotes]: I'm working my way through Silverman and Tate's Undergraduate Introduction to Elliptic Curves.  I haven't yet been able to study complex analysis, so it comes as no surprise that I'm having a tough time with that portion of the book right now.  
Let $\omega_1, \omega_2 \in \mathbb{C}$ be two complex numbers which are $\mathbb{R}$-linearly independent and let:
$$L = \mathbb{Z}\omega_1 + \mathbb{Z}\omega_2 = \{n_1\omega_1 + n_2\omega_2 : n_1, n_2 \in \mathbb{Z}\}$$
Let 
$\wp(u) = \frac{1}{u^2} + \sum\limits_{\omega \in L, \omega \neq 0} \left(\frac{1}{(u-\omega)^2} - \frac{1}{\omega^2}\right)$ 
Show that $\wp$ is a doubly periodic function, that is, show that 
$$\wp(u + \omega) = \wp(u)$$
If you are able, please give me a shove in the right direction.  Thank you!
Dear Answerers: Thank you, I have been able to figure it out.  Yes, convergence was quite tricky and I was trying to make this particular question much more difficult than it actually was.  Thank you!

REPLY [7 votes]: It is easy to show that $\wp'(z)$ is doubly periodic because replacing $z$ by $z + \omega_k$ in $\wp'(z)$ does not change the series because it is doubly infinite. Now, the periodicity of $\wp'(z)$ implies that $\wp'(z + \omega_k) - \wp'(z) = 0$, that is, $\wp(z + \omega_k) - \wp(z) = A = \text{const}$, but $\wp(z)$ is even, so upon setting $z = -\omega_k/2$ we deduce that $A = 0$.