How to show that the Weierstrass function is meromorphic in whole of $\mathbb{C}$ The Weierstrass function is defined as:
$$\wp(z) = \frac{1}{z^2}+ {\sum\limits_{\omega\in\Gamma'}} \left (\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\right).$$
Fix $r>0$. For $\left|z\right| \leq r,$ we split the sum in two parts:
\begin{align}
\wp(z) = \frac{1}{z^2}+{\sum_{\substack{\omega\in\Gamma', \\ \left|\omega \right|\leq 2r}}} \left (\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\right) + {\sum_{\substack{\omega\in\Gamma', \\ \left|\omega \right|>2r}}} \left (\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\right).
\end{align}
Clearly $z=0$ is a pole of $\wp$. Also, the first part of the summation meromorphic function. Now we will show that the second part of the summation is bounded above and therefore cannot have any pole. Note that $$ \frac{1}{(z-\omega)^2}-\frac{1}{\omega^2} = \frac{z(2\omega - z)}{\omega^2 \left(z-w\right)^2}. $$
Assuming $\left|\omega\right| > 2\left|z\right|$,
$$\left| 2\omega - z \right| \leq \left| 2\omega \right| + \left| z \right| < \frac{5}{2}\left| \omega \right|,$$
$$ \left| z-\omega \right| \geq \left| \omega \right| - \left| z \right| > \frac{\left| \omega \right|}{2}.$$
This implies $$ \left| \dfrac{1}{(z-\omega)^2}-\dfrac{1}{\omega^2} \right| \leq \dfrac{\left| z \right| \times \dfrac{5}{2} \left| \omega \right|}{\left| \omega \right|^2 \times \dfrac{\left| \omega \right|^2}{4}} = \dfrac{10\left|z\right|}{\left|\omega\right|^3} \leq \dfrac{10r}{\left|\omega\right|^3}$$
We know that $\sum \frac{1}{\omega^3} < \infty$. Hence $\wp$ is a meromorphic function in $|z|\leq r$.
Now how to show that $\wp$ is meromorphic in $\mathbb{C}$?
I have seen some proofs and I am not able to understand how do they proceed after this step.
 A: For completeness sake, although the question is from years ago, I'll comment what I think is the missing piece in OP's answer.
OP's proof shows the uniform convergence of the function in every disk centered in $0$. What's left to say is that if a function converges uniformly in every compact of an open set then you have a meromorphic function in that set. Given that you have uniform convergence in a set of compacts covering the whole plane (the disks of radius $r$ in this case), you have uniform convergence in every compact just by taking some disk covering it.
Worth mention that this whole process is basically the path to proof Mittag-Lefler theorem via Runge's theorem in an open subset of the complex plane. In this theorem you take some Laurent tails on some discrete set of points and you want to build a meromorphic function. To put it another way, this theorem proofs the existence of a meromorphic function exactly with the poles you want. For that aim, you'd like to use the sum of these tails (called principal parts) in some way so you take an increasing sequence of compacts covering the open set. In each compact, you'd like to take the sum of the principal parts holomorphic in that compact minus some rational function sufficiently close to get uniform convergence (the existence of this rational function is guaranteed by Runge's theorem and in OP's case it's enough with the first term of the Taylor polynomial).
Then in each compact some you separate the function in two parts, just as OP did: one holomorphic which converges uniformly and the meromorphic which is just a finite sum of poles. So your function is a meromorphic function in the compact. The function is meromorphic in every compact of the sequence so it's meromorphic in the whole domain.
