TITLE: Showing Weierstrass elliptic function has periods
QUESTION [7 upvotes]: Let $$\wp(z)=\frac{1}{z^2}+\sum_{w \in \Lambda^*} \left[\frac{1}{(z+w)^2}-\frac{1}{w^2}\right]
$$
be the Weierstrass elliptic function with $\Lambda=\Bbb{Z}+\Bbb{Z}\tau$, $\Lambda^*=\Lambda-0$. I want to show that $\wp(z+w)=\wp(z)$ whenever $w \in \Lambda$, without using differentiation. [Stein, Complex Analysis p.279]

Suggested hint: For large $R$, $\wp(z)=\wp^R(z)+O(1/R)$ where
  $$\wp^R(z)=\frac{1}{z^2}+\sum_{0<|w|<R}
 \left[\frac{1}{(z+w)^2}-\frac{1}{w^2}\right]$$ Also observe that
  $\wp^R(z+1)-\wp^R(z)$ and $\wp^R(z+\tau)-\wp^R(z)$ are
  $O(\sum_{R-c<|w|<R+c} |w|^{-2})=O(1/R)$

But I can't understand the hint. I know that for $|z|<R$, $\sum_{|w|>2R}
 \left[\frac{1}{(z+w)^2}-\frac{1}{w^2}\right]$ is $O(1/|w|^3)$ (uniformly) so defines a holomorphic function in $|z|<R$. But is the above hint also correct? $O(1/R)$ is ambiguous and restriction of $z$ seems to needed. So if I understood $\wp^R(z)$ as $$\wp^R(z)=\frac{1}{z^2}+\sum_{0<|w|<2R}
 \left[\frac{1}{(z+w)^2}-\frac{1}{w^2}\right] \quad (|z|<R)$$ But still don't know how to compare $\wp^R(z+1)-\wp^R(z)$. What should I do?

REPLY [4 votes]: I tried a slightly different approach that I think exposes the main ideas a bit more clearly.
Define
$$
\Omega(h,k)=\left\{\begin{array}{}
0&\text{if }h=k=0\\
1&\text{otherwise}
\end{array}\right.\tag{1}
$$
and $w=h+k\tau$. Then, with $0/0=0$ whenever it occurs,
$$
\wp(z)=\sum_{h=-\infty}^\infty\sum_{k=-\infty}^\infty\left[\frac1{(z+w)^2}-\frac{\Omega(h,k)}{w^2}\right]\tag{2}
$$
Define
$$
\wp_R(z)=\sum_{h=-R}^R\sum_{k=-R}^R\left[\frac1{(z+w)^2}-\frac{\Omega(h,k)}{w^2}\right]\tag{3}
$$
Using $(2)$ and $(3)$, we get
$$
\begin{align}
\wp_R(z+\tau)-\wp_R(z)
&=\sum_{h=-R}^R\left[\frac1{(z+h+(R+1)\tau)^2}-\frac1{(z+h-R\tau)^2}\right]\\
&=\sum_{h=-R}^RO\left(\frac1{R^2}\right)\\
&=O\left(\frac1R\right)\tag{4}
\end{align}
$$
and
$$
\begin{align}
\wp_R(z+1)-\wp_R(z)
&=\sum_{k=-R}^R\left[\frac1{(z+R+1+k\tau)^2}-\frac1{(z-R+k\tau)^2}\right]\\
&=\sum_{h=-R}^RO\left(\frac1{R^2}\right)\\
&=O\left(\frac1R\right)\tag{5}
\end{align}
$$
Sending $R\to\infty$ in $(4)$ and $(5)$ yields
$$
\wp(z+\tau)=\wp(z)=\wp(z+1)\tag{6}
$$
Induction and $(6)$ shows that $\wp(z+w)=\wp(z)$ whenever $w\in\Lambda$.