Show that Weierstrass function is elliptic function. Prove that Weierstrass function is periodic with respect to lattice $L (L\subset \mathbb{C})$ .i-e $f(z+w,L)=f(z,L)$ ($w\in L$).
$f(z,L)=\frac{1}{z^2}+\sum_{0\ne w\in L}\Big(\frac{1}{(z-w)^2}-\frac{1}{(w)^2}\Big)$
 A: Is not difficult to see the periodicity of $f(z,L)$.
By definition  we have $f(z,L)=\frac{1}{z^2}+\sum_{0\ne w\in L}\Big(\frac{1}{(z-w)^2}-\frac{1}{(w)^2}\Big)$.
Let now $z+w_0$ with $w_0\in L-\{0\}$; then 
$$f(z+w_0,L)=\frac{1}{(z+w_0)^2}+\sum_{0\ne w\in L}\Big(\frac{1}{(z+w_0-w)^2}-\frac{1}{(w)^2}\Big).$$
Since the sum run over $L-\{0\}$, and the lattice is a subgroup of $\mathbb{C}$, then $-w_0\in L$, hence we can rewrite $f(z+w_0,L)$ in this way:
$$f(z+w_0,L)=\frac{1}{(z+w_0)^2}+\frac{1}{(z+w_0-w_0)^2}+\sum_{0\ne w\in L; w\ne w_0}\frac{1}{(z+w_0-w)^2}-\sum_{0\neq w\in L}\frac{1}{(w)^2};$$
clearly $z+w_0-w_0=z$ and the piece $\frac{1}{(z+w_0)^2}$ can be incorporate in the first sum, so
$$f(z+w_0,L)=\frac{1}{z^2}+\sum_{w\in L; w\ne w_0}\frac{1}{(z+w_0-w)^2}-\sum_{0\ne w\in L}\frac{1}{(w)^2}.$$
Finally, since $L$ is a group the following sum are the same
$$\sum_{w\in L-\{0\}} \frac{1}{w^2}=\sum_{w_0-w \in L-\{0\}}\frac{1}{(w_0-w)^2},$$
and hence 
$$f(z+w_0,L)=\frac{1}{z^2}+\sum_{w\in L; w\ne w_0}\Big(\frac{1}{(z+w_0-w)^2}-\frac{1}{(w_0-w)^2}\Big).$$
With abuse of notation we can call the different $w_0-w$ as $w$; clearly $w_0-w\ne 0$ imply $w\ne 0$ and rewrite the sum:
$$f(z+w_0,L)=\frac{1}{z^2}+\sum_{0\ne w\in L}\Big(\frac{1}{(z-w)^2}-\frac{1}{(w)^2}\Big)=f(z,L).$$
