If $|z_n-z_m|> 2$ for every $n\ne m$ then $\sum \frac{1}{z_n^3}$ converges 
Let $(z_n)$ be a sequence of non-zero complex numbers such that $\forall n,m, n\neq m\implies |z_n-z_m|> 2$
Prove that $\sum \frac{1}{z_n^3}$ converges.

I'm clueless with this problem. A lower bound on $|z_n|$ is obviously needed but I can't come up with anything useful. Any hint is appreciated.

Take a circle $\Gamma_k$ with radius $1$ centered at every $z_k$. These circles are disjoint, hence inside $\|z\|\leq R$ cannot be more than $(R+1)^2$ points from the set $\{z_1,z_2,\ldots\}$, since the area enclosed by $\|z\|\leq R+1$ has to be bigger than the sum of the areas of the circles $\Gamma_k$ whose centre lie inside $\|z\|\leq R$. In particular:
$$ \left|\sum_{z_i\in E_R}\frac{1}{z_i^3}\right|\leq \frac{(2R+1)^2}{R^3}\approx\frac{4}{R},\qquad E_R=\{z_i: R\leq \|z_i\|\leq 2R\}. $$
Sum those inequalities for $R=1,2,4,8,16,\ldots$ and the job is done.
We get that $\sum z_i^{-3}$ is convergent and
$$\left|\sum \frac{1}{z_i^3}\right|\leq \left(\min \|z_i\|\right)^{-3}+17.$$

I think that the tightest upper bound for $\left|\sum z_i^{-3}\right|$ comes from the lattice $2\cdot\mathbb{Z}[\omega]$ and takes the following form:
$$\left|\sum\frac{1}{z_i^3}\right|\leq \left(\min\|z_i\|\right)^{-3}+\frac{1}{8}\left|G_3(\omega)\right| $$
where $G_3(\omega)$ is an Eisenstein series.
