At a lecture a professor stated us the following
Proposition. Consider the multivariate complex power series $$ f(z)=\sum_{\nu \in \mathbb{N}_0^n} a_\nu z^\nu $$ (where $z^\nu:=z_1^{\nu_1}\cdots z_n^{\nu_n}$ and $\mathbb{N}_0=\{0,1,2,\dots\}$). Suppose that there exists $S\in [0,+\infty)$ and $z^0\in\mathbb{C}^n$ with $z_j^0\neq 0$ for all $j$ such that $|a_\nu(z^0)^\nu|\leq S$ for all $\nu\in\mathbb{N}^n_0$. Then the series converges uniformly on compacts inside $D_{z^0}(0):=D_{|z_1^0|}(0)\times\cdots\times D_{|z_n^0|}(0)$, the complex open $n$-polydisk of polyradius $(|z_1^0|,\dots,|z_n^0|)$, centered at the origin.
As indication of the proof he only said that "we should compare with the geometric series." I'm not an analysis guy and I'm not used to comparisons tests techniques; I don't know exactly what does he mean by this. My guess is that I should somehow use the ratio comparison test, or the limit comparison test, with the expressions $S/|(z^0)^\nu|$ (since $z^0_j\neq 0$). But I'm not sure how to apply it.
Any comments are appreciated :)
Okay, the trick I wasn't seeing is just doing $$ |a_\nu z^\nu|=|a_\nu(z^0)^\nu|\left|\frac{z^\nu}{(z^0)^\nu}\right| \leq S\left|\frac{z^\nu}{(z^0)^\nu}\right|. $$