Abel's lemma for multiple power series

Remarks

For $z \in K_\rho$ one has $\abs{c_\alpha z^\alpha} \le \abs{c_\alpha w^\alpha} \rho^{\abs{\alpha}} \le M \rho^{\abs{\alpha}}$, and $\sum_{\alpha \in \N^n} \rho^{\abs{\alpha}} = (1 - \rho)^{-n}$ is a product of $n$ geometric series; the Weierstrass M-test concludes. If all $w_j \ne 0$, every compact subset of the polydisc $\set{\abs{z_j} < \abs{w_j}}$ lies in some $K_\rho$, so the series converges absolutely there, uniformly on compact subsets. In particular, convergence at one point $w$ with all $w_j \ne 0$ already gives convergence on this whole open polydisc. Example: for $\sum_{k \ge 0} (z_1 z_2)^k$ and $w = (2, 1/2)$, all terms have modulus $1$, so the series converges on $\set{\abs{z_1} < 2, \abs{z_2} < 1/2}$. The same series also converges on $\set{\abs{z_1} < 1/2, \abs{z_2} < 2}$, so its domain of convergence $\set{\abs{z_1 z_2} < 1}$ is larger than any single polydisc given by the lemma.

Going deeper

For a student

The lemma is the first step in describing where a power series converges. It shows that the domain of convergence is a union of polydiscs centred at $0$, hence a complete Reinhardt domain. Combined with Hölder's inequality it gives the logarithmic convexity of the domain of convergence. It also shows that the sum is a uniform limit of polynomials on compact sets, hence holomorphic.