If $f\in\mathbb{C}[[z_1,\dots,z_n]]$ such that $f(0)\neq 0$ converges absolutely in a nbhd of $0$, does $\frac{1}{f}$ also converge in a nbhd of $0$?

Question

Suppose $f\in R:=\mathbb{C}[[z_1,\dots,z_n]]$ is a formal power series in several variables such that $f(0)\neq 0$. As such, $f$ has a formal inverse $\frac{1}{f}$ in $R$. If it happens that $f$ converges absolutely in a neighborhood of $0$, is it also true that $\frac{1}{f}$ converges absolutely in a neighborhood of $0$?

I know that this is true from complex analysis in a single variable. However, of the texts covering complex analysis in several variables (Cartan, Krantz, Narasimhan) that I've consulted, none address the region of convergence for the reciprocal of a power series which is absolutely convergent in a neighborhood of the origin.

Motivation: I'm curious because I want to know why the ring $S$ of power series in several variables which are absolutely convergent on a neighborhood of $0\in\mathbb{C}^n$ is actually a local ring. In the larger ring $R$, the invertible elements are precisely those with nonzero constant term, i.e., those elements outside the maximal ideal $(z_1,\dots,z_n)$ in $R$. But it's not obvious to me that these inverses will still satisfy the condition that they converge absolutely in a nbhd of $0$, so that $(z_1,\dots, z_n)$ in $S$ is the unique maximal ideal to show $S$ is local. I'm approaching this from a commutative algebra viewpoint, and unfortunately don't really know any complex analysis of several variables.

Answer 1

This has a simple proof using the fact that a function is holomorphic (defined in terms of complex-differentiability) iff it can locally be represented as a power series. Consider $f$ as a holomorphic function on a neighborhood of $0$. Since $f(0)\neq 0$, the function $1/f$ is also holomorphic in a neighborhood of $0$, and so can be represented with an absolutely convergent power series in a neighborhood of $0$. Formal multiplication of convergent power series corresponds to multiplying the corresponding functions, and so formally multiplying the power series of $1/f$ and the power series of $f$ must give $1$. That is, the power series of $1/f$ must be the formal inverse of the power series of $f$, and in particular that formal inverse converges in a neighborhood of $0$.

Answer 2

$$f(z)=\sum_{a\in \Bbb{N}^n}c_a \prod_{j=1}^n z_j^{a_j}$$ Wlog assume that $f(0)=1$.

Take $r>0$ small such that $$\sum_{a\ne 0\in \Bbb{N}^n}|c_a| \prod_{j=1}^n r^{a_j}\le 1/2$$ You'll get that for $\sup_j |z_j| \le r$

$$\frac1{f(z)} = \sum_{k\ge 0} (1-f(z))^k=\sum_{k\ge 0}(-\sum_{a\ne 0\in \Bbb{N}^n}c_a \prod_{j=1}^n z_j^{a_j})^k $$ Since everything converges absolutely we can expand the $k$-th powers and change the order of summation to get that $\frac1{f(z)}$ is given by an absolutely convergent power series on $\sup_j |z_j| \le r$.