When does a multivariate power series define an entire function?
This answer cites passages from Lecture notes on several complex variables by H.P. Boas. The author provides information about the region of convergence of multivariate power series.
Intro: There are many crucial differences between single- and multivariate complex analysis. For instance
The theory of one-dimensional power series bifurcates into the theory of entire functions (when the series has infinite radius of convergence) and the theory of holomorphic functions on the unit disc (when the series has a finite radius of convergence—which can be normalized to the value $1$). In higher dimensions, the study of power series leads to function theory on infinitely many different types of domains.
In contrast to the one-variable case holomorphic functions of several variables never have isolated singularities and never have isolated zeroes.
In the following we want to characterize the domains of convergence. But before that we have to introduce a little bit of notation.
We consider the normed vector space $\mathbb{C}^n$ with the Euclidian norm: $$\|(z_1,\ldots,z_n)\|_2=\sqrt{|z_1|^2+\cdots+|z_n|^2}.$$ A point $(z_1,\ldots,z_n)$ in $\mathbb{C}^n$ is denoted by a single letter $z$.
If $\alpha$ is an $n$-dimensional vector all of whose coordinates are nonnegative integers, then $z^\alpha$ means the product $z_1^{\alpha_1}\cdots z_n^{\alpha_n}$; the notation $\alpha!$ abbreviates the product $\alpha_1!\cdots\alpha_n!$; and $|\alpha|$ means $\alpha_1+\cdots+\alpha_n$. In this multi-index notation, a multivariable power series can be written in the form $\sum_{\alpha}c_{\alpha}z^{\alpha}$ as abbreviation for $\sum_{\alpha_1=0}^\infty\cdots\sum_{a_n=0}^\infty c_{\alpha_1\ldots\alpha_n}z^{\alpha_1}\cdots z^{\alpha_n}$.
Since the value of a series depends in general on the order of summation, and there is no canonical ordering of $n$-tuples on nonnegative integers when $n>1$, we restrict attention to absolute convergence. The terms of an absolutely convergent series can be reordered arbitrarily without changing the value of the sum (or the convergence of the sum).
Domain of convergence: Reinhardt domain
The domain of convergence of a power series means the interior of the set of points at which the series converges absolutely. For example, the power series $$\sum_{n=1}^\infty z_1^nz_2^{n!}$$ converges absolutely on the union of three sets in $\mathbb{C}^2$: the points $(z_1,z_2)$ for which $|z_2|<1$ and $z_1$ is arbitrary; the points $(0,z_2)$ for arbitrary $z_2$; and the points $(z_1,z_2)$ for which $|z_2|=1$ and $|z_1|<1$.
The domain of convergence is the first of these three sets, for the other two sets contribute no additional interior points.
Being defined by absolute convergence, every convergence domain is multicircular: if a point $(z_1,\ldots,z_n)$ lies in the domain, then so does the point $(\lambda_1z_1,\ldots,\lambda_nz_n)$, when $1=|\lambda_1|=\cdots=|\lambda_n|$.
Moreover, the comparison test for absolute convergence of series shows that the point $(\lambda_1z_1,\ldots,\lambda_nz_n)$ remains in the convergence domain when $|\lambda_j|\leq 1$ for each $j$. Thus every convergence domain is a union of polydiscs centered at the origin. (A polydisc means a Cartesian product of discs, possibly with different radii.)
A multicircular domain is often called a Reinhardt domain. Such a domain is called complete if whenever a point $z$ lies in the domain, the whole polydisc $\{w:|w_1|\leq |z_1|,\ldots,|w_n|\leq |z_n|\}$ is contained in the domain. The preceding discussion can be rephrased as saying that every convergence domain is a complete Reinhardt domain.
But more is true. If both $\sum_{\alpha}|c_\alpha z^{\alpha}|$ and $\sum_{\alpha}|c_\alpha w^{\alpha}|$ converge, then Hölder's inequality implies that $\sum_{\alpha}|c_\alpha||z^{\alpha}|^t|w^{\alpha}|^{1-t}$ converges when $0\leq t\leq 1$. This property of a Reinhardt domain is called logarithmic convexity.
The preceding discussion shows that a convergence domain is necessarily a complete, logarithmically convex Reinhardt domain. The following theorem of Hartogs says that this geometric property characterizes domains of convergence of power series.
Theorem: A complete Reinhardt domain in $\mathbb{C}^n$ is the domain of convergence of some power series if and only if the domain is logarithmically convex.
The relationship of holomorphy of multi-variate functions with Reinhard domains of convergence of multi-variate power series is elaborated in the following pages. The next theorem from section 2.5 clarifies the existence of power series and covers also entire multivariate power series.
Theorem: (Cartan–Thullen). The domain of convergence of a multivariable power series is a domain of holomorphy. More precisely, for every domain of convergence there exists some power series that converges in the domain and that is singular at every boundary point.
Hint: Chapter I: Entire functions in Several Complex Variables, vol. III by L.I. Ronkin might be useful for further reading. The introductory section provides a lot of additional references.
The $n$-dimensional analogue of convergence circles in $\mathbb{C}$ are polydiscs, i.e. products of discs $B_{r_1}(z_1) \times ... \times B_{r_n}(z_n) \subseteq \mathbb{C}^n$. If $r_1=...=r_n$ this is just the $r$-ball around $z$ w.r.t. the $\infty$-norm on $\mathbb{C}^n$.
I claim:
$\sum_{\alpha\in\mathbb{N}^n} a_\alpha z^\alpha$ is absolutely convergent for all $z\in\mathbb{C}^n$ iff $R^{-1} := \limsup\limits_{k\to\infty} \left( \sum_{|\alpha|=k} |a_\alpha| \right)^{1/k} = 0$.
Proof: $\impliedby$ is an easy exercise. One can indeed show absolute convergence in the interior of every polydisc with radius $<R$: $\sum_\alpha |a_\alpha z^\alpha| \leq \sum_k (\sum_{|\alpha|=k} |a_\alpha|) \|z\|_\infty^k$ is convergent because for $\|z\|_\infty<r_1<r_2<R$ all but finitely many summands are eventually smaller than $r_2^{-k} r_1^k<(\frac{r_1}{r_2})^k$.
$\implies$ follows similarly to the one-dimensional case: Fix any $w$ with $|w_i|\geq R$ for all $1\leq i\leq n$ and assume $\sum a_\alpha w^\alpha$ converges absolutely.
Then $a_\alpha w^\alpha$ goes to zero for $|\alpha|\to\infty$. Now we have $|a_\alpha w^\alpha| \geq |a_\alpha| R^{|\alpha|}$. If the LHS goes to zero, this must mean $|a_\alpha|R^{|\alpha|} \to 0$ as well.
Therefore if $\epsilon>0$, then $|a_\alpha|\leq \epsilon R^{-|\alpha|}$ for $|\alpha|\gg 0$. This implies $\sum_{|\alpha|=k} |a_\alpha| \leq \binom{n+k-1}{k} \epsilon R^{-k}$ where the binomial is exactly the number of multiindices $\alpha\in\mathbb{N}^n$ with $|\alpha|=k$ by a standard exercise in combinatorics. The binomial can also be written as $\binom{n+k-1}{n}$ which is a polynomial in $k$ of degree $n$. Therefore the $k$-th root of it goes to 1 with $k\to\infty$.
This shows $\limsup\limits_{k\to\infty} (\sum_{|\alpha|=k} |a_\alpha|)^{1/k} \leq R^{-1}$ which proves the claim.
Looking a bit closer at this proof, one notices that there is actually a more precise statement hidden in there which is also similar to the one-dimensional case:
If $\sum a_\alpha z^\alpha$ converges absolutely at some point $z=w$ with $|w_i|\geq r_i>0$, then the power series converges absolutely on the open polydisc $B_{r_1}(0) \times ... \times B_{r_n}(0)$.
For $n=1$ this means that the region of convergence of the power series is necessarily wedged between $B_r(0)$ and $\overline{B_r(0)}$ for some critical radius $r\in[0,\infty]$, because discs in the plane are nicely contained in each other.
For $n>1$ these containment properties no longer hold. Instead the region of convergence is now somewhere in between the less simple sets $\bigcup_{r\in R} B_{r_1}(0) \times...\times B_{r_n}(0)$ and $\bigcup_{r\in R} \overline{B_{r_1}(0) \times...\times B_{r_n}(0)}$ for some critical set of radii $R\subseteq [0,\infty]^n$.
One can think of $R$ as some hyperbola, for example $R=\{(r_1,r_2) | r_1 r_2=1\}$ is the set of critial radii for $\sum_n (z_1 z_2)^n$. More specifically: The set $A$ of points where the power series converges absolutely is log-convex, that is $L:=\{(\log|z_1|, ..., \log|z_n|) \mid z\in A\}$ is a convex subset of $[-\infty,+\infty)^n$. And the set of critical radii is the boundary of $L$.