# Convergence of $\sum_{\alpha\in \Bbb N^n} z^\alpha$ in $\{z\in \Bbb C^n: |z_i| < 1 \text{ for all } 1\le i\le n\}$

The convergence of power series in several complex variables is defined in the following way:

Let $$\{c_\alpha\}_{\alpha\in\Bbb N^n} \subset\Bbb C$$. The power series $$\sum_{\alpha\in \Bbb N^n}c_\alpha (z-a)^\alpha = \sum_{\alpha\in \Bbb N^n} c_\alpha (z_1-a_1)^{\alpha_1} \ldots (z_n-a_n)^{\alpha_n}$$ in $$n$$ variables $$z:= (z_1,z_2,\ldots,z_n)$$ centered at $$a\in \Bbb C^n$$ converges at $$w\in \Bbb C^n$$ if there exists $$C> 0$$ such that $$\sum_{\alpha\in F} |c_\alpha| |(w-a)^\alpha| \le C$$ for all finite subsets $$F\subset\Bbb N^n$$.

I am trying to understand the convergence behavior of $$\sum_{\alpha\in \Bbb N^n} z^\alpha$$ in the unit polycylinder $$P^n_1(\mathbf 0) := \{z\in \Bbb C^n: |z_i| < 1 \text{ for all } 1\le i\le n\}$$. This is Example $$1.5.7$$ of the book.

Let $$z\in P^n_1(0)$$ and $$F\subset \Bbb N^n$$ be finite. There exists $$q\in [0,1)$$ such that $$|z_j| \le q$$ for all $$1\le j\le n$$. Hence, $$\sum_{\alpha\in F} |z^\alpha| = \sum_{\alpha\in F} |z_1|^{\alpha_1} \ldots |z_n|^{\alpha_n} \le \color{blue}{\sum_{\alpha\in F} q^{\alpha_1 + \ldots + \alpha_n} \le \sum_{j=0}^\infty q^j} = \frac{1}{1-q}$$

Why is the inequality in blue true? Certainly, there may exist distinct $$\beta,\gamma\in F$$ such that $$\sum_{i=1}^n \beta_i = \sum_{i=1}^n \gamma_i$$. In such a case, I don't see how to get the above inequality.

Perhaps the homogeneous expansion may be useful at some point: $$\sum_{\alpha\in \Bbb N^n} c_\alpha (w-a)^\alpha = \sum_{j=0}^\infty \left(\sum_{|\alpha| = j} c_\alpha (w-a)^\alpha \right)$$

In fact, using the above homogeneous expansion (once convergence issues are settled) the author goes on to say that

$$\color{blue}{\sum_{k=0}^\infty \left(\sum_{|\alpha| = k} z^{\alpha} \right) = \sum_{k_1 = 0}^\infty \ldots \sum_{k_n = 0}^\infty z_1^{\alpha_1} \ldots z_n^{\alpha_n}}$$

Why is this true?

Notation. For $$\alpha\in \Bbb N^n$$, $$|\alpha|$$ denotes $$\alpha_1 + \ldots + \alpha_n$$.

• I think $\frac{1}{1-j}$ should read $\frac{1}{1-q}$? Commented Jan 30, 2022 at 13:33
• @aschepler Thanks! Fixed that. Commented Jan 30, 2022 at 13:53
• If all the $|z_j|< 1$ then $\prod_{j=1}^n \frac1{1-|z_j|}=\sum_{\alpha\in \Bbb{N}^n} |z^\alpha|$ so that $\sum_{\alpha\in \Bbb{N}^n} z^\alpha$ converges to $\prod_{j=1}^n\frac1{1-z_j}$ independently of the order of summation. If some $|z_j|\ge 1$ then the terms of $\sum_{\alpha\in \Bbb{N}^n} z^\alpha$ don't converge to $0$ so the series diverges whatever the order of summation. Commented Jan 30, 2022 at 13:55
• @reuns Thanks, that's a good way to look at it. My questions are different though! Commented Jan 30, 2022 at 13:59
• Not quite. By comparaison with the geometric series it works the same way for any power series inside the domain of convergence (on the boundary everything can happen). The text you are following is just bad. Commented Jan 30, 2022 at 14:00

You're correct - it is not true that

$$\sum_{\alpha \in F} q^{\alpha_1+ \cdots + \alpha_n} \leq \sum_{j=0}^\infty q^j$$

Take for example $$n=5, z=\left(\frac 12, \frac 12, \frac 12, \frac 12, \frac 12\right), q=\frac 12$$, and the elements of $$F$$ are the 5 elements $$\alpha_k \in \mathbb{N}^5$$ with $$|\alpha_k|=1$$. $$\sum_{\alpha \in F} q^{\alpha_1+ \cdots + \alpha_n} = \frac 52$$ $$\sum_{j=0}^\infty q^j = \frac{1}{1-q} = 2$$

I think how the proof should go from there is:

Given a natural number $$j$$, the number of elements $$\alpha \in \mathbb{N}^n$$ such that $$|\alpha| = j$$ is the number of ways to write $$j$$ as the sum of $$n$$ natural numbers $$\alpha_1 + \cdots + \alpha_n = j$$. This count is the same as the number of ways to write $$j+n$$ as the sum of $$n$$ positive integers. If we picture $$j+n$$ dots arranged in a line, this count is the same as the number of ways to choose $$n-1$$ of the $$j+n-1$$ positions in between dots as places to divide the dots into groups. So

$$\big| \{ \alpha \in \mathbb{N}^n : |\alpha| = j\} \big| = {{n+j-1} \choose {n-1}} = {{n+j-1} \choose j}$$

$$\sum_{\alpha \in F} q^{\alpha_1+\cdots+\alpha_n} \leq \sum_{\alpha \in \mathbb{N}^n} q^{|\alpha|} = \sum_{j=0}^\infty {{n+j-1} \choose j} q^j = \frac{1}{(1-q)^n}$$

so the sum $$\sum_{\alpha \in F} |z^\alpha|$$ converges at each $$z \in P_1^n(0)$$.

• Could you please explain $$\sum_{\alpha \in F} q^{\alpha_1+\cdots+\alpha_n} \leq \sum_{j=0}^\infty {n+j \choose j} q^j = \frac{1}{(1-q)^{n+1}}$$ in more detail? Thanks for your answer! Commented Feb 1, 2022 at 16:17
• Edited with more steps and a correction. Commented Feb 1, 2022 at 18:17
• Thanks a lot, I understand it now! Commented Feb 2, 2022 at 9:48