How do you really define the convergence for power series of the form $\sum_{k_1, \ldots, k_n = 0}^{\infty}c_{k_1} \ldots c_{k_n}(z_1-a_1)^{k_1} \ldots (z_n-a_n)^{k_n}$ rigorously?

My guess is that you first define the partial sums of those series, then you say that the series converges if the partial sum converges. However, there are several ways that you can define the partial sums. So another guess is that absolute convergence will guarantee different definitions of partial sums will actually be equal while pushing to infinity. So the final guess is like in the one-variable case, convergence in some radius of convergence will result in absolute convergence.

Another way I would like to think of it is directly through counting measures. I might think of it as integration with respect to product counting measures. So I will not need to consider partial sums. Then surely absolute convergence would mean $L^1$.

  • 4
    $\begingroup$ Stick to absolute convergence... $\endgroup$ Commented Sep 27, 2017 at 4:11
  • $\begingroup$ ....seconded... $\endgroup$
    – user335907
    Commented Sep 27, 2017 at 4:58
  • $\begingroup$ Your guess that in a single variable a power series converges absolutely within the radius of convergence is correct, and is critically important. We thus get the multivariable result for free on the interior of polydisks. One needs care at the boundary in multivariable, but the same is true in one variable. $\endgroup$ Commented Oct 28, 2022 at 14:49

1 Answer 1


Here there is a definition that may satisfy you. I'm gonna proceed with full generality. Let $I$ be any set, and let $f:I\to\mathbb{C}$ be a complex-valued function. We say that the series $\sum_{j\in I}f(j)$ is summable with value $c\in\mathbb{C}$ if for every $\varepsilon>0$ there is a finite set $J_0\subset I$ such that for every finite set $J\subset I$ with $J_0\subset J$ we have $$ \left|c-\sum_{j\in J}f(j)\right|<\varepsilon. $$ (In your case, $I=\mathbb{N}^k$. If you're keen with filter theory, I think this definition is just convergence in the filter of cofinite sets.)

If we turn the set $I$ into a measure space equipping it with its power set as $\sigma$-algebra and the counting measure, then we have the following

Lemma. $\sum_{j\in I}f(j)$ is summable if and only if $f:I\to\mathbb{C}$ is integrable, and on this case, $$ \sum_{j\in I}f(j)=\int_Ifdn. $$

Some consequences of this integral formulation:

  1. Using this equivalent definition and Lebesgue theory, it follows for free that $\sum_{j\in I}f(j)$ is summable if and only if it is absolutely summable, and, thus, that if it's summable then it is also unconditionally convergent (this can be also seen using dominated convergence).

  2. You can also verify as well that if the series is summable, then the subset of indices where $f$ is non-zero, $\{j\in J:f(j)\neq 0\}$, must be countable, so it supposes no restriction to consider $I$ always countable. To show this, since $\int fdn=\int\operatorname{Re}fdn+i\int\operatorname{Im}fdn$, it suffices to show it for the real-valued case, and since $\int fdn=\int f_+dn-\int f_-dn$, in turn it suffices to show it for the non-negative case. Note that $\{j\in J:f(j)\neq 0\}=\bigcup_{n=1}^{+\infty}A_n$, where $A_n=\{j\in J:f(j)>\frac{1}{n}\}$. How big can $A_n$ be if the series is summable?

  3. If $I=\mathbb{N}^k$, from Fubini's theorem it follows that if the series is summable, then we have $$ \sum_{(n_1,\dots,n_k)\in \mathbb{N}^k}f(n_1,\dots,n_k)=\sum_{n_1=1}^{+\infty}\cdots\sum_{n_k=1}^{+\infty}f(n_1,\dots,n_k). $$ Also, you can apply Fubini-Tonelli theorem to reason that the series from the LHS is summable iff the series from the RHS converges absolutely.

If you want to prove the lemma, you should try to prove first the case where $f$ is real and non-negative, then after tackle the case where $f$ takes always real values, and reduce it to the first case; and finally, address the case where $f$ takes complex values and reduce it to the real case. For the case "$f$ real and non-negative," the following equality is useful: $$ \sup\left\{ \int_Isdn: 0\leq s\leq f,\; s\text{ simple} \right\} =\sup\left\{ \int_Isdn: 0\leq s\leq f,\; s\text{ simple},\; s(j)\in\{0,f(j)\}\;\forall j\in J \right\} $$ (try to prove it).


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