Associativity of infinite products

Question

It is well-known that if $\sum_{n=1}^\infty a_n$ is an absolutely convergent complex series and $\mathbb N$ is partitioned as $J_1,J_2,\dots$, then the series $\sum_{j\in J_n}a_j$ for all $n$ and $\sum_{n=1}^\infty\sum_{j\in J_n}a_j$ are both absolutely convergent, and $\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\sum_{j\in J_n}a_j$.

I am trying to make sense of the same identity for infinite products. The convention I'm using is the following

The product $\prod_{n=1}^\infty z_n$ is called convergent if there exists an $n_0$ such that the partial products $\prod_{j=n_0}^nz_j$ converge to a non-zero number as $n\to\infty$. Write $z_n=1+w_n$, the product is said to converge absolutely if $\prod_n(1+|w_n|)$ converges. It is a theorem that an absolutely convergent product is convergent and that $\prod_n(1+|w_n|)$ converges if and only if $\sum_n|w_n|$ does.

Suppose $\prod_{n=1}^\infty z_n$ converges absolutely and $\Bbb N=\bigcup_{n=1}^\infty J_n$ is a partition, I want to show that $$\prod_{n=1}^\infty z_n=\prod_{n=1}^\infty\prod_{j\in J_n}z_j,$$ with the RHS converging absolutely. By the characterisation of absolute convergence, we know that $\sum_{n=1}^\infty|z_n-1|$ converges, so the series $\sum_{j\in J_n}|z_j-1|$ converge for all $n$ and hence all the products $\prod_{j\in J_n}z_j$ converge absolutely. What I'm having trouble with is showing $\prod_{n=1}^\infty\prod_{j\in J_n}z_j$ converges (absolutely or not) and showing it is equal to the original product. By the characterisation of absolute convergence, it would be enough to show that $\sum_{n=1}^\infty|(\prod_{j\in J_n}z_j)-1|$ converges, but this has no discernible relation to the series $\sum_{n=1}^\infty\sum_{j\in J_n}|z_j-1|$, which we do know converges by the "associativity" for absolutely convergent series.

Answer 1

Take logs so you are only working with sums. At the end you can exponentiate to get the results you need in terms of products. Therefore, in place of products $\prod_{i=1}^\infty (1+w_i)$ you are working with sums $\sum_{i=1}^\infty \log(1+w_i)$. We don't yet know that this sum $\sum_{i=1}^\infty \log(1+w_i)$ converges absolutely. But if we did, then you would have that $\sum_{i=1}^\infty\sum_{j\in J_i}\log(1+w_j)=\sum_{i=1}^\infty \log(1+w_i)$ where everything converges absolutely as sums. At this point, after exponentiating everything, it would be a short argument to show the result you are after.

---

It turns out that if the product converges absolutely, then the sum does too:

One has

$$\log(1+|w|)\le |w|$$

$$\implies \log\left(\prod_{i=1}^n(1+|w_i|)\right)\le \sum_{i=1}^n |w_i|\le \sum_{i=1}^\infty |w_i|$$

so that if $\sum w_i$ converges absolutely, $\prod (1+w_i)$ converges absolutely.

The converse is also true: suppose that $\prod (1+w_i)$ converges absolutely. Then $\prod(1+|w_i|)$ converges, and hence, for sufficiently large $N$, $|w_i|<1$ for all $i>N$. If $|w|<1$ then

$$\frac{1}{2}|w|\le \log (1+|w|).$$ Therefore,

\begin{align*} \sum_{i>N} |w_i|&\le 2\sum_{i>N}\log(1+|w_i|)\\ &=2\log\left(\prod_{i>N}(1+|w_i|)\right). \end{align*}

So, $\sum |w_i|$ is convergent, so $\sum w_i$ is absolutely convergent.

Now that you have this, you can leverage all the theorems for absolutely convergent sums and apply them to absolutely convergent products. In particular, the results in your question.

---

Suppose that $\prod (1+w_i)$ converges absolutely. Take the logarithm of the partial products of $\prod_{i=1}^\infty\prod_{j\in J_i}(1+|w_i|)$ to get

$$\log\left(\prod_{i=1}^N\prod_{j\in J_i}(1+|w_i|)\right)=\sum_{i=1}^N\sum_{j\in J_i}\log(1+|w_i|)\le \sum_{i=1}^\infty\sum_{j\in J_i}|w_i|$$

the RHS of which is finite by well-known results cited in your question. Therefore,

$$\prod_{i=1}^N\prod_{j\in J_i}(1+|w_i|)$$

is bounded, hence $\prod_{i=1}^\infty\prod_{j\in J_i}(1+|w_i|)$ is convergent. So, $\prod_{i=1}^\infty\prod_{j\in J_i}(1+|w_i|)$ is absolutely convergent.

Answer 2

Summary: absolute convergence of the infinite product of a sequence $(z_n)_{n\in\mathbb{N}}$ of non-zero complex numbers is equivalent to multipliability of that sequence in $\mathbb{C}^*;$ and, as a special case of the general concept of summability in complete Hausdorff commutative topological groups, this concept behaves 'nicely', including obeying the infinite associative law.

In more detail (but without copying out the whole of Bourbaki!):

[1, Chapter III, $\S5$]

Proposition 2. In a complete group $G,$ every subfamily of a summable family is summable.

Theorem 2 (Associativity of the sum). Let $(x_\iota)_{\iota\in I}$ be a summable family in a complete group $G,$ and let $(I_\lambda)_{\lambda\in L}$ be any partition of $I.$ If $s_\lambda$ denotes $\sum_{\iota\in I_\lambda}x_\iota,$ then the family $(s_\lambda)_{\lambda\in L}$ is summable and has the same sum as the family $(x_\iota)_{\iota\in I}.$

We can thus write the associativity formula for sums: $$ \sum_{\lambda\in L}\bigg(\sum_{\iota\in I_\lambda}x_\iota\bigg) = \sum_{\substack{\iota\in\bigcup I_\lambda \\ \lambda\in L}}x_\iota, $$ which is valid whenever the family $(I_\lambda)$ is a partition of its union and the right-hand side is defined.

[2, Chapter VIII, $\S3$]

Theorem 1. The family $(1+u_\iota)_{\iota\in I}$ is multipliable in $\mathbb{C}^*$ if and only if the family $(|u_\iota|)$ is summable in $\mathbb{R}.$

Definition 1. An infinite product of complex numbers, with general factor $1+u_n,$ is said to be absolutely convergent if the product with general factor $1+|u_n|$ is convergent (or, equivalently, if the series with general term $|u_n|$ is convergent).

In the notation of the question: suppose $\prod_jz_j$ is absolutely convergent. By Theorem 1 and Definition 1, the family $(z_j)_{j\in\mathbb{N}}$ is multipliable in $\mathbb{C}^*.$ For all $n,$ by Proposition 2, the family $(z_j)_{j\in J_n}$ is multipliable in $\mathbb{C}^*.$ By Theorem 2, the family $(\prod_{j\in J_n}z_j)_{n\in\mathbb{N}}$ is multipliable in $\mathbb{C}^*,$ and $$ \prod_{n=1}^\infty\bigg(\prod_{j\in J_n}z_j\bigg) = \prod_{j\in\mathbb{N}}z_j = \prod_{j=1}^\infty z_j. $$

Bibliography:

[1] Nicolas Bourbaki, Elements of Mathematics: General Topology, Part 1 (Springer, 1989; original French edition published by Masson, Paris, 1971).

[2] Nicolas Bourbaki, Elements of Mathematics: General Topology, Part 2 (Springer, 1989; original French edition published by Masson, Paris, 1974).