$\prod_{n=0}^\infty(1+x_m)=\sum_{S\in\mathcal{P}_f(\mathbb{W})} \prod_{m\in S}{x_m}$

Question

I am self studying products and had some success exploiting (slight variants) of the infinite analog of $$ \prod_{n=0}^m (1+x_n)=\sum_{S\subset [m]}\prod_{i\in S} x_i $$ $[m] =\{0,1,2,\dots, m\}$ to make a claim about the zeta function as well in some other places. And I just want to suss out the details here: What happens as we take $m\to \infty$?

Question: Is it the case that:
$$\prod_{n=0}^\infty(1+x_n)=\sum_{S\in\mathcal{P}_f(\mathbb{W})} \prod_{j\in S}{x_j}$$

Where $\mathcal P_f(\mathbb W)$ is the set of finite subsets of the non-negative natural numbers. It feels like this has just gotta be the case. We have been warned to avoid the "obvious trap": which is to say "Oh yeah, that's obvious."

I can't quite see through the details of showing this formally. Thanks for any help. A good push in the right direction would be welcome. I wouldn't mind if someone wanted to simply prove it outright.

Answer 1

I assume that $\mathbb{W}$ denote the set $\mathbb{Z}_{\geq 0}$ of nonnegative integers. I shall prove that, for real numbers $x_0,x_1,x_2,\ldots$, we have $$\prod_{m\in\mathbb{W}}\,\left(1+x_m\right)=\sum_{S\in \mathcal{P}_f(\mathbb{W})}\,\prod_{s\in S}\,x_s\,,\tag{*}$$ provided that $\prod\limits_{m\in\mathbb{W}}\,\left(1+x_m\right)$ converges unconditionally to a nonzero limit.

By unconditional convergence to a nonzero limit, I mean that, for any renumeration $\{t_0,t_1,t_2,\ldots\}$ of $\mathbb{W}$, we have that $$\lim_{N\to\infty}\,\prod_{m=0}^N\,\left(1+x_{t_m}\right)$$ converges to a nonzero limit. I forgot about unconditional convergence in my comment, and it should be clear why we need this extra condition. However, if $x_0,x_1,x_2,\ldots$ can be complex numbers, then a good condition is that the product $\prod\limits_{m\in\mathbb{W}}\,\big(1+|x_m|\big)$ converges (i.e. $\prod\limits_{m\in\mathbb{W}}\,(1+x_m)$ converges absolutely). See also this thread. If $x_0,x_1,x_2,\ldots$ are real numbers, then $\prod\limits_{m\in\mathbb{W}}\,\left(1+x_m\right)$ converges unconditionally to a nonzero limit if and only if $x_m\neq -1$ for all $m\in\mathbb{W}$ and $\prod\limits_{m\in\mathbb{W}}\,\left(1+x_m\right)$ converges absolutely.

Suppose that $\prod\limits_{m\in\mathbb{W}}\,\left(1+x_m\right)$ converges unconditionally and $$\lim_{N\to\infty}\,\prod_{m=0}^N\,\left(1+x_m\right)=L\,,$$ where $L\neq 0$. First of all, (*) is clearly true if $x_m\geq 0$ for every $m\in\mathbb{W}$. Our strategy is to consider $m\in\mathbb{W}$ such that $x_m\geq 0$ and $m\in\mathbb{W}$ such that $x_m<0$.

As the product converges unconditionally, we have $$\prod_{m\in\mathbb{W}^+}\,\left(1+x_m\right)=L^+\text{ and }\prod_{m\in \mathbb{W}^-}\,\left(1+x_m\right)=L^-\,,$$ for some $L^+>0$ and $L^-\in\mathbb{R}\setminus\{0\}$ (with $L=L^+L^-$). Here, $$\mathbb{W}^+:=\left\{m\in\mathbb{W}\,\big|\,x_m\geq 0\right\}\text{ and }\mathbb{W}^-:=\left\{m\in\mathbb{W}\,\big|\,x_m<0\right\}\,.$$

Now, you can handle the products $\prod\limits_{m\in\mathbb{W}^+}\,\left(1+x_m\right)$ and $\prod\limits_{m\in\mathbb{W}^-}\,\left(1+x_m\right)$ separately. For $\prod\limits_{m\in\mathbb{W}^+}\,\left(1+x_m\right)$, we clearly have a version of (*): $$\prod_{m\in\mathbb{W}^+}\,\left(1+x_m\right)=\sum_{S\in\mathcal{P}_f(\mathbb{W}^+)}\,\prod_{s\in S}\,x_s\,.$$

For $\prod\limits_{m\in\mathbb{W}^-}\,\left(1+x_m\right)$, we may without loss of generality assume that $x_m>-1$ for all $m\in\mathbb{W}^-$ (otherwise, note that there are finitely many $m$ such that $x_m<-1$, and we can remove them). That is, $L^->0$, and so $$\frac{1}{L^-}=\prod_{m\in\mathbb{W}^-}\,\left(\frac{1}{1+x_m}\right)>\prod_{m\in\mathbb{W}^-}\,\left(1-x_m\right)>0\,.$$ Thus, $\prod\limits_{m\in\mathbb{W}^-}\,\left(1-x_m\right)$ converges, and we get a version of (*) for $\prod\limits_{m\in\mathbb{W}^-}\,\left(1-x_m\right)$: $$\prod_{m\in\mathbb{W}^-}\,\left(1-x_m\right)=\sum_{S\in\mathcal{P}_f(\mathbb{W}^-)}\,\prod_{s\in S}\,(-x_s)\,.$$ This shows that $\sum\limits_{S\in\mathcal{P}_f(\mathbb{W}^-)}\,\prod\limits_{s\in S}\,x_s$ converges absolutely, whence $$\prod_{m\in\mathbb{W}^-}\,\left(1+x_m\right)=\sum\limits_{S\in\mathcal{P}_f(\mathbb{W}^-)}\,\prod\limits_{s\in S}\,x_s\,.$$