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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.

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    $\begingroup$ The answer is yes, if the product converges to a nonzero limit. The identity you are trying to prove doesn't work, at least, in some cases where the product converges to $0$. For example, try $x_m=-\frac{1}{m+1}$ for $m=0,1,2,\ldots$. Then, you will end up with a summand $-\sum_{m=0}^\infty\,\frac{1}{m+1}$ on the right-hand side. $\endgroup$ – Batominovski Jul 24 '18 at 17:28
  • $\begingroup$ @Batominovski. That would be fine though right (in some sense)? In the case you have written we can say that both sides diverge. (One side to infinity and the other to zero) . Oh, usually we say that a product "diverges" to zero I thought. $\endgroup$ – Mason Jul 24 '18 at 17:29
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    $\begingroup$ In fact, you should be able to prove this under suitable convergence hypotheses - for instance, if $x_n$ can be bounded by $n^{-\alpha}$ for sufficiently large $\alpha$, then you should be able to give bounds on the remainders of partial sums on the RHS and show that they converge to zero. $\endgroup$ – Steven Stadnicki Jul 24 '18 at 17:32
  • $\begingroup$ @StevenStadnicki So then the project is becomes: figure out what are the needed convergence hypotheses? $\endgroup$ – Mason Jul 24 '18 at 17:34
  • $\begingroup$ Would you mean $\mathcal P_f(\mathbb N)$, by any chance? $\endgroup$ – Did Jul 24 '18 at 17:38
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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 $$\prod_{m\in\mathbb{W}^-}\,\left(1+x_m\right)=\sum\limits_{S\in\mathcal{P}_f(\mathbb{W}^-)}\,\prod\limits_{s\in S}\,x_s\,.$$

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  • $\begingroup$ Perhaps not necessary, but it is difficult to establish convergence, at least when you are dealing with an infinite sum of complex numbers, without absolute convergence. $\endgroup$ – Batominovski Jul 24 '18 at 18:25
  • $\begingroup$ Then "good" means something like: A logical place to start if one was to make efforts in saying anything about this... $\endgroup$ – Mason Jul 24 '18 at 18:26
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    $\begingroup$ No, that's what it is implied by having an unconditionally convergent product to a nonzero limit. If the positive-side product blows up, you can rearrange $\mathbb{W}$ (see the hidden box) so that the limit does not exist. Likewise, the negative-side product will go to $0$. $\endgroup$ – Batominovski Jul 24 '18 at 18:39
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    $\begingroup$ Yes, $\epsilon$-$\delta$ will work (but that comes directly from the fact that $\prod\limits_{m\in\mathbb{W}}\,(1+x_m)=L$). And I shall give you a good renumeration of $\mathcal{P}_f(\mathbb{W})$: $$\emptyset,\{0\},\{1\},\{0,1\},\{2\},\{0,2\},\{1,2\},\{0,1,2\},\{3\},\{0,3\},\{1,3\},\{2,3\},\ldots\,.$$ (Apparently, this is the same as your renumeration.) $\endgroup$ – Batominovski Jul 24 '18 at 19:00
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    $\begingroup$ And to make you see why it makes so much sense that we need the unconditional convergence condition, try $x_{2m}=+\frac{1}{m+2}$ and $x_{2m+1}=-\frac{1}{m+2}$ for $m=0,1,2,\ldots$. Then, the product $\prod\limits_{m=0}^N\,(1+x_m)$ goes to $\frac12$ as $N\to\infty$, but not unconditionally. And for this product, it is very unclear how to evaluate $\sum\limits_{S\in\mathcal{P}_f(\mathbb{W})}\,\prod\limits_{s\in S}\,x_s$, as this sum does not converge unconditionally. Pick a wrong renumeration of $\mathcal{P}_f(\mathbb{W})$ and you will get nothing close to the desired limit $\frac12$. $\endgroup$ – Batominovski Jul 24 '18 at 19:10

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