Let ${(a_n)}_{n=1}^{\infty}$ be a sequence of complex numbers. Let $\sigma:\mathbb{N}\rightarrow \mathbb{N}$ be a bijective map.
Is it possible then that $\prod\limits_{n=1}^{\infty} (1+|a_n|)=\prod\limits_{n=1}^{\infty} (1+|a_{\sigma(n)}|)=:r<\infty$, but that at the same time
$X:=\prod\limits_{n=1}^{\infty} (1+a_n)$ and $Y:=\prod\limits_{n=1}^{\infty} (1+a_{\sigma(n)})$ are convergent against two different real numbers $s$ and $t$?
If this cannot be the case, then why not?
Thanks for the help!
Switching to logarithms, it is easy to check that the conditional convergence of $$ \prod_{n\geq 1}(1+b_n) $$ is equivalent to the conditional convergence of $$ \sum_{n\geq 1}b_n. $$ So the first line of the problems gives that $$ \sum_{n\geq 1}a_n $$ is absolutely convergent, so for any rearrangement $\sigma$ we have: $$ \sum_{n\geq 1} a_n = \sum_{n\geq 1}a_{\sigma(n)} $$ giving that: $$ \prod_{n\geq 1}(1+a_n) = \prod_{n\geq 1}(1+a_{\sigma(n)}), $$ so the answer to your question is negative.