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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!

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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.

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  • $\begingroup$ Thank you very much for your answer. I understand, why both of the two products $X:=\prod\limits_{n=1}^{\infty} (1+a_n)$ and $Y:=\prod\limits_{n=1}^{\infty} (1+a_{\sigma(n)})$ are conditionally convergent. But I am unable to see why their limits have to be equal... $\endgroup$ – Stein Chen Mar 7 '15 at 18:49

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