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We know if $\sum_{n=1}^\infty|z_n|$ converges then $\sum_{n=1}^\infty z_n$ converges absolutely. (kind of trivial)

I wonder whether it holds for infinite products, that is, if $\prod_{n=1}^\infty|z_n|$ converges then $\prod_{n=1}^\infty z_n$ converges absolutely.

Is the statement above true or not?

Thanks a lot!

I know $\prod_{n=1}^\infty|z_n|$ converges doesn't imply $\prod_{n=1}^\infty z_n$ converges.

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  • $\begingroup$ First line, second sum: take off the absolute value. And no: it's not a trivial claim. $\endgroup$
    – DonAntonio
    Commented Mar 5, 2014 at 4:59
  • $\begingroup$ Did you mean the statement is false? Is there any counterexample? $\endgroup$
    – Falang
    Commented Mar 5, 2014 at 5:04
  • $\begingroup$ @DonAntonio, can you please have look at another post? math.stackexchange.com/questions/699928/… $\endgroup$
    – Falang
    Commented Mar 5, 2014 at 5:06
  • $\begingroup$ @DonAntonio, thank you very much! $\endgroup$
    – Falang
    Commented Mar 5, 2014 at 5:06
  • $\begingroup$ No, the statement just doesn't make much sense as it is: it should be, imo: if the series $\;\sum |z_n|\;$ converges then so does the series $\;\sum z_n\;$ , and then we say this last one converges absolutely. The claim is the first part above, and its proof is not trivial (though it isn't hard, either) $\endgroup$
    – DonAntonio
    Commented Mar 5, 2014 at 5:08

1 Answer 1

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No. Take $z_n=-1$, so that $\prod_{n=1}^\infty |z_n|$ trivially converges to $1$. But $z_n=1+(-2)$, and the definition of $\prod_{n=1}^\infty (1+(-2))$ converging absolutely is that $\prod_{n=1}^\infty (1+|{-}2|) = \prod_{n=1}^\infty 3$ converges, which it does not.

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  • $\begingroup$ I'm not sure why this makes sense. $|1+(-2)|\neq |1|+|-2|$. It would be better to say that $\prod_{n\in\mathbb N} -1$ does not converge since it oscillates. $\endgroup$
    – Ian Coley
    Commented Mar 5, 2014 at 5:33
  • $\begingroup$ $\Re z_n$ should be positive according to the definition of absolute convergence. $\endgroup$
    – Falang
    Commented Mar 5, 2014 at 5:36
  • $\begingroup$ The definition of $\prod y_n$ converging absolutely is not that $\prod|y_n|$ converges; rather, it is that $\prod(1+|y_n-1|)$ converges. A necessary condition is that $y_n$ tends to $1$, which means that one may assume that $\Re y_n$ is positive if one wishes. $\endgroup$
    – Greg Martin
    Commented Mar 5, 2014 at 6:48

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