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Does anyone know of an example of an uncountable convergent sum $\sum_{s \in S} x_s$ of real numbers where uncountably many $x_s$ are non-zero? Ideally I'd like to be able to compute $\sum_{s \in S}x_s$ as well. I know that an uncountable convergent sum of positive numbers must have at most countably many non-zero terms. Thus $\sum_{s \in S} x_s$ must not be absolutely convergent.

Most of my attempts have revolved around trying to mimic an alternating series and prove convergence in a similar way to the proof of the alternating series test.

Here is one series which I thought might work: Let $S$ be the set of finite subsets of $[0,\infty)$. Then for $s \in S$ define $x_s = (-1)^{|s|}/|s|$. I've tried mimicing the proof othe alternating series test but to no avail. A variation of the above series which I think might work is $x_s = (-1)^{|s|}/\sup s$ but I've also had no success here. However, in both cases I've been able to show that if one of the series converges, then its limit must be $- \log 2$ (because $\sum_{n=1}^\infty (-1)^n/n$ is a sub-series).

Austin Shiner
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  • To have a conditionally but not absolutely convergent sum (as you certainly will have here), you need to specify some ordering for the sum. You cannot simply sum over a set a not-absolutely-convergent summation – Jake Mirra Aug 10 '21 at 19:38
  • Why would we need an order? Does the usual definition of uncountable sum not work? I.e. $\sum_{s \in S}x_s = L$ iff for any $\epsilon > 0$ there exists a finite set set $F_0 \subseteq S$ such that for any finite set $F \subseteq S$ containing $F_0$, we have $| \sum_{s \in F} x_s - L| < \epsilon$. Or perhaps are you saying that a not-absolutely convergent sum can never converge using this definition because of reordering issues? – Austin Shiner Aug 10 '21 at 20:22
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    Correct, a non-absolutely convergent sequence will not converge under that alternate definition, either. – Thomas Andrews Aug 10 '21 at 20:35

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No. Any convergent sum has at most countably many non-zero terms.

Suppose on the other hand that $S$ is uncountable and $x_s$ is a non-zero real for every $s\in S$. For $n=1,2\dots$ let $$E_n=\{s\in S:s>1/n\}$$and $$F_n=\{s\in S:x_s<-1/n\}.$$Since $$S=\bigcup_{n=1}^\infty(E_n\cup F_n)$$and $S$ is uncountable there must exist $m$ such that $E_m\cup F_m$ is infinite; wlog say $E_m$ is infinite. So given $\epsilon>0$ and a finite set $F\subset S$ there exists $F'\subset S$ with $F\subset F'$ and $$\sum_{s\in F'}x_s-\sum_{s\in F}x_s\ge\epsilon,$$showing that $\sum_Sx_s$ is not convergent.

David C. Ullrich
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