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I want to prove that I can do "whatever I want" with the arrangement of an absolutely convergent series, and that its value will always remain unchanged. But the theorem I've seen uses a limited definition of "rearrangement":

$(b_n)$ is a rearrangement of $(a_n)$ when $b_n = a_{f(n)}$ for all $n$, where $f$ is a bijection from $\mathbb{N}$ to $\mathbb{N}$.

This definition does not handle many useful cases.

For instance, we might split the sequence $(a_n)$ into two subsequences $(a_{0,n})$ and $(a_{1,n})$, and we would expect that $$ \sum_{n = 0}^\infty a_n = \sum_{n = 0}^\infty a_{0,n} + \sum_{n = 0}^\infty a_{1,n}. $$ But this is not a rearrangement as defined above. And even though it is easy to prove this case, we would still not have covered nested sums. We might split the sequence into infinitely many subsequences $(a_{0,n}), (a_{1,n}), (a_{2,n}), \dots$, and we would expect that $$ \sum_{n = 0}^\infty a_n = \sum_{m = 0}^\infty \sum_{n = 0}^\infty a_{m,n}. $$

I'm sure we could prove this as well, but there would still be more to prove. What about triple or quadruple nesting? Or even worse, what about infinite nesting? $$ \sum_{n = 0}^\infty a_n = \cdots \sum_{n_2 = 0}^\infty \sum_{n_1 = 0}^\infty \sum_{n_0 = 0}^\infty a_{n_0, n_1, n_2, \cdots} $$ (This would need a more precise formulation. See the comments.) And even if we prove that that infinite nesting worked, there is probably an even deeper type of nesting that I can't even comprehend. It feels like no matter what I prove, there's always more cases left that I haven't proved. So my question is this:

Is there a proof that you can rearrange absolutely convergent series in completely full generality, regardless of any type of nesting of infinite sums?

I hope that this is a well-defined question. I can't even figure out how to ask it in a perfectly precise mathematical way, which is probably a big part of why I haven't been able to solve it.

Polygon
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  • There's a problem with your "infinitely nested" sum. There are $|\mathbb{N}|^{|\mathbb{N}|}=|\mathbb{R}|$ sequences from $\mathbb{N}$ to $\mathbb{N}$. And the sum runs over all such sequences, but there is only a countable number of elements in the original sum – Carlyle Aug 03 '23 at 16:50
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    If $a_n \ge 0$ then the simplest way to think of rearrangement is to define $\sum a_n=\sup_{I}\sum_{n \in I} a_n$ where $I$ runs through all finite sets of natural numbers; this essentially takes care of any rearrangment since the partial sums of such are always expressed in finite terms; for general real or complex terms, one usually uses the notion of nets and defines the sum of an absolutely convergent series as the net limit of $b_{I}=\sum_{n \in I} a_n$ where again $I$ goes through the finite sets (which is a directed family) and that again takes care of rearragments, nested sums etc – Conrad Aug 03 '23 at 16:57
  • @Carlyle This is why it needs a more precise definition. We could specify that for any $a_{n_0,n_1,n_2,...}$, only finitely many $n_i$'s are nonzero. This would be in line with how sums are typically carried out - computing the inner sum before incrementing the outer one, so there will never be a point where all $n_i$'s have been incremented. Thus there are as many $a_{n_0,n_1,...}$'s as there are finite lists of natural numbers, which is countable. – Polygon Aug 03 '23 at 16:58
  • I think what you are after is the following: let $\mathbb{N}=\sum_{k=1}^NA_k$ be a decomposition into nonempty subsets, where $N\in \mathbb{N}\cup {\infty}.$ Then $$\sum_{n=1}^\infty a_n=\sum_{k=1}^N\sum_{n\in A_k}a_n$$ provided that the internal sum is made according to the order of $n.$ For example the standard rearrangement corresponds to a decomposition into singletons, $N=\infty.$ – Ryszard Szwarc Aug 03 '23 at 19:27
  • I think this is closely related to transfinite ordinals. When we split the sum into two sums, we are effectively bijectively mapping each natural number to an ordinal less than $2\omega$, and then summing the terms "in that order". For the twice-nested sums, we are mapping to ordinals less than $\omega^2$, and for infinite nesting, we are mapping to ordinals less than $\omega^\omega$. I want to show that carrying out the sums in any such order gives the same result. But I don't know enough about ordinals to be able to give a proof. – Polygon Feb 25 '24 at 17:04
  • @Conrad can you elaborate on how the supremum of all finite sums helps us deal with arbitrarily nested infinite sums? I see this answer also mentions the supremum of all finite sums, but it also mentions a lot of concepts I am unfamiliar with. I think I'm in over my head here. – Polygon Mar 03 '24 at 20:06
  • If you have a family of numbers $a_i, i \in I$ you can always compute $S_J=\sum_{i \in J}a_i$ when $J$ is a finite set; but now the set of finite subsets of $I$ with inclusion is a net (in other words for every two finite sets there is a finite set that majorizes them, eg their union) so it makes sense to talk about $lim_J S_J$ (if it exists it is the unique number $S$ st for any $\epsilon>0$ there is finite $J_{\epsilon}$ st $|S_J-S| < \epsilon$ for all $J_{\epsilon} \subset J$ - the net property ensures limit is unique if exists; that is then by definition the unique $\sum_{i \in I}a_i$ – Conrad Mar 04 '24 at 02:52
  • if the numbers are nonnegative that limit always exists though possibly $\infty$ and is the supremum of all finite subsums, while if $a_i$ are just complex, the limit exists iff $\sum |a_i| < \infty$ in the above sense; a few remarks are that if the limit exists, $a_i$ can be non zero only for $i$ in a countable set and then however you index/and or split that countable set, any conceivable notion of limit of $\sum a_i$ (eg nested infinite subsets) is unique and is the above – Conrad Mar 04 '24 at 02:56
  • @Conrad thank you for explaining! I followed everything except for your last remark. I'm not seeing how the limit of all finite sets translates to many nested limits. I understand that there will always be a finite set whose sum is arbitrarily close to arbitrarily many limits, but I can't mentally make the leap to say that it works for every nested sum imaginable. – Polygon Mar 05 '24 at 01:12
  • Nested limits must be countable to make sense so you can always get within $\epsilon$ by getting within $\epsilon/2^k$ of the $k$ step - write a nested sum and think how to do the above – Conrad Mar 05 '24 at 01:58

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