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Let there be given some conditionally convergent infinite series $S$. Then let $R$ be some real number, and $Q_k$ a rearrangement of $S$ such that the sum is equal to $R$.

Is $Q_k$ unique? In other words, is there some other rearrangement $W_k$ for which the sum of $S$ is also $R$?

I believe the answer is that $Q_k$ is in fact unique, because the cardinality of the set of permutations of the natural numbers (i.e. the positions of each element of the series $S$), is the same as the cardinality of the real numbers; and therefore the mapping from the real numbers to the rearrangements should be $1-1$. Is this line of reasoning valid, or is this result well known?

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Double checking the spelling of your titles tends to be a good idea! :) – Mariano Suárez-Alvarez Aug 30 '10 at 22:16
$Q_k$ certainly isn't unique. Given any rearrangement, simply switch the order of finitely many terms and you have a (generally) different series that converges to the same number. – Ryan Budney Aug 30 '10 at 22:21
Triple check :) – Mariano Suárez-Alvarez Aug 30 '10 at 22:21
Similarly, you could compose any rearrangement with the product of infinitely many "small" transpositions $(1 2)(3 4)(5 6)\cdots$ and you'd have another series converging to the same thing. – Ryan Budney Aug 30 '10 at 22:24
up vote 8 down vote accepted

The line of reasoning is not valid:

If you find a particular $Q_k$, then I can find infinitely many other rearrangements that give the same value: Pick any finite subset $X$ of $\mathbb{N}$ and let $\sigma$ be any permutation of $X$. Then the re-rearrangement where we start with $Q_k$ and just mix up the subscripts in $X$ according to $\sigma$ will yield the same sum.

The point is that only rearranging a finite number of terms doesn't change the sum.

Your reasoning breaks down where you say "and therefore the mapping should be 1-1". Just because two sets have the same cardinality doesn't mean any function you think of should be 1-1. Having the same cardinality just means "there is some very specific choice of function which is 1-1 and onto" - it says nothing at all about a generic function you pick between two sets.

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I didn't refresh my browser in time to see Ryan Budney's comments. What's the accepted practice for dealing with my answer in this case? – Jason DeVito Aug 30 '10 at 22:41
No worries. Your answer is nice. – Ryan Budney Aug 30 '10 at 22:47
Thank you, I appreciate that. – Jason DeVito Aug 30 '10 at 22:52

A permutation $i \to \pi(i)$ preserves convergence of sums ($\Sigma a_{\pi(i)}$ converges whenever $\Sigma a_i$ does) if and only if it maps [1,N] to a bounded number of intervals of integers, with the bound allowed to depend on the permutation but not on N. This places strong constraints on how much "motion" the permutation can create.

In this respect a rearrangement has to be nearly trivial in order to never change the sum.

(This is considering the problem for all convergent series simultaneously. A single sum might be compatible with additional permutations.)

EDIT: permutations satisfying the condition preserve the sum of the series and not only the convergence of the partial sums. That is, $\Sigma a_i = \Sigma a_{\pi(i)}$.

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