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(From wikipedia) The Riemann series theorem (also called the Riemann rearrangement theorem), named after 19th-century German mathematician Bernhard Riemann, says that if an infinite series is conditionally convergent, then its terms can be arranged in a permutation so that the series converges to any given value, or even diverges.

Let $A=(a_i)_{i=1}^{\infty}$ be a conditionally convergent series, so that $\sum_{i=1}^{\infty} a_i$ converges and $\sum_{i=1}^{\infty} |a_i|$ diverges.

Riemann's theorem says that, for any real number $r$, there is a permutation $\pi_r$ of $\mathbb{N}$ such that $\sum_{i=1}^{\infty} a_{\pi_r(i)}$ converges to $r$.

My question is "Given $A$ and $r$, how many permutations of $A$ are there that give $r$?"

It seems clear to me that there are at least $\aleph_0$ such permutations, since you can go beyond $r$ as far as you want in as many places as you want, and then do the same going down.

But, since there are an uncountable number of permutations of $\mathbb{N}$ (since, engaging in notation-abuse, $\aleph_0! \ge 2^{\aleph_0}$), are there any real numbers $r$ for which there are an uncountable number of permutations whose rearranged sum converges to $r$?

I would also like to know what the answer is to the same question about divergent rearrangements.

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My feeling is that thinking in terms of cardinalities isn't going to give you a very interesting answer, simply because there aren't a lot of cardinalities it could be to start with. It would be interesting to find a more precise measure. –  Potato Jul 18 '13 at 23:56

1 Answer 1

up vote 2 down vote accepted

We can produce continuum many. For simplicity of exposition, assume initially that the terms are all different.

Take any rearrangement that gives sum $r$, where $r$ may be $+\infty$ or $-\infty$. Permute the first and second entries if you feel like it, and then, if you feel like it, the $11$th and $12$th, and the $21$st and $22$nd, and so on forever. The sum is unaffected. We obtain $2^{\aleph_0}$ permutations in this way.

If the terms are not all different, a small modification is needed. We say there is a transition at $n$ if $a_n\ne a_{n+1}$. Every tenth transition, do as in the case where the terms are all distinct.

Remark: For completeness, we explain why the sums are all equal to $r$. Most partial sums agree, except (in the simple case) possibly the first and second, the $11$th and the $12$th, and so on. But the terms go to $0$, so the partial sums converge to $r$.

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