(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.
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 $r$.