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Say I have an infinite series $\sum_{i\ge 1}a_i$ that is conditionally convergent. Is it true that $$a_1+a_2+a_3+a_4+a_5+\cdots = a_2+a_1+a_4+a_3+a_6+\cdots?$$

That is, swapping terms $a_{2i}$ and $a_{2i+1}$ for all $i\ge 1$. In particular, does the (conditional) convergence of the $\text{LHS}$ imply the (conditional) convergence of the $\text{RHS}$?

My question is motivated by trying to prove that the $\text{RHS}$ converges when $(a_n)$ is the alternating harmonic sequence.

A positive answer doesn't seem far fetched. I know of the theorem of Riemann stating that for a conditionally convergent series $\sum a_i$, and any real $\ell$, there is a permutation $\sigma\colon\mathbb N\to\mathbb N$ such that $\sum a_{\sigma(i)}=\ell$.

Is there a related result giving restrictions we can we place on $\sigma$ to ensure that $\sum a_{\sigma(i)}=\sum a_i$?

Thanks!

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    $\begingroup$ You are just swapping consecutive terms? But then the even numbered partial sums are unchanged, and the odd numbered partial sums differ from the old ones by a number which is going to $0$. $\endgroup$ – lulu Jan 10 '17 at 15:54
  • $\begingroup$ @lulu in that case, just the adjacent ones, yes. And yes I see that now, thanks! $\endgroup$ – Szmagpie Jan 10 '17 at 15:58
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    $\begingroup$ I'd want to write it out, but I imagine that any "bounded rearrangement " would work the same way. That is, a rearrangement where no term moved more than $B$ for some fixed bound. Again, the partial sums would be mostly the same up to some collection of not more than $B$ elements, and all the terms are going to $0$. $\endgroup$ – lulu Jan 10 '17 at 16:03
  • $\begingroup$ Yes, always refer back to the way the partial sums behave. $\endgroup$ – Simply Beautiful Art Jan 10 '17 at 16:33
  • $\begingroup$ @lulu That seems right to me. The partial sums differ by at most $B$ terms for all $n$, and the magnitude of the difference must converge to $0$ since $S_n$ converges! $\endgroup$ – Szmagpie Jan 10 '17 at 17:21

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