Riemann series theorem and bounded permutations? Let $(a_1, a_2, a_3, ...)$ be a sequence of real numbers, with $\sum_{n=1}^{\infty} a_n = L$ conditionally convergent. We know by Riemann series theorem that for any real $M$, there exists a permutation $\pi$ such that
$$
\sum_{n=1}^{\infty} a_{\pi(n)} = M
$$
I noticed there seems to be a pattern that all of these permutations seem to follow to allow for convergence to some different value other than $L$, or to fail to converge. Namely, that they are able to send $a_n$ an arbitrarily far distance away from its original place in the series. In other words, it seems all permutations that allow for this to happen are "not bounded", if we define bounded here as follows:
A permutation $\pi$ is bounded if $$\exists M\in \mathbb{N}, \forall i \in \mathbb{N}, |\pi(i) - i| < M$$
Is my intuition correct here? I've been trying to prove it, but my efforts are handwavy at best. If this is true, is this the strongest condition we can place on $\pi$ such that $\sum_{n=1}^{\infty} a_n = L \Rightarrow \sum_{n=1}^{\infty} a_{\pi(n)} = L$?
This answer I found seems to be related, but I'm not sure I  understand what it means to "map [1, N] to a bounded number of intervals of integers, with the bound allowed to depend on the permutation but not on N".
 A: You are correct, the permutation needs to be unbounded in order for it to change the final sum!
To prove this, it's probably easier to show that if a permutation is bounded, then the sum cannot change. I will only give an idea of how to prove the statement, but I hope it is enough to show how the formal proof could go.

Take some $N$ and look at the values $$\sum_{n=1}^N a_n$$ and $$\sum_{n=1}^N a_{\pi(n)}.$$
The limits of these values as $N\to\infty$ are, of course, the respective infinite sums. Now, if $\pi$ is bounded, then you know that, for every $n$, you have $\pi(n)\leq n+M$, which means
$$\{\pi(1), \pi(2),\pi(3),\dots, \pi(M-N)\}\subseteq \{1,2,3,\dots,N\}$$
and, because an inverse of a bounded permutation is also bounded, you also have
$$\{1,2,3,\dots,M-N\}\subseteq\{\pi(1),\pi(2),\dots,\pi(N)\}$$
which means that the two partial sums can only really differ in the last $M$ terms. Since the sum of the last two terms goes to $0$ as $N$ is big, the two partial sums must converge to the same number.
