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Let $\sigma: \mathbb{N}\to\mathbb{N}$ be bijective such that there is a sequence $(n_k)_{k\ge 0}$ in $\mathbb{N}$ satisfying $|\sigma(n_k)−n_k|\to\infty$ for $k\to\infty$.

Question: Is there a (conditionally) convergent series $\sum_{n\ge 0}a_n$ of real numbers $a_n$ such that $\sum_{n\ge 0}a_{\sigma(n)}$ converges and $\sum_{n\ge 0}a_{\sigma(n)}\neq \sum_{n\ge 0}a_n$?

Remark: If $\sigma: \mathbb{N}\to\mathbb{N}$ is bijective and there is a constant K such that $|\sigma(n)−n|\le K$ for all $n\ge 0$ then $\sum_{n\ge 0}a_{\sigma(n)}= \sum_{n\ge 0}a_n$ for each convergent series $\sum_{n\ge 0}a_n$. I.e. the condition on $\sigma$ above is necessary for changing the value of a rearrangement. So the question asks if it is also sufficient (if not $\sigma$ must satisfy a stricter growth condition).

I posted the question some time ago on SO, but didn't get an answer: https://math.stackexchange.com/questions/4027851/growth-condition-for-a-rearrangement

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    $\begingroup$ Not necessarily. In general, the situation is governed by the following proposition (appearing as an exercise in some old French textbook, as I learned recently, but also not hard to prove from scratch). If $\sigma$ is a permutation of $\mathbb N$, then the necessary and sufficient condition for the series $\sum_n a_{\sigma(n)}$ to converge to the same sum as $\sum_na_n$ every time the latter series converges is that there exists $M$ such that for every $n$, $\sigma(\{1,2,\dots ,n\})$ is a union of $\le M$ intervals of consecutive integers. $\endgroup$
    – fedja
    Commented Mar 6, 2021 at 17:50
  • $\begingroup$ @fedja: Great! Thank you very much for your answer. $\endgroup$
    – tj_
    Commented Mar 6, 2021 at 18:30

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