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I'm interested in the following combinatorial problem: What is a necessary and sufficent condition on a permutation $\sigma : \mathbb{N} \rightarrow \mathbb{N}$, so that there exist a summable sequence of real numbers $a_n$ for which $\Sigma a_{\sigma(n)}$ converges but $\Sigma a_{\sigma(n)} \neq \Sigma a_n$.

Call an interval of $\mathbb{N}$ a set of consecutive integers. A necessary condition, which follows from applying the Cauchy criterion in both ways is the following: There does not exist a sequence (indexed by $i$) of sets of intervals $\{A^i_1,A^i_2....,A^i_{m_i}\}$, and $\{B^i_1,B^i_2....,B^i_{n_i}\}$ s.t $\{n_i\}$ and $\{m_i\}$ are bounded, $\sigma(\cup_{r=1}^{m_i} A^i_r)=\cup_{r=1}^{n_i} B^i_r$ for all $i$, and the sequence $\{\cup_{r=1}^{m_i} A^i_r\}$ forms an increasig filtration of $\mathbb{N}$.

To see why this condition is necessary: Suppose such sequences of sets of intervals existed, and suppose $a_n$, $a_{\sigma(n)}$ are both convergent series. Let $\epsilon > 0$,and denote by $N_i$ the maximum of $\cup_{r=1}^{m_i} B^i_r$. By the cauchy criterion for $a_n$, there exist an $N>0$ s.t the sum of $a_n$ over any finite interval in $[N,\infty]$ is smaller in absolute value than $\epsilon$. Let $K$ be large enough s.t $\{1,..N\} \subset \cup_{r=1}^{n_i} B^i_r$. Then for all $i>K$ we see that $\Sigma^{N_i}_{n=1}a_n-\Sigma_{\cup_{r=1}^{n_i} B^i_r}a_n$ is a sum over at most $k-1$ intervals in $[N,\infty]$ and hence is less than $k\epsilon$ in absolute value. Hence taking the limit over $i$ we see that $\Sigma_{\cup_{r=1}^{n_i} B^i_r}a_n \rightarrow \Sigma^{\infty}_{n=1}a_n$. Similarly we can see that $\Sigma_{\cup_{r=1}^{m_i} A^i_r}a_{\sigma(n)} \rightarrow \Sigma^{\infty}_{n=1}a_{\sigma(n)}$. But $\Sigma_{\cup_{r=1}^{n_i} B^i_r}a_n=\Sigma_{\cup_{r=1}^{m_i} A^i_r}a_{\sigma(n)}$ and hence the two series have the same value.

I'm unsure of wheter the above condition is sufficent however.

Edit: Also, I'm aware of this question which answers when a permutation sends all convergent series to convergent series with the same sum. However, it doesn't quite answer this question as the complement to that set includes many permutations which send convergent series to non convergent series, but which don't send any convergent series to a convergent series with a different value. From my testing, it seems that being able to alter the value of a series while still keeping it convergent is a far more delicate issue than just being able to make a series not converge.

Et-
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    In your last paragraph, you talk about "permutations which send convergent series to non convergent series, but which don't send any convergent series to a convergent series with a different value". Are you certain that such permutations really exist? – Will Brian Aug 15 '23 at 12:35
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    OK, such permutations do exist. To get one, you can take a partition of $\mathbb N$ into longer and longer intervals, with the lengths of the intervals growing very fast. Then find a permutation $\sigma$ that never maps any number outside of the interval it starts in, but does jumble up each of the intervals internally. Such a permutation could make a conditionally convergent series diverge by oscillation, but could never make it converge to a different value (because the finite partial sums of the rearrangement equal those of the original series infinitely often, at the end of each interval). – Will Brian Aug 15 '23 at 13:03
  • @WillBrian Yep, most of my toy examples also fall into this category. This is also part of why I find this question much more subtle than the convergent to divergent one. If you want to find a sequence that turns a convergent one to a divergent one, you can build it "part by part", sprinkling in zeroes between each section of the sequence. Here however, such a "part by part" construction can't work, as you need to maintain the mixing throughout the whole construction, all while carefully balancing both sequences. – Et- Aug 15 '23 at 14:13
  • Also, it's worth noting that my necessary hypothesis immediately implies (by the answer I linked) that such a permutation also sends some convergent series to divergent series. – Et- Aug 15 '23 at 14:14
  • For $A\subset \mathbb N$ denote $\nu(A)$ the number of maximal intervals $I\subset A$. Then your condition on $\sigma$ may be rephrased: if ${A^i}i$ is an increasing filtration of $\mathbb N$ then $$\lim{i\to\infty}\nu(A^i)+\nu(\sigma(A^i))=\infty$$ (Do you agree? it should be essentially a change of notation).

    Question: would it be equivalent to ask it only for the particular case $A^i:=[i]={1,\dots,i}$, that is $$\lim_{i\to\infty} \nu(\sigma([i])=\infty$$ ?

    – Pietro Majer Sep 01 '23 at 17:53
  • I think the latter condition $\lim_{i\to\infty}\nu(\sigma([i])=\infty$ is not sufficient (for the existence of the two convergent series with different values), so if it is equivalent, the answer to your question would be negative. – Pietro Majer Sep 01 '23 at 19:02
  • On the other hand, denote $\delta(A)$ the maximum length of the intervals $I\subset A$. Then it seems to me that a sufficient condition (but maybe not necessary) would be $$\delta(\sigma([i])=o(i)\qquad (i\to\infty).$$ – Pietro Majer Sep 01 '23 at 19:21
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    @PietroMajer For the first question, I don't think the conditions are equivalent. I can't write right now a full construction, but I think you can make a permutation that shifts between first opening holes (acting on an interval, up to translation, by x->2x) and later filling them, always increasing the sizes of the intervals involved in each step. If you do this in a pattern of open 1 -> open 2 -> fill open 1 -> open 3 -> fill open 2 -> open 4... you will get a permutation that fits your criteria but doesn't fit mine (you can write a filtration with $v(A^i)=2, v(\sigma(A^i))=1$) – Et- Sep 01 '23 at 20:17
  • Also, I think your second condition is not a necessary one. Consider the permutation that changes the sequence $1,-\frac{1}{2}, \frac{1}{3}...$ to $1,\frac{1}{3},-\frac{1}{2},\frac{1}{5}...$ then the first interval of $\sigma(3n)$ will be of length $2n$, even though the permutation does change the value of summation. – Et- Sep 01 '23 at 20:47
  • Yes, the second condition should not be necessary. One may also insert many zeros into a series $\sum_i a_i\neq \sum_i a_{\sigma(i)}$ and create a new permutation that does not satisfy that condition, and still changes the value of summation. – Pietro Majer Sep 01 '23 at 21:17

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