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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 $M_i$ the maximum of $\cup_{r=1}^{m_i} A^i_r$. 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 $\{1,..N\} \subset \cup_{r=1}^{m_i} A^i_r$. Then for all $i>K$ we see that $\Sigma^{M_i}_{n=1}a_n-\Sigma_{\cup_{r=1}^{m_i} A^i_r}a_n$ 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 $\Sigma_{\cup_{r=1}^{m_i} A^i_r}a_n \rightarrow \Sigma^{\infty}_{n=1}a_n$. Similarly we can see that $\Sigma_{\cup_{r=1}^{n_i} B^i_r}a_{\sigma(n)} \rightarrow \Sigma^{\infty}_{n=1}a_{\sigma(n)}$. But $\Sigma_{\cup_{r=1}^{m_i} A^i_r}a_n=\Sigma_{\cup_{r=1}^{n_i} B^i_r}a_{\sigma(n)}$ 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.

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.

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 $\mathbb{N}$.

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.

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 $M_i$ the maximum of $\cup_{r=1}^{m_i} A^i_r$. 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 $\{1,..N\} \subset \cup_{r=1}^{m_i} A^i_r$. Then for all $i>K$ we see that $\Sigma^{M_i}_{n=1}a_n-\Sigma_{\cup_{r=1}^{m_i} A^i_r}a_n$ 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 $\Sigma_{\cup_{r=1}^{m_i} A^i_r}a_n \rightarrow \Sigma^{\infty}_{n=1}a_n$. Similarly we can see that $\Sigma_{\cup_{r=1}^{n_i} B^i_r}a_{\sigma(n)} \rightarrow \Sigma^{\infty}_{n=1}a_{\sigma(n)}$. But $\Sigma_{\cup_{r=1}^{m_i} A^i_r}a_n=\Sigma_{\cup_{r=1}^{n_i} B^i_r}a_{\sigma(n)}$ 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.

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 $\mathbb{N}$.

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

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 $\mathbb{N}$.

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.