How much can we rearrange a series?

Solution 1:

A solution was pointed out by @fedja in the comment. I give here a complete proof. A necessary and sufficient condition on $\pi$ is :

$(C)$ There is a constant $M>0$ such that for every $n \in \mathbb N$, $\pi(\{0,\ldots , n\})$ is a union of at most $M$ intervals of consecutive integers

Proof Let us write $[m,n] = \{m,m+1,\ldots ,n-1 \}$ for any integers $n\in\mathbb N$.

1. Let $M>0$ be as in $(C)$. For $n$ big enough we can write : $$\pi([0,n]) = [0,b^1_n] \cup [a^2_n,b^2_n] \cup \ldots \cup [a^M_n,b^M_n]$$

with $b_1^n < a^{2}_n \leq b^2_n <a^3_n\leq \ldots$ (since $[k,k] = \emptyset$ there is some ambiguity, but we can always find such sequences).

Then $\lim_{n\to \infty} b_n^1 = +\infty$ and therefore : $$\begin{array}{rcl} \displaystyle\sum_{k=1}^n a_{\pi(k)} &=& \displaystyle\sum_{k=0}^{b^1_n} a_k + \sum_{i=2}^M \sum_{k= a^i_n}^{b^i_n} a_{k} \\ &\displaystyle\overset{ n\to\infty}{\longrightarrow}&\displaystyle \sum_{k=0}^\infty a_k \end{array}$$

2. Assume that for every $M>0$ there is a $n\in\mathbb N$ such that $\pi([0,n])$ is a disjoint union of $>M$ (separated) intervals of consecutive integers. Let us build $(a_n) \in {\mathbb R}^{\mathbb N}$ such that $\sum_{n=0}^\infty a_n= 0$ and $\sum_n a_{\pi(n)}$ does not.

We can find a strictly increasing sequence of integers $N_k$ such that :

• $\pi([0,N_k])$ is a disjoint union of $>k$ intervals of consecutive integers.
• the first interval of $\pi([0,N_{k+1}])$ contains $0$ and $\pi([0,N_k])$.

Then for $n\in\mathbb N$, let : $$a_n = \left\{ \begin{array}{cl} \frac{1}{k} & \text{if } n \text{ is the first integers of one of the intervals of } \pi([0,N_k]) \\ -\frac{1}{k} &\text{if } n+1 \text{ is the first integers of one of the intervals of } \pi([0,N_k]) \\ 0 & \text{else} \end{array}\right.$$

This way, we have : \begin{align} \lim_{N\to \infty} \sum_{n=0}^N a_n = 0 \\ \forall k\in \mathbb N, \sum_{n=0}^{N_k} a_{\pi(n)} \geq 1 \end{align}

Remarks and further questions

Another related question is to find permutations $\pi$ such that $\sum a_n$ and $\sum a_{\pi(n)}$ converge, their sum is the same. This is for example the case if $\pi([0,n]) = [0,n]$ infinitely often, which is compatible with $(\neg C)$.