# The set of all permutations of indices such that the new series converges to the same limit forms a group?

Let $\sum_{i = 1}^{\infty} a_i = s \in \mathbb{C}$ be a convergent series of complex numbers. Then the set of all permutations $\sigma \in\operatorname{Perm}(\mathbb{N})$ such that $\sum_{i=1}^{\infty} a_{\sigma(i)} = s$ forms a group? Seems nontrivial to prove if it is true.

• Is it a group if we allow different limits and just say it need to converges ? – Dominic Michaelis Oct 6 '13 at 19:04
• That's a good question – AlgebraicGeometryStudent Oct 6 '13 at 19:06
• @DominicMichaelis: There is a tangential generalization of this to permutations which preserve the convergence of arbitrary conditionally convergent series (but not guaranteeing that upon rearrangement the limit is the same). This can be shown to form a submonoid, but not a subgroup of $S_{\mathbb{N}}$. See these two references: projecteuclid.org/DPubS/Repository/1.0/… matwbn.icm.edu.pl/ksiazki/cm/cm69/cm69210.pdf – Alex R. Oct 6 '13 at 19:36
• We clearly are not going to have that it is closed by composition. It is just tedious to write down an example. The question of preserving convergence of arbitrary convergent series is very different. – OR. Oct 6 '13 at 19:40
• @ABC I don't think it is clear that they aren't closed under composition – Dominic Michaelis Oct 6 '13 at 19:42

Here is a reasonably simple counterexample. Let $\sigma\colon\mathbb{N}\to\mathbb{N}$ be the following permutation of the natural numbers: $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & \cdots \\ \hline \sigma(n) & \color{blue}2 & \color{blue}4 & \color{blue}6 & \color{red}1 & \color{blue}8 & \color{blue}{10} & \color{blue}{12} & \color{red}3 & \color{blue}{14} & \color{blue}{16} & \color{blue}{18} & \color{red}{5} & \cdots\\ \hline \end{array}$$ with three even numbers for every odd number. We will demonstrate a convergent series with the property that $\sigma$ does not change its value, but $\sigma^2$ does.

Let $\{a_n\}_{n=1}^\infty$ be the sequence $$-1,2,-1,\;-\frac12,\frac22,-\frac12,\;-\frac13,\frac23,-\frac13,\;-\frac14,\frac24,-\frac14,\;\ldots$$ and let $\{b_n\}_{n=1}^\infty$ be the sequence whose odd terms are zero, and whose even terms satsify $b_{2n} = a_n$. Then the series $$\sum_{n=1}^\infty b_n \;=\; 0 - 1 + 0 + 2 + 0 - 1 \;+\; 0 - \frac12 + 0 + \frac22 + 0 - \frac12 + 0 \;+\; \cdots$$ converges to $0$. Applying the permutation $\sigma$ to the terms of this series gives $$\sum_{n=1}^\infty c_n \;=\; \color{blue}{-1+2-1} \color{red}{+ 0} \color{blue}{- \frac12 + \frac22 - \frac12} \color{red}{+ 0} \color{blue}{- \frac13 + \frac23 - \frac13} \color{red}{+ 0} + \cdots$$ which also converges to $0$. However, applying $\sigma$ again yields $$\sum_{n=1}^\infty d_n \;=\; \color{blue}{2 + 0 + \frac22} \color{red}{- 1} \color{blue}{+ 0 + \frac23 + 0} \color{red}{-1} \color{blue}{+\frac24 + 0 + \frac25} \color{red}{- \frac12} + \cdots.$$ We claim that this series does not converge to $0$.

To prove this, observe that the $8N$'th partial sum of this series is $$\sum_{n=1}^{8N} d_n \;=\; \left(\sum_{k=1}^{3N} \frac{2}{k} \right) - \left(2\sum_{k=1}^{N} \frac{1}{k} \right) \;=\; 2H_{3N} - 2H_N,$$ where $H_i$ denotes the $i$'th harmonic number. It is known that $$H_n \;=\; \log(n) + \gamma + o(1)$$ where $\gamma$ is the Euler-Mascheroni constant, and therefore \begin{align*} \sum_{n=1}^{8N} d_n \;&=\; 2H_{3N} - 2H_N \\ \;&=\; 2\bigl(\log(3N) + \gamma + o(1) \bigr) - 2\bigl(\log(N) + \gamma + o(1)\bigr) \\[2ex] \;&=\; \log(9) + o(1) \end{align*} Since the terms of the series converge to zero, it follows that $$\sum_{n=1}^\infty d_n \;=\; \log(9).$$

Edit:
By the way, I should mention how I made this example up. Suppose we are given a permutation $\sigma$ and a series $\sum_{n=1}^\infty s_n$ with the following properties:

1. The distance $|\sigma(n) - n|$ that $\sigma$ moves terms is unbounded.

2. The series is conditionally convergent.

Under these conditions, $\sigma$ will almost always change the value of the sum (or change the convergent series to a divergent one).

So all I did was pick the simplest permutation $\sigma$ that I could think of, and then I created a conditionally convergent series with $0$'s in the right places to be unchanged by $\sigma$.

Under these conditions, it will usually be the case that $\sigma^2$ changes the value of the sum. The only hard part is making sure that the original series converges, and that the original sum and the final sum can be evaluated explicitly.

To answer the weaker question. What structure do such transformations form when $\sum_{i=1}^{\infty} a_{\sigma(i)}$ simply has to converge?

The set of all sequences $(a_i)$ such that $\sum_{i=1}^{\infty} a_{\sigma(i)}$ converges forms a vector space. Proof, let $\lambda$ be a scalar and $a, b$ be two sequences with this property, each of whose series converge under $\sigma$ to $s$ and $t$ respectively. Then for all $\epsilon \gt 0$ we have for finite $N$ $$| \sum_{i=1}^{N} (\lambda a_{\sigma(i)} + b_{\sigma(i)}) - \lambda s + t| \leq \lambda|\sum_{i=1}^N a_{\sigma(i)} - s| + |\sum_{i=1}^{N} b_{\sigma(i)} - t|$$

The right side can be made as small as we want for sufficiently large $N$ so we do indeed have a vector space.

Let the vector space induced by the index permutation $\sigma$ be called $V_{\sigma}$. Another index permutation $\pi$ can be decomposed into a possibly infinite sequence of transpositions $\pi = \tau_1 \circ \tau_2 \circ \dots$ (Proof?)

So if $(a_i) \in V_\sigma$, then $(a_{\tau_1(i)}) \in V_{\sigma}$ since $(a_{\tau_1(i)}) = (a_i) + (0,\dots, a_{\tau_1(i)} - a_i, \dots, a_i - a_{\tau_1(i)}, \dots, 0)$, the second term being absolutely convergent and so clearly in $V_{\sigma}. \$ Similarly $(a_{\tau_1\cdot\tau_2(i)}) \in V_{\sigma}$, and so on...

If $V_{\sigma}$ is closed, then it contains all its limit points. Assuming it's closed...

I give up. Too many assumptions without proof.

• I believe this is related to the Levy Steinitz theorem. – Alex R. Oct 6 '13 at 20:10
• My last comment above shows that it is not going to be a group. I would bet it is not going to be even a semigroup, but for the moment you can change the question to ask if it could be a semigroup. – OR. Oct 6 '13 at 20:52
• @ABC asking if it could be a semigroup is equivalent to asking if it is a monoid as the trivial permutation is surely the identity – Dominic Michaelis Oct 8 '13 at 10:56
• And furthermore in what topology it shall be closed ? – Dominic Michaelis Oct 8 '13 at 10:59