Let $C$ be the set of sequences of real numbers whose associated series is convergent. I define a permutation $f$ of the naturals to be convergent-invariant if, for every sequence $a_n$ in $C$, the series of the sequence $a_{f(n)}$ converges, and to the same sum. Certainly, any permutation that moves only finitely many natural numbers is convergent-invariant. I was surprised to learn, however, that there are other convergent-invariant permutations. So, now, I define a bounded permutation to be a permutation that does not move natural numbers arbitrarily far. For example, the permutation that moves $0$ to $1$, $1$ to $0$, $2$ to $3$, $3$ to $2$, $4$ to $5$, $5$ to $4$, etc is bounded, because it does not move any natural number greater than a distance of $1$. So, my question now is, is being a bounded permutation equivalent to being a convergent-invariant permutation? If not, does either direction of the implication hold?
Is every permutation of naturals that is convergent-invariant, a bounded permutation
permutationssequences-and-series
Related Solutions
Whenever we have a series, $$\sum_{i=1}^{\infty} a_i,$$ we "automatically" get two sequences out of that series:
- The sequence of terms, which is $a_1,a_2,a_3,\ldots$; and
- The sequence of partial sums, which is $s_1,s_2,s_3,\ldots$, where $$\begin{align*} s_1 &= a_1\\ s_2 &= a_1+a_2\\ s_3 &= a_1+a_2+a_3\\ &\vdots\\ s_n &= \sum_{i=1}^n a_i = a_1+a_2+\cdots + a_n. \end{align*}$$
When we talk about "convergence of the series", we are really talking about convergence of the sequence of partial sums: the series $\sum a_i$ converges if and only if the sequence $(s_n)$ converges. That is, your definitions about "series" are really about "sequence of partial sums", and so you have the usual relationship:
In particular, $$\sum_{i=1}^{\infty}a_i\text{ converges}\Longleftrightarrow \{s_i\}_{i=1}^{\infty}\text{ converges}\Longrightarrow \{s_i\}_{i=1}^{\infty}\text{ is bounded}\Longleftrightarrow \sum_{i=1}^{\infty}a_i\text{ is bounded}$$ (where "is bounded" is as per your definition above); but it is possible for $\{s_i\}_{i=1}^{\infty}$ to be bounded, and not convergent, so one can have a series $\sum_{i=1}^{\infty}a_i$ that is bounded (i.e., the sequence of partial sums is bounded) but does not converge.
A simple example of this is $\sum_{i=1}^{\infty} (-1)^n$. The partial sums are $s_{2k+1} = -1$ and $s_{2k}=0$ for every $k$, so the sequence of partial sums is: $$-1,\ 0,\ -1,\ 0,\ -1,\ldots$$ which is bounded but not convergent. So the series is bounded but not convergent.
The relevant theorem for sequences, as you are no doubt aware, is:
Theorem. If $\{b_n\}$ is a monotone sequence, then $\{b_n\}$ converges if and only if it is bounded.
How does that translate for series? When is the sequence of partial sums monotone?
$\{s_i\}$ is increasing if and only if $s_n\leq s_{n+1}$ for all $n$, if and only if $s_{n+1}-s_n\geq 0$ for all $n$; but $s_{n+1}-s_n = a_{n+1}$. So:
The sequence of partial sums of $\displaystyle \sum_{i=1}^{\infty}a_i$ is increasing if and only if all the terms $a_i$ are nonnegative. The sequence of partials sums is strictly increasing if and only if all the terms $a_i$ are positive.
Likewise,
The sequence of partial sums of $\displaystyle \sum_{i=1}^{\infty}a_i$ is decreasing if and only if all the terms $a_i$ are nonpositive. The sequence of partial sums is strictly decreasing if and only if all the terms $a_i$ are negative.
So we conclude:
Theorem. Let $\displaystyle \sum_{i=1}^{\infty}a_i$ is a series in which every term $a_i$ is nonnegative. Then the series converges if and only if it is bounded (in the sense that the sequence of partial sums is bounded).
It's pretty hard to answer this question when 'familiar' isn't defined further.
Every finite group (and thus every permutation group) has a composition series, which is unique in the sense that the length and composition factors of any two composition series are the same up to permutation and isomorphism. If you define familiar groups to be simple groups, then these quotients are the familiar pieces you're looking for.
Defining familiar as simple is a stretch, though. You'd be hard pressed to find someone who found the Held group especially recognizable.
The other problem is that two nonisomorphic groups can have the same composition factors. The factors do help you break the group into smaller, more recongizable chunks, but they don't tell you how those chunks interact.
There are so many variations on combinations of smaller groups that it is difficult to imagine how we could recognize them all without group presentations. Another way of breaking down a group into recognizable groups is by looking at its Sylow subgroups and seeing how they interact (this is local group theory). But for this, we need to understand $p$-groups enough to call them recognizable. My answer to this question should give you an idea of the magnitude of what we're dealing with just in the world of $p$-groups.
Take for example $\text{SmallGroup}(16,3)$. There is no other name for that group, as far as I know. It is isomorphic to $(\mathbb{Z}_4\times\mathbb{Z}_2)\rtimes \mathbb{Z}_2$, which shows us it can be split into those recognizable pieces. This decomposition uses semidirect products, however, which is basically the same as the relations / presentation concept we are trying to avoid.
So in summary, there are many a lot of different ways to divide a group up into pieces that are as familiar as possible, to gain understanding about it and how it works. However, to achieve a full description of the group, most of the time we need to use specific relations in a group presentation to show the way those pieces fit together.
Best Answer
No; while bounded permutations preserve convergence, convergence-preserving permutations need not be bounded. For instance, consider the (unbounded) permutation that swaps $(1,2)$, $(4,8)$, $(16,32)$, etc., and leaves all other positions alone. This permutation preserves convergence. Proof: partial sums $\sum_{n=1}^{N}a_n$ up to $N=2-3,8-15,32-63,\ldots$ are unchanged, and any other partial sum is changed by $|a_x-a_y|$, where $x$ is the previous power of $2$ and $y$ is the next one. By the assumption that $\sum_n a_n$ converges, this difference goes to zero as $N\rightarrow\infty$. So the limit of the partial sums is the same after the permutation as it was before.