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\begin{document}

\title{Rearranging Series}
\author{Chris Arettines, Dakota Blair, Josiah Sugarman}
\maketitle


\begin{hyp*}
Let $X$ be an abelian topological group and $\sigma \in S_\natural$.
Let $a = \{a_i\}_{i<\omega} \subset X$ be a sequence.
Then the series corresponding to $a$ converges and
\[
\sum_{i<\omega} a_i = \sum_{i<\omega} a_{\sigma(i)}
\]
if and only if
$\sigma$ has some easily stated property.
\end{hyp*}

\begin{remark*}
Any permutation $\sigma$ such that $|i - \sigma(i)|$ is bounded preserves every series.
\end{remark*}

\begin{remark*}
The permutation
\[
\sigma = \prod_{i<\omega}(b_{2i} b_{2i-1})
\]
preserves every series where $\{b_i\}_{i<\omega}$ is an increasing sequence
such that $b_{2n} - b_{2n-1} > 2$ for infinitely many $n$.
\end{remark*}

\begin{proof}
For any $N$ such that $b_{2n - 1} < N < b_{2n}$ we have that
\begin{align*}
\sum_{i=0}^N a_i - \sum_{i=0}^N a_{\sigma(i)}
& = 
\sum_{i \in \Dom \sigma \restrict N} a_i - \sum_{i \in \Dom \sigma \restrict N} a_{\sigma(i)}\\
& = 
\sum_{i \in \Dom \sigma \restrict N} a_i - a_{\sigma(i)}\\
\sum_{i=0}^N a_i - \sum_{i=0}^N a_{\sigma(i)}
& = 
\sum_{i = 0}^N a_{b_{2i}} - a_{b_{2i-1}} = 0.
\end{align*}
Therefore every partial sum is the same and so the series is the same.
\end{proof}

\begin{remark*}
The previous permutation can be chosen
such that $|i - \sigma(i)|$ is not bounded.
\end{remark*}

\begin{remark*}
There exists a permutation which is the product of transpositions,
but does not preserve every series.
\end{remark*}

\begin{proof}
We build the permutation inductively, that is
for each $n$ we will construct $\tau_n$,
either a transposition or the identity,
such that
\[
\sigma = \prod_{i=1}^\infty \tau_n
\eqand
\sum_{n=1}^\infty a_{\sigma(n)} = 0.
\]
Let $a_n = (-1)^{n+1}/n$.
Define $\tau_0 = e$.
At the $N$th stage assume that we have defined $\tau_n$ for $n \leq N$.
Define
\[
\sigma_n = \prod_{i=1}^N \tau_i
\eqand
c_n = \sum_{i=1}^n b_n
\]
%Further let $d_n$ be the last index such that $\sign c_{d_n} = -\sgn c_{d_n - 1}$.
Finally let $f_N$ be the last index such that
$\sigma_N \restrict \natural \minus f_N$ is the identity.
If $N < f_n$ then define
$\tau_{n} = e$ for $N \leq n \leq f_n$
and move to the $f_n + 1$ stage.

Then compute $c_N = \sum_{n=1}^{N} a_{\sigma_N(n)}$.
If $c_N > 0$ then let $\tau_3$
 
 while $\sum_{n=1}^N a_n > 0$
\end{proof}

\begin{remark*}
There exists a permutation which has unbounded cycle length
and preserves all series.
In particular
\[
\prod_{i=1}^\infty \left({i \choose 2},{{i\choose 2} + 1}, \ldots ,
{{i+1} \choose 2}-1\right)
\]
\end{remark*}

\begin{proof}[***]
Given a sequence $\{a_n\}$, and a triangular number $t$
\[
\sum_{i=1}^t a_n = \sum_{i=1}^t a_{\sigma(n)}.
\]
Therefore if the permuted series converges,
it converges to the same value as the original series.
Let $\epsilon > 0$, $A_n = \sum_{i=1}^n a_n$ and $A'_n = \sum_{i=1}^n a_{\sigma(n)}$. 
By the Cauchy criterion there is an $N$ such that
$|A_n - A_m| < \epsilon$ for all $n, m > N$.
Then $N$ is between two triangular numbers, $t$ and $N_0$.
Then $N_0$ works for the Cauchy property.

[This is not a proof since we could have easily done something
similar with cycles of length $2^n$ which is demonstrably not Cauchy.
\end{proof}

\begin{remark*}
There exists a permutation with an infinite cycle that preserves all series,
namely
\[
\sigma = (\cdots 53124 \cdots)
\]
\end{remark*}

We are also interested in sets $B \subset \natural$,
identified with their characteristic function $B \in 2^\omega$, such that
\[
F(B) = \sum_{n=0}^\infty {(-1)^{B(n)}}{n}
\]

We use the following notation:
\begin{align*}
\eta_B(x) & = | B \intersect [0,x] | = \sum_{i=0}^{\floor{x}} B(n)\\
\delta(B) & = \lim_{x \to \infty} \frac{\eta_B(x)}{x}
\end{align*}

\begin{remark*}
It is necessary that $\delta(B) = \frac{1}{2}$.
\end{remark*}

\begin{proof}[***]
WLOG $\delta(B) < \frac{1}{2}$.
WLOG $B(N) = 0$.
Then for any sufficiently large $N$ we have that for every $M > 0$
$|B_M| < M$ where $B_M = B \intersect [N,N+2M]$.
Then for every $1 \in B_M$ there is a preceeding $0$.
Then we may underestimate the sum by making all these $1$s
equal in magnitude to the location of their preceeding $0$.
Then in the interval $[N,N+2M]$ these values cancel out
but there are a positive number of unmatched $0$s.
This series will be bounded below by $\sum_{n=1}^\infty \frac{1}{nM}$
it will diverge.
\end{proof}


\begin{remark*}
It is not sufficent that $\delta(B) = \frac{1}{2}$.
\end{remark*}

\begin{proof}
Consider the set $B = \{b \in \natural | b \text{is even or a prime}\}$.
Then $\delta(B) = 1/2$ but $F(B)$ diverges.
\end{proof}

\end{document}
Let $B \subset \natural$
$\eta_B(x) = | B \intersect [0,x] | = \sum_{i=0}^{\floor{x}} B(n)$
$\delta(B) = \lim_{x \to \infty} \frac{\eta_B(x)}{x}
