Rearrangements

Definition

Let $\{k_n\}$ be a sequence in which every positive integer appears once and only once. Suppose $r:\mathbb{N}\to \mathbb{N}$, a bijection. Putting

$$a_n^{\prime }=a_{r(n)}$$

for $n=1,2,3,\cdots$, we say that $\sum a_n^{\prime }$ is a rearrangement of $\sum a_n$.

Quite literally, we are just moving around the values of $a_n$.

Theorem (Convergence)

Let $\sum a_n$ be absolutely convergent. Then every rearrangement $\sum a_n^{\prime }$ also converges to the same limit as $\sum a_n$.

Proof: Let

1. $A_n=a_1+\cdots +a_n$
2. $A_n^{\prime }=a_1^{\prime }+\cdots +a_n^{\prime }$ Since $\sum a_n$ converges, then ${\sum }_{n=1}^{\infty }{a}_n={\lim }_{n\to \infty }{A}_n$. We will show that

$$\forall \epsilon >0,\exists N\in \mathbb{N},\forall n\ge N,|A_n-A_n^{\prime }|<\epsilon$$

This implies $(A_n^{\prime })$ converges to ${\lim }_n{A}_n$ with an $\epsilon /2$ proof. Since $\sum |a_n|$ converges, $\exists N\in \mathbb{N}$ such that $\forall m,\ell \ge N$,

$$\begin{array}{r}\sum _{k=\ell }^m|a_k|<\epsilon /2\stackrel{m\to \infty }{\Rightarrow }\sum _{k=\ell }^{\infty }|a_k|\le \epsilon /2<\epsilon ⟹\sum _{k=\ell }^{\infty }|a_k|<\epsilon \end{array}$$

Since $r$ is a bijection, then $\exists m\in \mathbb{N}$ such that

$$\{1,2,\dots ,N-1\}\subset \{r(1),r(2),\dots ,r(m)\}$$

Indeed, $r$ is surjective, so $M$ exists and $r$ is injective, so $M\ge N-1$. Let $n>M$. Then

$$\begin{array}{rl}|A_n-A_n^{\prime }|& =|a_1+a_2+\cdots +a_n-a_1^{\prime }-a_2^{\prime }-\cdots -a_n^{\prime }|\\ & \le \sum _{k=N_1}^{\infty }|a_k|<\epsilon \end{array}$$

and so it converges.

Since we remove some terms, it’s sum will be bounded by the absolute convergence, and this will imply convergence.