Show that for all $x\in\mathbb R$ there is a permutation $\rho : \mathbb N \rightarrow \mathbb N$ such that $\sum^{\infty}_{n=1}x_{\rho(n)}=x.$

Question

Assume that $\sum x_n$ is convergent but not absolutely convergent. Show that for all $x\in\mathbb R$ there is a permutation $\rho : \mathbb N \rightarrow \mathbb N$ such that $\sum^{\infty}_{n=1}x_{\rho(n)}=x.$

I don't really know where to start on this one.

I'm also having difficulty wrapping my head around the fact that simply changing the order of the terms within the sum can change the sum itself. Of course I accept it as true but I can't quite understand why it would be the case and so an explanation of that would also be appreciated.

Thank you

A picture of the proof given in Apostol's book is in this question. — , Aug 08 '16 at 22:37
This is the Riemann rearrangement theorem, whose proof you can find in numerous elementary real analysis textbooks. — , Aug 08 '16 at 22:38

score 5 · Answer 1 · answered Aug 08 '16 at 22:33

The idea is this: separate your sequence into the negative and the positive terms. Then start picking the positive terms, in index order, until their sum is larger than $x$. Then pick negative terms until the total sum of all picked terms is smaller than $x$. Then pick positive terms again, and so on.

Here is an informal proof: Since the positive terms and the negative terms both sum to their respective infinities, you will, in any single step in the above, pass $x$ so that you can change sign. Since you pick them in order (within their respective sign set), every element is eventually chosen. And since the original sequence converges, the terms eventually get smaller, so this reordered sequence also converges. For the same reason, the series cannot converge to anything other than $x$.

score 0 · Answer 2 · answered Aug 08 '16 at 22:36

Let $a_0,a_1\dots a_n\dots$ be the positive elements in order of appearance, and let $b_0,b_1\dots b_n$ be the negative elements in order of appearance.

Then we define $z_n$ inductively. $z_n$ is going to be the first value of the form $a_j$ that hasn't been used yet if $\sum_{i=0}^n z_i<x$ and $z_n$ is going to be the first value of the form $b_j$ that hasn't appeared yet if $\sum_{i=0}^n z_i\geq x$.

Let $w$ be the first value such that $\sum_{n=0}^w z_n\geq x$. Then we have $|z_n-x|\leq |z_n|$ for all $n\geq w$. This implies $\lim\limits_{n\to \infty}|x-\sum_{i=1}^{n}z_n|=0$ (because $\lim\limits_{n\to \infty}|x_n|=0$ as $x_n$ is a converging sequence).

Show that for all $x\in\mathbb R$ there is a permutation $\rho : \mathbb N \rightarrow \mathbb N$ such that $\sum^{\infty}_{n=1}x_{\rho(n)}=x.$

2 Answers2