# Riemann Rearrangement Theorem

An *infinite sequence* $\{s_{n}:\space n \ge 1\}$ that has a limit $\displaystyle\lim_{n\rightarrow\infty}s_{n}$ is said to *converge* or be *convergent*. The sequence that is not convergent is said to *diverge* or be *divergent*.

A *series* is an expression $\displaystyle\sum_{n}a_{n},$ where $\{a_{n}\}$ is an infinite sequence. With every series we associate the sequence $\{s_{n}\}$ of its partial sums:

$s_{n} = a_{1} + a_{2} + ... + a_{n}$

The series $\displaystyle\sum_{n}a_{n}$ is said to be *convergent* if the sequence of its partial sums is convergent. The limit of the partial sums, if it exists, is called the *sum* of the series. The series that is not convergent is called *divergent*. In most cases, there is no point in assigning any sum to a divergent series. Sometimes, when all the terms $a_{n}$ are possitive, a divergent series is said to diverge to infinity: $\sum a_{n} =\infty;.$

If the series $\sum |a_{n}|$ is convergent, the series $\sum a_{n}$ is said to be *absolutely convergent*, or *converge absolutely*. A convergent series which is not absolutely convergent converges *conditionally*.

The alternating harmonic series $1 - 1/2 + 1/3 - 1/4 + 1/5 - \ldots$ is conditionally, but not absolutely, convergent.

There is a huge difference between absolute and conditional convergences. The terms of an absolutely convergent series can be reshuffled without changing the sum of the series. In fact,

Let $p$ be a permutation of the set $\mathbb{N}$ of positive integers. If $\sum a_{n}$ is absolutely convergent, then so is $\sum a_{p(n)}$ and their sums coincide: $\sum a_{n} = \sum a_{p(n)}.$

The situation is entirely different with the series that are only conditionally convergent. A conditionally convergent series $\sum a_{n}$ may be considered as a mixed conbination of two divergent series: one, say $\sum b_{n},$ with all terms positive, the other, say $\sum c_{n},$ with all terms negative. The first of these deverges to infinity, the other one diverges to minus infinity. For this reason, reshuffling the terms of a conditionally convergent series, it is possible to obtain any sum whatsoever.

### Theorem

Let $\sum a_{n}$ be a conditionally convergent series of real numbers and let $a\in \mathbb{R}$ be real. Then the exists a permutation $p:\space\mathbb{N}\rightarrow\mathbb{N}$ such that $\sum a_{p(n)} = a,$ conditionally.

### Proof

First note that the terms of a convergent series - regardless whether the convergence is absolute or conditional - tend to $0:$ $a_{n}\rightarrow 0$ as $n\rightarrow\infty.$ This is so because $a_{n} = s_{n+1} - s_{n}$ and the sequences ${s_{n}}$ and ${s_{n+1}}$ having the same limit.

Suppose, for the definiteness' sake, that $a$ is positive: $a \gt 0.$

Since the series $\sum b_{n}$ of positive terms in $\sum a_{n}$ is divergent, there exists first index $k$ such that

$B = b_{1} + b_{2} + ... + b_{k} \gt a.$

Since the series $\sum c_{n}$ of negative terms in $\sum a_{n}$ is divergent, there is first index $m$ such that

$BC = B + c_{1} + c_{2} + ... + c_{m} \lt a.$

We continue in the same manner to claim the existence of smallest indices $u$ and $v$ such that

$BCB = BC + b_{k+1} + b_{k+2} + ... + b_{u} \gt a.$

and

$BCBC = BCB + c_{m+1} + c_{m+2} + ... + c_{v} \lt a.$

And so on. The sequence $B,$ $BC,$ $BCB,$ $BCBCB,$ ... so obtained converges to $a.$ The reason for that is that, as we have observed, the terms $a_{n}$ (and trivially also $b_{n}$ and $c_{n})$ tend to $0,$ implying that, with every step in the construction, the difference between the resulting sum and $a$ becomes smaller.

A slight modification of the proof shows that, given a closed segment $[a, b],$ it is possible to reshuffle a conditionally convergent series $\sum a_{n}$ so that the closure of the set of partial sums ${s_{n}}$ will contain $[a, b].$

|Contact| |Front page| |Contents| |Algebra| |Up|

Copyright © 1996-2018 Alexander Bogomolny

67101798