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.


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.


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

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


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

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 $[a, b].$

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Copyright © 1996-2017 Alexander Bogomolny


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