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I have recently learned about Riemann's rearrangement theorem, and I have some questions regarding the theorem.

Let $$\sum a_n = a_1 + a_2 + a_3 + a_4 + \cdots$$ be conditionally convergent. If we switch two of the terms, for example $a_2$ and $a_3$, we get the new series $$a_1+a_3+a_2+a_4+\cdots.$$ Both the new and the old series have the same sum, and this will be the case no matter which two terms we switch. Then my (obviously wrong) claim is this: A rearrangment is just repeated switching of terms.

Why is my claim wrong? I can imagine that it has something to do with the fact that for many rearrangements one has to switch two terms infinitely many times. If that is the case, I guess that one can do any finite amount of switching without altering the sum. Is this correct?

I know that it was Riemann who first proved this theorem, but I can't find out when and where. Did he write a paper on it, and if he did, where can I find a copy of it?

I am also grateful for any related information that could be of interest.

Riemann's rearrangement theorem: If an infinite series is conditionally convergent, then its terms can be arranged in a permutation so that the series converges to any given value, or even diverges. (Wikipedia)

I appreciate all help.

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@Elvind: Certainly correct. Formally, let $s_n$ be the sum of the first $n$ terms, and let $s_n'$ be the sum of the first $n$ terms of the rearranged sequence. If there is only a finite amount of switching, then $s_n'=s_n$ for any large enough $n$, so the limits of $(s_n)$ and $(s_n')$, if they exist, are the same. –  André Nicolas May 23 '11 at 17:27
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3 Answers

up vote 18 down vote accepted

Unlike finite symmetric groups, the infinite symmetric group $\text{Sym}(\mathbb{N})$ is not generated by transpositions. The problem is that an infinite composition of transpositions doesn't make sense: consider the composition of $(1 \; 2)$ followed by $(2 \; 3)$ followed by $(3 \; 4)$ and so on to infinity. Where does $1$ go? So yes, rearranging only finitely many terms necessarily doesn't affect the value of the series, but that doesn't imply anything about rearranging infinitely many terms.

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Thanks for the answer! I'm having a hard time understanding the first sentence since I'm not that familiar with groups, but the rest of it makes sense. –  please delete me May 23 '11 at 21:33
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@Elvind: The first sentence says that any permutation (=rearrangement) of a finite set can be described in the form “Switch the two elements a,b; then switch the two elements c,d; then … ; then switch y,z.” (i.e. as a finite list of steps, as a product of transpositions) — and so if we know that switching any two elements leaves the sum unchanged, then we can conclude that the total rearrangement leaves it unchanged. In the infinite case, though, one can’t express all permutations as such products of transpositions, so one can’t repeat the same reasoning. –  Peter LeFanu Lumsdaine May 23 '11 at 22:19
    
I've been thinking. If we start with the sequence $1,2,3,4,5,...$ and do the transpositions you propose, will we not get $2,1,3,4,5,... \to 3,1,2,4,5,... \to 4,1,2,3,5,...$ and so on? I don't see the problem with determining where 1 goes. Or am I missing something? –  please delete me May 25 '11 at 10:10
    
@Peter: Thank you for the clarification. –  please delete me May 25 '11 at 10:10
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@Eivind: I meant the reverse order of compositions. In any case there's also a problem doing it that way around — what ends up in first position? –  Zhen Lin May 25 '11 at 10:18
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The theorem appears in the paper where Riemann also defines the Riemann integral: Ueber die Darstellbarkeit einer Function durch eine trigonometrische Reihe. It's at the beginning of section 3 (the integral comes in section 4). You can find it Riemann's collected works, which is freely available online, for example here; see p. 221 (which is p. 236 in the file).

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Thank you very much! –  please delete me May 23 '11 at 21:36
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I can imagine that it has something to do with the fact that for many rearrangements one has to switch two terms infinitely many times. If that is the case, I guess that one can do any finite amount of switching without altering the sum. Is this correct?

Yes and yes. The subgroup of $\text{Aut}(\mathbb{N})$ generated by transpositions is the group of bijections which fix all but finitely many points. There are countably many such bijections, but uncountably many bijections $\mathbb{N} \to \mathbb{N}$ (exercise), so this subgroup is not the entire group.

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Thanks! I don't understand all of it at the moment, but I will look into it. –  please delete me May 23 '11 at 21:38
    
I think that you can take any $\sigma\in AUT(\mathbb{N})$ such that $\max\{|i-\sigma(i)|<\infty:i\in\mathbb{N}\}$ and it will leave any conditionally convergent sequence fixed. –  Baby Dragon Jul 12 '13 at 21:14
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