Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The analytic proofs of the Riemann rearrangement theorem are easy to understand but they don't explain why commutativity breaks down when you go from finite sums to infinite sums of real numbers. I suspect that this involves the action of the symmetric group on infinite sets vs finite sets of real numbers. But, so far I haven't come across a proof that only uses tools from Group Theory. Does such a proof exist?

share|cite|improve this question
I often think it is wrong to think of an infinite sum as a "sum." It never was a sum. I doubt there is a group-theoretic proof. – Thomas Andrews Jul 23 '14 at 2:56
up vote 6 down vote accepted

First: commutativity of finite sums. $\mathbb{R}$ is a field, and as such, addition is commutative. This means that for two elements $a,b\in\mathbb{R}$, $a+b=b+a$. However, it is possible to extend this result. Consider $a_1,\ldots,a_n\in\mathbb{R}$, and a permutation $\pi$ of $\{1,\ldots,n\}$. Then, by induction, one can easily show that

$$ \sum_{i=1}^n a_i = \sum_{i=1}^n a_{\pi(i)}. $$

Next: infinite sums. While we generally think of an infinite sum as a "sum," it is technically a limit of a sequence of real numbers. In particular, if $a_i\in\mathbb{R}$ for $i\in\mathbb{N}$, then the symbol

$$ \sum_{i=1}^\infty a_i \qquad\text{is defined as}\qquad \lim_{n\to\infty} \sum_{i=1}^n a_i $$

assuming the limit exists. Furthermore, since the limit ignores any finite initial segment of a sequence, we have the result that

$$ \sum_{i=1}^\infty a_i = \sum_{i=1}^\infty a_{\pi(i)} $$

whenever $\pi$ is a permutation of $\mathbb{N}$ which fixes all but finitely many elements. You can prove this by applying our result about finite sums being invariant under rearrangement.

Finally: With this in mind, you cannot have a group theoretic proof of the Riemann rearrangement theorem because it is a theorem about infinite sums which are analytic objects in essence. In particular, the group structure of $\mathbb{R}$ (under addition) doesn't encode its analytic structure (completeness).

share|cite|improve this answer
Thank you for the detailed explanation. I think your last point is essential. It wasn't emphasized in my course on Analysis. – Aidan Rocke Jul 23 '14 at 3:43
@Aidan, to expound on this last point further: $\mathbb{R}$ is a group under addition regardless of its topology. Since limits are a topological property, by changing the topology we may find that the only convergent series are those whose terms are eventually zero (so the partial sums are constant, e.g., use the discrete topology or the co-countable topology). In the same vein, we may change the topology so that any given series converges simultaneously to every real number (e.g., use the trivial topology). – J. Loreaux Jul 23 '14 at 13:06

You can't expect a purely group theoretic proof since the result is fundamentally about convergence. Some analytic element is thus necessary. In fact it is not at all a group-theoretic result.

Moreover, I don't know which proof you refer to but the one I know (and I presume it's basically the only proof (up to irrelevant detail)) explains perfectly well why it possible to rearrange the terms of a conditionally convergent series to obtain any desired sum. Informally, a series is conditionally convergent if it converges but its positive terms sum up to $\infty$ while its negative terms sum up to $-\infty$. That means that you have an unbounded supply to work with given in the form of ever tinier bits of numbers, both positive and negative. To approach a prescribed sum all you do is start summing up positive element until you just pass the desired sum. Then you add negative elements until you just pass the desired sum. You continue in this way. You are guaranteed not to get stuck (i.e., exhaust all terms in the original series) since you do have an infinite supply to work with. You are guaranteed to converge to the desired sum since the infinite supplies come in ever tinier bits.

share|cite|improve this answer
Indeed, the permutations necessary are of a very specific type - splitting the natural numbers into the ones where the sequence is positive and negative, then "re-braiding" them. It's not particularly interesting to compose two such permutations, for example. – Thomas Andrews Jul 23 '14 at 3:05
@ThomasAndrews, such permutations are called shuffles and the way they compose is (in the finite case, at least) very, very interesting and useful in many contexts :-) – Mariano Suárez-Alvarez Jul 23 '14 at 4:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.