display | more...

Put simply, Riemann's rearrangement theorem argues that a conditionally convergent series, if rearranged, can be made to converge to any number. To put it into more exact language with less jargon, suppose we have a sequence of real numbers a(n) (a is a function from the naturals to the reals), such that the series ∑a(n) is convergent to some number -∞ < S < +∞, but such that the series of absolute values ∑|a(n)| diverges to +. Then for any real number -∞ ≤ S' ≤ -, there exists a permutation p(n) of the natural numbers (that is, p is a one-to-one map from the naturals onto the naturals) such that the series ∑a(p(n)) converges to S'.

The example classically associated with this theorem is the alternating harmonic series, a(n) = (-1)n/n. Every freshman calculus students knows that ∑a(n) = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ... = ln 2, and that ∑|a(n)| = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... (the harmonic series) diverges. So how can I rearrange this series to give some arbitrary desired sum S'? Pretty simple: keep all the positive terms in order (1, 1/3, 1/5, 1/7, ...) in one column and all the negative terms in order (-1/2, -1/4, -1/6, ...) in the other column and pick one element at a time to add to the series in the following way. If the current partial sum is smaller than S' add the biggest unused positive term (from the top of the first column, and scratch it off), otherwise add the smallest (that is, most negative) unused negative term (and scratch it off). If you keep doing that ad infinitum, I claim, you get the desired rearranged series. To prove this claim, I have to show two things. First, that all terms are indeed used (i.e. that this is a legal rearrangement per the rules set above): if a particular, say positive, term 1/n was never reached to be used then that means that all terms in my series past a given point are negative (since 1/n is always used before smaller positive terms), but it's easy to see that the partial sums in that case would decrease unboundedly and at some point must become smaller than S', at which point a positive term must be used --> contradiction. Similarly if a particular negative term is never used. Second, I must show that the sequence of partial sums approaches S': but that is clear, since for a given (odd) number n after both 1/n and -1/(n+1) have been used up, the partial sums never strays farther away than 1/n away from S'. Voila. (This recipe doesn't work to create a rearrangement that makes the series diverge to +or -∞, but I'll leave those tasks as an exercise for the reader.)

Riemann's rearrangement theorem teaches us of the treachery of intuition. It teaches us that while it makes sense to ask for the sum of an ordered sequence a(n) of numbers, it apparently doesn't make sense to ask for the sum of an unordered set A of numbers, it is not well defined. Rather, it is not well defined if for some arrangement of the set (a: N → A, assuming the set is countable) the series ∑a(n) is conditionally convergent. If, however, the set of numbers is absolutely convergent under some arrangement, it turns out it will converge to the same sum under any rearrangement, so it does make sense to ask for the sum of all the numbers in such a set. A more tangible example of the implications of Riemann's rearrangement theorem is the uniform charge density paradox.

Log in or register to write something here or to contact authors.