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The basic idea is pretty simple. If Σa_i is conditionally convergent — that is, Σa_i converges but Σ|a_i| does not — then there are an infinite number of both
by jfarmer 13y ago
The basic idea is pretty simple. If Σa_i is conditionally convergent — that is, Σa_i converges but Σ|a_i| does not — then there are an infinite number of both positive and negative a_i. See [1] below for a hint on how to prove this; it's straightforward.
Given that, let's say we want to rearrange the a_i in Σa_i to converge to some arbitrary real number M. Take all the positive a_i in order and call that set A_+ and take all the negative a_i in order and call that set A_-. From [1] we know that both A_+ and A_- have an infinite number of elements.
You then construct a new series like so. Take the fewest elements from A_+ (in the order they appear in the original series) such that their sum is greater than M. Next, add to that sum the fewest number of elements from A_- (also in order) such that this new sum is smaller than M. Go back to taking elements from A_+ until you're just larger M and then back to taking elements from A_- until you're just under M. Repeat this ping-pong maneuver ad infinitum to construct a new series.
You can prove that this new series converges (conditionally) to M.
[1]: Here's a hint. What can we say about Σ|a_i| if we know that Σa_i converges but has only a finite number of negative terms?
- xyzzyz 13y ago> If Σa_i is conditionally convergent then there are an infinite number of both positive and negative a_i. That's true, but that's not enough for your proof -- e.g. \sum (-1)^n/n^2 also has infinitely many positive and negative terms, but converges absolutely. What you need is that sum of all elements in at least one of A_+, A_- (thus also the sum of all elements in the second one) diverges (the order you take the sum in doesn't matter here, as you surely know).
- jfarmer 13y agoI left out lots of other details, too, because it was meant to be an outline for the original commenter, not a proof. In any case, \sum_{i=1}^{\infty} \frac{(-1)^i}{i^2} is not conditionally convergent, so it is not a counterexample to my original statement. Here I take conditionally convergent to mean "the series converges, but it does not converge absolutely." :) What's more, if Σa_i is conditionally convergent then the sums of both A_+ and A_- diverge. You're right that one has to use this fact in the full proof.