Re:The sum of all the positive integers is not -1/12
by xyzzyz on Saturday January 18, 2014 @07:55AM
(#7080889)
> 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).
Re:The sum of all the positive integers is not -1/12
by jfarmer on Saturday January 18, 2014 @04:27PM
(#7082302)
I 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}
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.
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?