> 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).
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}
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.
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).