# Why do we clarify between Absolute and Conditional Convergence?

First, can an alternating series diverge? Or does it always converge since you're adding then subtracting? If it must converge, then some of the following may be moot.

Next, to review, if S is alternating, then we can examine |S|. We know that S < |S| since S is alternating and subtracts alternate terms. So, if |S| converges, we know that S must also converge, since S < |S|.

Case 1: S = converges........ |S| = converges..... Ergo, S is absolutely convergent.

Case 2: S = converges..... |S| = diverges...... Ergo, S is conditionally convergent.

Case 3: S = unknown |S| = diverges...... What can we conclude about S ?

Also, if you know S is convergent, why go further with |S| in the first place? what is the point of examining |S| and adding the label S = "absolutely convergent"? It seems the entire point of absolute convergence, is to make a conclusion on the original S, if that is unknown.

• Consider $1+1-1+1-1+1-1+\dots$ which is alternating and diverges. – Nate Eldredge Apr 5 '18 at 13:50
• When the series of absolute terms is divergent the original series could be divergent too, or convergent. Above you were given one that both are divergent and $\sum_n(-1)^n/n$ is not absolutely convergent but it does converge. For testing convergence of non-absolutely convergent series you can use theorem like the alternating series one, or stronger ones like Abel's. When series converge but not absolutely Riemann's theorem happens. – user547557 Apr 5 '18 at 14:06
• Right, so alternating can diverge. – JackOfAll Apr 5 '18 at 14:16

Nothing. $S$ is not absolutely convergent, and that's it.