Two questions, and the order matters BC only

Given a series with terms of mixed sign, ask in this order:

1. Does $\sum |a_n|$ converge? If yes, $\sum a_n$ is absolutely convergent, and it converges. Stop.
2. If not, does $\sum a_n$ itself converge? If yes, it is conditionally convergent. If no, it simply diverges.

The order is not a convention; it saves work. The first question is asked about a series of positive terms, which unlocks every test in 10.4 through 10.6 and 10.8. The second question is the awkward one, because a series of mixed signs is eligible for almost nothing except the alternating series test.

Three definitions, so the vocabulary is unambiguous:

1. Absolutely convergent: $\sum |a_n|$ converges. Then $\sum a_n$ converges too.
2. Conditionally convergent: $\sum a_n$ converges but $\sum |a_n|$ does not.
3. Divergent: $\sum a_n$ does not converge, whatever $\sum |a_n|$ does.

Absolute convergence implies convergence. The reason is a comparison: $0 \le a_n + |a_n| \le 2|a_n|$, so if $\sum |a_n|$ converges then $\sum (a_n + |a_n|)$ converges by 10.6, and subtracting the convergent $\sum |a_n|$ leaves $\sum a_n$ convergent.

The converse is false, and 10.7 already built the counterexample. $\sum \frac{(-1)^{n+1}}{n}$ converges to $\ln 2$, and $\sum \frac{1}{n}$ diverges. So a convergent series' absolute counterpart may do anything at all.

That asymmetry is worth stating as a working rule: you may always transfer good news upward from $\sum |a_n|$ to $\sum a_n$, and never downward. Every version of CA10 is a downward transfer, whether it appears as "it converges, so it converges absolutely" or as "the absolute series diverges, so the series diverges".

One consequence worth carrying to 10.13. Because the ratio test is defined on absolute values, whenever it returns $L < 1$ it has established the stronger statement directly, which is why it is the tool of choice for power series.

An absolutely convergent series behaves the way finite sums do: you may reorder its terms freely and the sum is unchanged. A conditionally convergent one does not. Rearranging $\sum \frac{(-1)^{n+1}}{n}$ can be made to produce any value you like, which is a genuinely startling fact and the reason the two cases carry different names.

The intuition is that in a conditionally convergent series the positive terms alone diverge and the negative terms alone diverge, and the finite sum is an artefact of the order in which they are allowed to cancel. In an absolutely convergent series the positive and negative parts each converge on their own, so no ordering can matter.

For this course the practical consequences are smaller but real: the classification is what the free response asks for, and it is what determines the behaviour at the endpoints of an interval of convergence in 10.13.

A worked pass through the flow on $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}$:

1. Absolute series: $\sum \frac{1}{\sqrt{n}}$ is a $p$-series with $p = \frac{1}{2} \le 1$, so it diverges. Not absolutely convergent.
2. The series itself: $b_n = \frac{1}{\sqrt{n}}$ decreases and tends to $0$, so the alternating series test applies and the series converges.
3. Verdict: conditionally convergent.

Two failure modes to avoid at step 2. First, both alternating conditions have to be checked, since a series can reach step 2 and still fail there. Second, failing the alternating series test is not the same as diverging: if $b_n$ does not decrease you have learned nothing and need another argument, whereas if $b_n \not\to 0$ the series genuinely diverges by 10.3.

And remember the branch students forget the flow has. $\sum (-1)^{n} \frac{n}{n+1}$ fails at step 1 and fails at step 2, because its terms do not tend to zero. It is neither absolutely nor conditionally convergent: it is divergent, and the nth term test settles it before either question is reached.