The test runs one way only BC only

The test is one line:

$$\text{If } \lim_{n \to \infty} a_n \neq 0 \text{ (or does not exist), then } \sum a_n \text{ diverges.}$$

Why it is true is short enough to be worth seeing, because the argument is where the one-directionality comes from. If $\sum a_n$ converges to $L$, then $S_n \to L$ and also $S_{n-1} \to L$, so

$$a_n = S_n - S_{n-1} \longrightarrow L - L = 0.$$

So convergence forces the terms to vanish. The test is the contrapositive of that: if the terms do not vanish, the series cannot have converged. Nothing in the argument runs backwards. Knowing $a_n \to 0$ tells you that $S_n$ and $S_{n-1}$ are getting close to each other, which is a long way from either of them getting close to a number.

Set the possibilities out explicitly, because the false one is the one that feels most natural:

1. The series converges $\Rightarrow a_n \to 0$. True. This is the argument above.
2. $a_n \not\to 0 \Rightarrow$ the series diverges. True. This is the contrapositive of the first, and it is the test.
3. $a_n \to 0 \Rightarrow$ the series converges. False. This is the converse of the first, and converses of true statements are not true by default.

The counterexample is the harmonic series $\sum \frac{1}{n}$, whose terms tend to zero and whose partial sums grow without bound. That claim is proved in 10.5, where the integral test settles it in two lines; for now it is enough that it is a fact, and that it is not an exotic one. Plenty of series have vanishing terms and no sum.

A useful way to hold the distinction: $a_n \to 0$ is the entry requirement, not the qualification. Every convergent series meets it. So does every divergent series that this test cannot touch, which is most of them.

Despite concluding so little, this is the test to try first on every series you meet, for three reasons:

1. It is nearly free. One limit of the term, which you can usually take by inspection.
2. When it fires, you are finished. No other test is needed and none would be cheaper.
3. When it does not fire, it has narrowed nothing but cost nothing. You proceed to a real test having spent one line.

It fires more often than students expect, because the terms only have to fail to reach zero. Three shapes worth recognising immediately:

1. A rational term with matching degrees. $\sum \frac{3n^{2}+1}{n^{2}+5}$ has $a_n \to 3$. Divergent.
2. A term with no limit at all. $\sum \cos n$ and $\sum (-1)^{n}$ both have terms that never settle, and "does not exist" is covered by the test just as "nonzero" is.
3. A term that grows. $\sum \frac{n}{\ln n}$ and $\sum n!$ diverge for the plainest possible reason.

Choosing a test is a skill in its own right, and the failures are of two kinds: reaching for a test whose hypotheses the series does not satisfy, and grinding through a hard test when the series' shape makes an easy one decisive. A workable order of attack:

1. Do the terms tend to zero? If not, you are done. If so, continue, having learned nothing.
2. Is it geometric, or a $p$-series? Both are settled on sight by 10.2 and 10.5, with no work.
3. Does it look like one of those? Then a comparison test, from 10.6.
4. Does it alternate? Then the alternating series test, from 10.7.
5. Are there factorials or $n$th powers? Then the ratio test, from 10.8.

The habit worth building now is to spend one line on step 1 before anything else, every single time. The commonest waste in this unit is a careful comparison argument applied to a series whose terms tend to $5$.