The nth Term Test for Divergence BC only

If $\sum a_n$ converges to $L$ then $S_n \to L$ and $S_{n-1} \to L$, so $a_n = S_n - S_{n-1} \to 0$: convergence forces the terms to vanish. The nth term test is the contrapositive, that $a_n \not\to 0$ (including the case where the limit fails to exist) forces divergence, and nothing in the argument runs the other way. The converse, that $a_n \to 0$ implies convergence, is false, and the harmonic series is the standing counterexample with vanishing terms and unbounded partial sums, a fact 10.5 proves with the integral test. Knowing $a_n \to 0$ says only that consecutive partial sums are close to each other, which is not the same as either being close to a number.

The test is still the correct first move on every series, because it costs one limit, finishes the problem outright when it fires, and costs nothing when it does not. It fires whenever the terms merely fail to reach zero: a rational term with matching degrees such as $\frac{3n^{2}+1}{n^{2}+5} \to 3$, a term with no limit at all such as $\cos n$ or $(-1)^{n}$, or a term that grows. After that the order of attack is to recognise a geometric or $p$-series on sight, then compare against one, then check for alternation, then reach for the ratio test when factorials or $n$th powers appear. The commonest waste of effort in the unit is a careful comparison argument spent on a series whose terms were never heading to zero.