A series converges when its partial sums settle BC only

Write $\sum_{n=1}^{\infty} a_n$ and you have named three things at once. Keeping them apart is most of this unit:

1. The sequence of terms $a_n$. A list: $a_1, a_2, a_3, \dots$. It is what you are adding, not the sum.
2. The sequence of partial sums $S_n$. A second list, built from the first: $S_1 = a_1$, $S_2 = a_1 + a_2$, and in general $S_n = \sum_{k=1}^{n} a_k$. Each entry is a finite sum, so nothing infinite has happened yet.
3. The sum of the series. A single number, and only sometimes: it is $\lim_{n \to \infty} S_n$ when that limit exists.

For $a_n = \left(\frac{1}{2}\right)^{n-1}$ the terms are $1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots$ and the partial sums are $1, \frac{3}{2}, \frac{7}{4}, \frac{15}{8}, \dots$. The terms are heading to $0$; the partial sums are heading to $2$. Those are two different limits of two different lists, and only the second one is the sum of the series.

The sentence "the series goes to zero" has no meaning, because it attaches a limit belonging to the terms to an object that is a sum. When you catch yourself writing it, the question to ask is which of the three lists you are actually talking about.

There is only one definition, and it is about $S_n$:

$$\sum_{n=1}^{\infty} a_n = L \quad \text{means} \quad \lim_{n \to \infty} S_n = L.$$

The series converges if that limit exists and is finite, and diverges otherwise. Divergence has more than one flavour: the partial sums may grow without bound, as with $1 + 2 + 3 + \cdots$, or they may simply fail to settle, as with $1 - 1 + 1 - 1 + \cdots$, whose partial sums alternate between $1$ and $0$ forever. Neither has a sum.

Two consequences worth holding onto:

1. Every question about a series is a question about $S_n$. Every test in this unit is a shortcut for deciding whether $S_n$ settles, invented because computing $S_n$ in closed form is usually impossible.
2. An infinite sum is a limit, not an addition. You never add infinitely many numbers. You add $n$ of them and ask what happens as $n$ grows, which is why the familiar rules of finite addition need re-earning here.

Here is the error this unit is built around. It is true that if $\sum a_n$ converges then $a_n \to 0$. The converse is false, and the counterexample is not exotic.

The harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$ has terms $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots$, which certainly tend to zero. Its partial sums do not settle. They start $1$, $\frac{3}{2}$, $\frac{11}{6}$, $\frac{25}{12} \approx 2.083$, and they pass every number you can name, eventually. Slowly: reaching $10$ takes over $12{,}000$ terms. But they never stop.

Compare the two directly. The geometric series above and the harmonic series have the same first two partial sums, $1$ and $\frac{3}{2}$. By $n = 4$ the harmonic has passed $2$, which the geometric never reaches. Both have $a_n \to 0$, and one converges while the other does not.

So $a_n \to 0$ is a gate, not a verdict. It can rule a series out, and it can never rule one in:

1. If $a_n \not\to 0$: the series diverges. This is the nth term test, and it is the only conclusion the test supports.
2. If $a_n \to 0$: you have learned nothing about convergence and must use another test. Topic 10.3 makes this the whole subject.

Almost no series has a partial sum you can write in closed form, which is why the rest of the unit is tests rather than computations. Two families are the exceptions, and they are worth meeting now because they make $S_n$ concrete.

Geometric. For $\sum_{n=0}^{\infty} ar^{n}$ the partial sum is $S_n = a\dfrac{1 - r^{n}}{1 - r}$. When $|r| < 1$, $r^{n} \to 0$ and $S_n \to \dfrac{a}{1 - r}$. Topic 10.2 is entirely about the three ways this gets misapplied.

Telescoping. For $\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)$ almost everything cancels: $S_n = 1 - \frac{1}{n+1}$, so the sum is $1$. Notice what made this possible. We found $S_n$ as a formula and took its limit, which is the definition applied directly rather than a test.

Both illustrate the same point. When you can write $S_n$ down, you do not need any test at all : you take a limit and you are finished. Every test from 10.3 onward exists for the overwhelming majority of series where you cannot.