Add a half, then a quarter, then an eighth, and keep going for ever. The sum never stops growing, yet it never passes 1. That is the idea behind one of the first results in the course that deals with infinity, and it deserves a proof rather than a formula to memorise.
Start with a finite sum
A geometric series multiplies by the same ratio $r$ at every step:
$$S_n = a + ar + ar^2 + \dots + ar^{n-1}.$$
Multiply every term by $r$ and the list shifts along by one place:
$$rS_n = ar + ar^2 + \dots + ar^{n-1} + ar^n.$$
Subtract, and everything in the middle cancels:
$$S_n - rS_n = a - ar^n \quad\Longrightarrow\quad S_n = \frac{a(1-r^n)}{1-r}, \qquad r \neq 1.$$
Now let the sum run for ever
The only part of $S_n$ that depends on $n$ is $r^n$. If $|r| < 1$, each multiplication by $r$ makes $r^n$ smaller in size, and $r^n \to 0$ as $n \to \infty$. So the finite sums settle on a single value:
$$S_\infty = \frac{a}{1-r}, \qquad |r| < 1.$$
For the opening example, $a = \tfrac{1}{2}$ and $r = \tfrac{1}{2}$, so $S_\infty = \dfrac{1/2}{1 - 1/2} = 1$.
Why the condition matters
If $|r| \ge 1$, the terms do not shrink, and the sums do not settle. With $r = 2$ they double without limit; with $r = -1$ they jump between $a$ and $0$ for ever. The formula would still produce a number, but that number would not be the sum of anything. Checking $|r| < 1$ is not a formality: it is the reason the answer exists.
One surprising consequence
The recurring decimal $0.999\ldots$ is the series $0.9 + 0.09 + 0.009 + \dots$, with $a = 0.9$ and $r = 0.1$. Its sum is
$$\frac{0.9}{1 - 0.1} = 1.$$
So $0.999\ldots$ and $1$ are the same number, written two ways. Many students find this hard to accept at first, and that is a good sign: it means they are taking infinity seriously.
Chapter 2 of BetterMath, Sequences & Series, builds this result step by step, with Your Turn practice and exam-style questions on convergence.