Geometric series theorem

Step 1 (write the partial sum in closed form). Multiply $S_N$ by $r$: $S_N=\sum_{n=0}^{N-1}ar^n$ and $rS_N=\sum_{n=0}^{N-1}ar^{n+1}=\sum_{n=1}^{N}ar^n$. Subtracting, almost every term cancels: $S_N-rS_N=a-ar^N$, so $(1-r)S_N=a(1-r^N)$.

Step 2 (solve for the partial sum). If $r\neq1$ this gives $S_N=a\dfrac{1-r^N}{1-r}$, an exact, finite formula for every $n$ — no limit has been taken yet.

Step 3 (take the limit). If $|r|<1$, then $r^N\to0$ as $N\to\infty$ (a power with base of absolute value less than $1$ shrinks to zero), so $S_N\to a\dfrac{1-0}{1-r}=\dfrac{a}{1-r}$, which is exactly $\sum_{n=0}^{\infty} ar^n = \dfrac{a}{1-r}$.

Step 4 (the converse: divergence). If $|r|\ge1$ and $r\neq1$, then $r^N$ does not tend to any finite limit (it oscillates or grows without bound), so $S_N$ has no limit and the series diverges; if $r=1$ then $S_N$ equals $Na$, which diverges to infinity since $a$ is nonzero. This covers every case, proving the 'if and only if'.