I was supposed to prove:
Theorem: Let $a \neq 0$, the geometric series $\sum_{n=1}^{\infty}(aq^{n-1})$
i) if $|q|< 1$, converges and has the result $S=\frac{a}{1-q}$.
ii) if $|q| \ge 1$, diverges.
I did it as follows:
Let the geometric series $\sum_{n=1}^{\infty}(aq^{n-1})$ and consider $S_n$ the n-th sum of terms of the series. Then,
$S_n=a+aq+aq^2+...+aq^{n-1}$ $\,$ and $\,$ $qS_n=aq+aq^2+aq^3+...+aq^{n-1}+aq^n$
$\implies$ $S_n-qS_n=(a+aq+aq^2+...+aq^{n-1})-(aq+aq^2+aq^3+...+aq^{n-1}+aq^n)$
$\implies$ $S_n-qS_n=a-aq^n$
$\implies$ $S_n(1-q)=a(1-q^n)$
$\implies$ $S_n=\frac{a(1-q^n)}{1-q}$
We have that $S=\lim_{n \to \infty}S_n$. This way, in this case, $S=\lim_{n \to \infty}\frac{a(1-q^n)}{1-q}=(\lim_{n \to \infty}\frac{a}{1-q})(\lim_{n \to \infty}(1-q^n))$. Also, $\lim_{n \to \infty}\frac{a}{1-q}=\frac{a}{1-q}$, since its a constant. Therefore,
i) If $|q|<1$, then $\lim_{n \to \infty}(1-q^n)=\lim_{n \to \infty}(1)-\lim_{n \to \infty}(q^n)=1-0=1$. So, $S=(\lim_{n \to \infty}\frac{a}{1-q})(\lim_{n \to \infty}(1-q^n))=\frac{a}{1-q} \cdot 1=\frac{a}{1-q}$, and the series converges.
ii) If $|q| > 1$, then $\lim_{n \to \infty}(1-q^n)=\lim_{n \to \infty}(1)-\lim_{n \to \infty}(q^n)=1-\infty=-\infty$. So, $S=(\lim_{n \to \infty}\frac{a}{1-q})(\lim_{n \to \infty}(1-q^n))=\frac{a}{1-q} \cdot (-\infty)=\pm\infty$, and the series diverges.
Is it valid? I would consider the case $q=1$ separately (it would give a series with summand a, which is known to diverge) because it doesn't fit here, since we would have $1-q=0$.
The proof of the first case is correct. As to the second case, $(q^n)$ doesn't have a limit if $q\leqslant -1$, so it's wrong to write $\lim q^n=+\infty$. Just say a product of a nonzero constant with a diverging sequence diverges.
In either case, you can factor out by $a$ and only study the series $\sum q^n$ since $a\neq 0$ (what you did is also right, just saying there's no need to carry $a$ around).