I want to prove that for $|q|<1$, the function $f(z):=\frac{(az;q)_\infty}{(z;q)_\infty}$ is analytic on the set $\{z:|z|<1\}$.
My approach: We consider the sequence of functions $\{f_n\}$ defined as $f_n(z):=\frac{(az;q)_n}{(z;q)_n}$ for every $n\in\mathbb N$. Clearly, this sequence converges to $f$ pointwise on $|z|<1$.
If we can prove that this convergence is uniform on any compact subset of $|z|<1$, then we can conclude, by Weierstrass theorem, that $f$ is analytic on $|z|<1$ because each $f_n$ is analytic on this set.
Comment: On a compact metric space, equicontinuity and pointwise convergence implies uniform convergence.