Geometric series in a topological ring

Question

Let $R$ be a topological ring and $q \in R$ be an element such that $1-q$ is invertible. Does it follow that $\sum_{n=0}^{\infty} q^n$ converge? In this case, it would clearly converge to $(1-q)^{-1}$, so the question is equivalent to: Does $\sum_{n=0}^{\infty} q^n$ converge to $ (1-q)^{-1}$?

Usually in topological rings one encounters the following situation: One wants to show that a certain element $1-q$ is invertible. When one knows that $\sum_{n=0}^{\infty} q^n$ converges, for example when $R$ is complete with respect to the $I$-adic topology for some ideal $I$ containing $q$, then it follows that this series provides an inverse to $1-q$. So my question is, in some sense, the converse of this.

Answer 1

Consider some topological fields such as $\Bbb R$ and $q=2$

Answer 2

It converges if you can prove $\lim_{n\to\infty}q^n=0$, because of the formula $$(1-q)^{-1}=1+q+\dots+q^{n-1}+(1-q)^{-1}q^n.$$