Double check z-transform ROC of $a^nu[n]$

Question

For the z-transform ROC of signal $a^nu[n]$, it has been computed to be $|z|>a$. For example (as I have found on Wikipedia), the signal $(\frac{1}{2})^nu[n]$'s ROC will be $|z|>\frac{1}{2}$, as $$\sum_{n=0}^{\infty}(\frac{1}{2})^nz^{-n}=\sum_{n=0}^{\infty}(\frac{0.5}{z})^n=\frac{1}{1-0.5(z)^{-1}}$$ To this point, it said for such equality to hold, $|z|>0.5$. But I am thinking is it possible that $|z|$ can also be less than $0.5$, as long as $z\neq1$ and so the denominator won't go to $0$. However, I then realize that if $|z|$ is less than $0.5$, this term $\sum_{n=0}^{\infty}|(\frac{0.5}{z})|^n$ will go to $\infty$. Could I know if my "realization" is correct: $|z|$ cannot be less than $a$ simply because the term $\sum_{n=0}^{\infty}|(\frac{a}{z})|^n$ will go to $\infty$. Thanks in advance.

Answer 1

You're right, the sum doesn't converge for values $z$ with $|z|\le \frac12$, that's why the equality only holds for $z$ inside the region of convergence ($|z|>\frac12$).