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

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.

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 ($$|z|>\frac12$$).