Set $\alpha = \limsup_{n\to\infty} |a_n|^{1/n}$. We show the series converges absolutely for $|z-z_0| < 1/\alpha$ and diverges for $|z-z_0| > 1/\alpha$.

Convergence for $|z-z_0| < 1/\alpha$:

Choose $\rho$ with $|z-z_0| < \rho < 1/\alpha$, so $\alpha < 1/\rho$. Since $\alpha = \limsup |a_n|^{1/n}$, there exists $N$ such that $|a_n|^{1/n} < 1/\rho$ for all $n \geq N$. Therefore $|a_n| < 1/\rho^n$ for $n \geq N$, giving

$|a_n(z-z_0)^n| < \left(\frac{|z-z_0|}{\rho}\right)^n$

for $n \geq N$. Since $|z-z_0|/\rho < 1$, the series $\sum (|z-z_0|/\rho)^n$ converges (geometric series), so $\sum a_n(z-z_0)^n$ converges absolutely by comparison.

Divergence for $|z-z_0| > 1/\alpha$:

Choose $\rho$ with $1/\alpha < \rho < |z-z_0|$, so $\alpha > 1/\rho$. Since $\alpha = \limsup |a_n|^{1/n}$, the inequality $|a_n|^{1/n} > 1/\rho$ holds for infinitely many $n$. For these $n$:

$|a_n(z-z_0)^n| > \left(\frac{|z-z_0|}{\rho}\right)^n \to \infty.$

The terms do not converge to zero, so the series diverges.

The boundary $|z-z_0| = 1/\alpha$:

Both convergence and divergence can occur on the boundary circle. For example, $\sum z^n/n^2$ ($R = 1$, converges on $|z|=1$) and $\sum z^n$ ($R = 1$, diverges on $|z|=1$) have the same radius but different boundary behavior. $\blacksquare$