Suppose \(\rho {\lt} 1\). Then we can find \(a\) with \(\rho {\lt} a {\lt} 1\) and \(N\in \mathbb {N}\) such that \(n\geq N\Rightarrow |a_n|^{1/n} {\lt} a\). So \(n\geq N\Rightarrow |a_n| {\lt} a^n\). As \(a {\lt} 1\), \(\sum _{n\geq N} a^n\) converges. So, as \(0\leq |a_n| {\lt} a^n\) for \(n\geq N\), \(\sum _{n\geq N} |a_n|\) converges. Thus \(\sum |a_n|\) converges as well.

On the other hand, suppose \(\rho {\gt} 1\). Then there are infinitely many \(n\) such that \(|a_n|^{1/n} {\gt} 1\). So, equivalently, there are infinitely many \(n\) such that \(|a_n| {\gt} 1\). But then \(\lim _{n\to \infty } a_n \neq 0\). So \(\sum a_n\) diverges.

We have \(\limsup |a_n(z-z_0)^n|^{1/n} = |z-z_0|\limsup |a_n|^{1/n} = |z-z_0|\rho \). If \(\rho = 0\), then \(\sum a_n(z-z_0)^n\) converges absolutely no matter what \(z\) is. So \(R = +\infty \). If \(\rho = +\infty \), then \(\limsup |a_n(z-z_0)^n|^{1/n} = +\infty \) unless \(z = z_0\). So \(R = 0\). On the other hand, if \(0 {\lt} \rho {\lt} +\infty \), then \(\limsup |a_n(z-z_0)^n|^{1/n} {\lt} 1 \Leftrightarrow |z-z_0| {\lt} 1/\rho \), and \(\limsup |a_n(z-z_0)^{1/n} {\gt} 1 \Leftrightarrow |z - z_0| {\gt} 1/\rho \). So \(\sum a_n(z-z_0)^n\) converges absolutely for \(|z - z_0| {\lt} 1/\rho \) and diverges for \(|z - z_0| {\gt} 1\). So \(R = 1/\rho \).