With $1/R=\limsup\abs{a_n}^{1/n}$, the series $\sum a_n(z-z_0)^n$ converges absolutely for $\abs{z-z_0}<R$ and diverges for $\abs{z-z_0}>R$.Suggest a slogan
Let \((a_n)_{n\ge0}\) be complex numbers, \(z_0\in\CC\), and \[ R\coloneqq\liminf_{n\to\infty}\abs{a_n}^{-1/n}=\frac{1}{\limsup_{n\to\infty}\abs{a_n}^{1/n}}\in[0,\infty], \] with the conventions \(1/0=\infty\) and \(1/\infty=0\). Then the power series \(\sum_{n\ge0}a_n(z-z_0)^n\)
converges absolutely on \(\ts{z\st\abs{z-z_0}<R}\),
converges uniformly on \(\ts{z\st\abs{z-z_0}\le r}\) for every \(r<R\), and
diverges on \(\ts{z\st\abs{z-z_0}>R}\).