The radius of convergence $R$ of $\sum c_kz^k$ is given by $R^{-1}=\limsup\abs{c_k}^{1/k}$.Suggest a slogan
Let \((c_k)_{k\ge0}\) be complex numbers, and define \(R\in[0,\infty]\) by \[ \frac1R\coloneqq\limsup_{k\to\infty}\abs{c_k}^{1/k}, \] with the conventions \(1/0=\infty\) and \(1/\infty=0\). Then the power series \(f(z)=\sum_{k\ge0}c_kz^k\) converges absolutely for \(\abs{z}<R\), uniformly on every closed disc \(\abs{z}\le r\) with \(r<R\), and diverges for \(\abs{z}>R\). So \(R\) is the radius of convergence of the series.
Moreover \(f\) is holomorphic on \(D_R\coloneqq\ts{z\st\abs{z}<R}\), and it can be differentiated term by term: \[ f'(z)=\sum_{k\ge1}kc_kz^{k-1},\qquad \abs{z}<R, \] where the differentiated series has the same radius of convergence \(R\).