Let \(\Omega\subseteq\CC\) be open, let \(z_0\in\Omega\), let \(f\) be analytic on \(\Omega\), and let \(n\geq1\). Then there is a function \(R_n\) analytic on \(\Omega\) such that for all \(z\in\Omega\), \[ f(z)=\sum_{k=0}^{n-1}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k+R_n(z)(z-z_0)^n. \]
Define \(R_n(z)\coloneqq\bigl(f(z)-\sum_{k=0}^{n-1}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k\bigr)(z-z_0)^{-n}\) for \(z\in\Omega\setminus\{z_0\}\). On a disc about \(z_0\) in \(\Omega\), \(f(z)=\sum_{k\ge0}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k\), so there \(R_n(z)=\sum_{k\ge n}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^{k-n}\) for \(z\neq z_0\). This power series has the same radius of convergence, so setting \(R_n(z_0)\coloneqq f^{(n)}(z_0)/n!\) makes \(R_n\) analytic at \(z_0\); away from \(z_0\) it is a quotient of analytic functions with nonvanishing denominator.