Let \(f(z)\) be a holomorphic function in the disk \(|z| < 2\). Show that \(\sum_{n=1}^{\infty} \frac{f^{(n)}(z)}{n!}\) converges uniformly in any compact subset of the unit disk \(|z| < 1\).
Solution
Let \(K\) be a compact subset of \(|z|<1\), and choose \(r<1\) with \(K\subset\{|z|\le r\}\). Pick \(\rho\) so that \[ 1<\rho<2-r. \] Then for every \(z\in K\), the circle \(|\zeta-z|=\rho\) lies in \(|\zeta|<2\). Set \[ M=\max_{|\zeta|\le r+\rho}|f(\zeta)|. \] Cauchy’s estimates give \[ \frac{|f^{(n)}(z)|}{n!}\le \frac{M}{\rho^n} \qquad(z\in K). \] Since \(\rho>1\), the geometric series \(\sum M/\rho^n\) converges. The Weierstrass \(M\)-test therefore shows that \[ \sum_{n=1}^\infty\frac{f^{(n)}(z)}{n!} \] converges uniformly on \(K\). As \(K\) was arbitrary, the convergence is uniform on every compact subset of the unit disk.
Indeed, Taylor’s theorem gives the sum explicitly as \[ f(z+1)-f(z), \] because the disk of convergence of the Taylor series of \(f\) about any \(z\in K\) has radius greater than \(1\).