A power series may be re-expanded about any point of its disk of convergence

Suppose \[ f(z)=\sum_{n=0}^\infty a_nz^n \] has radius of convergence \(R>0\), and let \(z_0\) satisfy \(|z_0|<R\). Then the sum \(f\) is holomorphic on \(D(0,R)\). Choose \[ 0<\rho<R-|z_0|. \] By Cauchy’s formula, for \(|z-z_0|<\rho\), \[ f(z)=\sum_{k=0}^\infty \frac{f^{(k)}(z_0)}{k!}(z-z_0)^k. \] Indeed, taking any circle \(|\zeta-z_0|=s\) with \(|z-z_0|<s<R-|z_0|\), expand \[ \frac1{\zeta-z} =\frac1{\zeta-z_0} \sum_{k=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^k \] uniformly on the circle and insert this into Cauchy’s integral formula. Thus \(f\) has a power-series expansion about every point of its original disk of convergence.