Let \(G_N(z) = \sum_{n=0}^{N} z^n\) and \(G(z) = (1-z)^{-1}\).
Carefully prove that \(G_N \to G\) as \(N \to \infty\), uniformly on compact subsets of \(\mathbb{D}\).
What is the power series expansion of \(G(z)\) at \(z = 2023\), and what is its radius of convergence?
Solution
For \(z\ne1\), \[ G_N(z)=\frac{1-z^{N+1}}{1-z}, \] so \[ G_N(z)-G(z)=-\frac{z^{N+1}}{1-z}. \] If \(K\Subset\mathbb D\), choose \(r<1\) with \(|z|\le r\) on \(K\). Also \(\delta=\min_{z\in K}|1-z|>0\). Hence \[ \sup_{z\in K}|G_N(z)-G(z)| \le \frac{r^{N+1}}{\delta}\longrightarrow0. \] Thus \(G_N\to G\) uniformly on compact subsets of \(\mathbb D\).
Put \(w=z-2023\). Then \[ G(z)=\frac1{1-z} =-\frac1{2022+w} =-\frac1{2022}\frac1{1+w/2022}. \] Therefore \[ \boxed{ G(z)=\sum_{n=0}^\infty \frac{(-1)^{n+1}}{2022^{n+1}}(z-2023)^n } \] for \(|z-2023|<2022\). The radius is \[ \boxed{2022}, \] the distance from the center \(2023\) to the unique singularity \(z=1\).