Let \(z\in \mathbb{C}\). If \(\lvert z\rvert<1\), then the series \(\sum_{n=0}^{\infty} z^n\) is absolute convergent . To be more precise, we have
\begin{equation*} \sum_{n=0}^{\infty} z^n=\frac{1}{1-z}. \end{equation*}Let \(z\in \mathbb{C}\). If \(\lvert z\rvert<1\), then the series \(\sum_{n=0}^{\infty} z^n\) is absolute convergent . To be more precise, we have
\begin{equation*} \sum_{n=0}^{\infty} z^n=\frac{1}{1-z}. \end{equation*}