# Absolute convergence of the Laurent series

Consider the power series (which is the Laurent series around $$z_0 = 0)$$ $$\sum_{n=-\infty}^{+\infty}a_nz^{-n} \tag{1}$$ where $$z \in \mathbb{C}$$. It's known that the ROC (region of convergence) of $$(1)$$ is an annulus $$r\lt|z|\lt R$$ and maybe some of the boundary points. In the signal processing, we usually define the ROC as the set of points for which $$(1)$$ converges absolutely. I think this definition, at most misses some of the boundary points of the original ROC. For example, take $$a_n = \begin{cases}\frac{1}{n}, n\ge1 \\ 0 , n\le 0\end{cases}$$ Easily, it can be shown that the series $$\sum_{n=1}^{+\infty}\frac{z^{-n}}{n}$$ converges absolutely for $$|z|\gt 1$$ and converges conditionally for $$z = -1$$. Also it diverges for $$|z|\lt1$$. So the difference between two ROCs is the circle $$|z|=1$$. Is it possible to prove this statement in general? Or is it possible to find examples such that the absolute convergence criterion misses other points as well (in addition to the boundary)?

• Indeed, this follows from the analogous statement for power series, since the Laurent series is the sum of a power series in $z$ and a power series in $z^{-1}$. Oct 10, 2021 at 18:29