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)?
Absolute convergence of the Laurent series
convergence-divergencelaurent seriespower seriessignal processingz transform
Best Answer
It is not possible. The series will always converge absolutely in the interior of the annulus and diverge outside. Only on the boundary you can have a more complex behavior. Such is the nature of power series. Check any Complex Analysis book for the proof.