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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)?

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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.

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    $\begingroup$ 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 $z^{-1}$. $\endgroup$ Oct 10, 2021 at 18:29
  • $\begingroup$ Convergence is also uniform in any compact subset of the annulus of convergence. $\endgroup$
    – GReyes
    Oct 10, 2021 at 18:50
  • $\begingroup$ Thanks. So the difference between ROCs (if there is any) should be the boundary points, right? $\endgroup$
    – S.H.W
    Oct 10, 2021 at 20:45
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    $\begingroup$ Right. Those are the only points where convergence depends on more subtle properties of your specific series. $\endgroup$
    – GReyes
    Oct 10, 2021 at 22:31

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