Region of convergence

Remarks

Let $r = \limsup_{n \to \infty} \abs{x_n}^{1/n}$ and $1/R = \limsup_{n \to \infty} \abs{x_{-n}}^{1/n}$, with the conventions $1/0 = \infty$ and $1/\infty = 0$. By the root test, the open annulus $r < \abs{z} < R$ lies in the region, and no point with $\abs{z} < r$ or $\abs{z} > R$ does. This is the radius of convergence argument applied to each half of the Laurent series. Points on the circles $\abs{z} = r$ and $\abs{z} = R$ may or may not belong to it. Example: $x_n = 1$ for $n \ge 0$ and $x_n = 0$ for $n < 0$ has region $\abs{z} > 1$; the point $z = 1$ fails because $\sum_{n \ge 0} 1$ diverges. Non-example: $z = 1/2$ is not in the region of $x_n = 2^n$ ($n \ge 0$), since $\sum 2^n 2^n$ diverges. Some texts add the point $z = \infty$ to the region of a sequence with $x_n = 0$ for $n < 0$, or $z = 0$ for a sequence with $x_n = 0$ for $n > 0$.