Suppose one is given a series of the form $\sum\limits_{n=0}^\infty a_n (z – \alpha)^{-n}$ where $a_n,z,\alpha \in \mathbb{C}$ ($z$ is our indeterminate). How would one determine the radius of convergence of this series? Would the root (or ratio) test work? I imagine substituting $u = (z – \alpha)^{-1}$, looking at $\sum\limits_{n=0}^\infty a_n u^n$ and then applying the root or ratio test is the right thing to do, but I'm not sure.
[Math] Laurent series radius of convergence
complex-analysis
Related Solutions
The radius of convergence is defined by the following equation: $$r^{-1}=\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|.$$
We have, hence: $$ \begin{align} r^{-1}&=\lim_{n \to \infty}\left|\frac{(\log(n+1))^2}{(\log(n))^2}\right|\\ &=\lim_{n \to \infty}\left|\left(\frac{\log(n+1)}{\log(n)}\right)^2\right|\\ &=\lim_{n \to \infty} \left(\frac{\log(n+1)}{\log(n)}\right)^2\\ &=\lim_{n \to \infty} (\log_{n}(n+1))^2\\ &=1.\\ r^{-1}=1 &\Rightarrow r=1. \end{align} $$
Addendum
Here's how you apply Bernoulli's rule: $$ \begin{align} \lim_{n \to \infty}\left(\frac{\log(n+1)}{\log (n)}\right)^2&=\lim_{n \to \infty}\frac{((\log(n+1))^{2})^{\prime}}{((\log(n))^2)^{\prime}}.\\ \text{Given } (f \circ g)^{\prime}(n)&=(f^{\prime}\circ g)(n)g^{\prime}(n),\\ ((\log(n+1))^{2})^{\prime}&=2(\log(n+1))(\log(n+1))^{\prime}.\\ (\log(n+1))^{\prime}&=\frac{1}{n+1}\cdot 1=\frac{1}{n+1}.\\ \text{Therefore, } ((\log(n+1))^{2})^{\prime}&=2(\log(n+1))\cdot \frac{1}{n+1}.\\ ((\log(n))^2)^{\prime}&=2(\log n)\cdot \frac{1}{n}.\\ \text{Hence, } \lim_{n \to \infty}\frac{((\log(n+1))^{2})^{\prime}}{((\log(n))^2)^{\prime}}&=\lim_{n \to \infty}\frac{2(\log(n+1))\cdot \frac{1}{n+1}}{2(\log n)\cdot \frac{1}{n}}\\ &=\lim_{n \to \infty}\frac{\log(n+1)}{\log(n)}\frac{n}{n+1}\\ &=\lim_{n \to \infty}\frac{\log(n+1)}{\log(n)}\lim_{n \to \infty}\frac{n}{n+1}.\\ \lim_{n \to \infty}\frac{n}{n+1}&=1.\\ \text{Thus, } \lim_{n \to \infty}\frac{\log(n+1)}{\log(n)}\lim_{n \to \infty}\frac{n}{n+1}&=\lim_{n \to \infty}\frac{\log(n+1)}{\log(n)}\\ &=\lim_{n \to \infty}\frac{(\log(n+1))^{\prime}}{(\log(n))^{\prime}}\\ &=\lim_{n \to \infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}\\ &=\lim_{n \to \infty}\frac{n}{n+1}\\ &=1. \end{align} $$
It is very useful to remember that the radius of convergence of power series in the complex plane is basically the distance to nearest singularity of the function. Thus if a function has poles at $i$ and $-i$ and you do a power series expansion about the point $3+i$, then the radius of convergence will be $3$ since that is the distance from $3$ to $i$.
Best Answer
Laurent series converge on the annulus $\left\{z\in \mathbb C| R_1 < |z − z_0| < R_2 \right\}$ where $0 \le R_1 < R_2 \le \infty$ when in the form
$$f(z)=\sum_{n=-\infty}^\infty a_n(z-z_0)^n$$
http://www.maths.manchester.ac.uk/~cwalkden/complex-analysis/complex_analysis_part6.pdf