Find the radius of convergence of $\sum\limits_{n=1}^{\infty}{\frac{(-1)^{n+1}}{n!}(z-1-i)^n}$

Question

Find the radius of convergence of $$\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{n!}(z-1-i)^n}$$ but when I tried the ratio test, I got $$\lim_{n\rightarrow\infty}\left|\dfrac{\frac{(-1)^{n+2}}{(n+1)!}}{\frac{(-1)^{n+1}}{n!}}\right|=\lim_{n\rightarrow\infty}\left|\dfrac{-1}{n+1}\right|=0$$ so does this mean that the radius of convergence is (sorry for wrong notation; you get it) $\frac{1}{0}=\infty$? This was given as an exercise but we haven't had any examples like this... Is this okay?

Answer 1

Why not directly use Cauchy-Hadamard formula when reasonably easy?:

$$\frac1R=\sqrt[n]{|a_n|}=\sqrt[n]{\left|\frac{(-1)^{n+1}}{n!}\right|}=\frac1{\sqrt[n]{n!}}\xrightarrow[n\to\infty]{}0\implies R=\infty $$