find the radius of convergence when $\sum_{n=0}^\infty e^\frac{1}{z^n}$

Question

find the radius of convergence when $$\sum_{n=0}^\infty e^\frac{1}{z^n}$$ I do not know if it convrges or diverges. How can I know? I tried to show that $\lim_{n\to \infty} \frac{|a_{n+1}|}{|a_{n}|}<1$ in order to use one of the conditions of converging. but I got: $e^r < 1$ which leads to $r<0$. Contradiction.

Answer 1

If it converges at some point $z,$ then $e^{\frac {1}{z^n}}\to 0\implies {\text{Re}(e^\frac {1}{z^n})}\to 0.$ If $z = re^{i\theta}, 0 \leq\theta < 2\pi$ then the latter is $e^{\frac{\cos n\theta}{r^n}}\to 0,$ but this must mean: $$\dfrac{\cos(n\theta)}{r^n}\to - \infty$$ . However, this is impossible since $\cos(n\theta)$ will be alternating signs, unless $\theta = 0, \dfrac{\pi}{2,}\dfrac{3\pi}{2}$. But in those case the series is obviously divergent since for example: $$e^{r^{-n}} > 1,$$ when $\theta = 0$ etc.