Radius of convergence of power series $\sum_{n=0}^{\infty} z^{n!}$

Question

$\sum_{n=0}^{\infty} z^{n!}$ Radius of convergence

I know Radius of convergence $(1/R)= \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$ where $a_n$ is nth term of $\sum a_n |z-a|^n$

But here z is interms of powers of n!

nth term = 1; with z-power as $z^{n!}$ .

$a_n = 1, a_{n-1}=0 \implies 1/R = 1/0 = \infty \implies R=0$

If we take $a_n = 0, a_{n-1}=1 \implies 1/R = 0/1 =0; \implies R= \infty$

Pls point me to where i am making mistake.

Answer 1

If $z=1$, the series diverges. Therefore, $R\leqslant1$.

But if $|z|<1$, $|z|^{n!}\leqslant|z|^n$. So, by the comparison test, $\sum_{n=0}^\infty z^{n!}$ converges absolutely.

So, $R=1$.

Answer 2

Hint:

The exact fomula is Hadamard's formula: $$\frac1 R=\limsup_{n\to \infty} |a_n|^{\tfrac1n}$$