Find radius convergence of power series

Question

Find the radius of convergence of the series

$$\sum\limits_{n=0}^\infty 3^nz^{n!}$$

My approach is as follow: $$R=\frac{1}{\lim\limits_{n\to\infty}\sup\sqrt[n]{a_n}}=\frac{1}{\lim\limits_{n\to\infty}\sup\sqrt[k!]{3^k}}=1.$$ Am I right? Any other approach would be apppriciated. Thanks in advanced.

Answer 1

OK, except I just noticed that you somehow switched to $k$ from $n$. Anyway since $$\lim_{n\rightarrow \infty} 3^{n/(n!)} = \lim_{n\rightarrow \infty} 3^{-(n-1)!} = 1$$

Your answer is correct.

As far as another approach, I would just use inspection: $|z^{n!}|$ grows incredably fast when ever $1 < |z|$ in order for the series to converge in that region, you would need coefficients that can keep it in check, at the least they need to bring the terms down to zero for even a chance at convergence. $3^n$ does not help, it only adds to the problem of divergence by making the terms yet larger in size.

Answer 2

A different approach based on the quotient test (applied to the terms of the series, not to the coefficients). $$ \frac{3^{n+1}z^{(n+1)!}}{3^nz^{n!}}=3\,z^{(n+1)(n!-1)}. $$ As $n\to\infty$, this converges to $0$ if $|z|<1$ and to $\infty$ if $|z|>1$.