Here is Theorem 3.39 in the book Principles of Mathematical Analysis by Walter Rudin, third edition:
Given the power seires $\sum c_n z^n$, put $$\alpha = \lim_{n\to\infty}\sup\sqrt[n]{\vert c_n \vert}, \ \ \ R = \frac{1}{\alpha}.$$ (If $\alpha = 0$, $R = +\infty$; if $\alpha = +\infty$, $R = 0$.) Then $\sum c_n z^n$ converges if $\vert z \vert < R$, and diverges if $\vert z \vert > 1$.
The proof makes use of the root test, which is Theorem 3.33. Now in Example 3.40 (b), Rudin states that the series $\sum {z^n \over n!}$ has $R = +\infty$, meaning $\alpha = 0$. How does this hold, especially in view of the fact that $\lim_{n\to\infty} \sqrt[n]{n} = \lim_{n\to\infty} \sqrt[n]{p} = 1$ for any $p > 0$, as has been stated and proved by Rudin in Theorem 3.20 (b) and (c)? That is, how to rigorously show (using only the machinery developed by Rudin until Theorem 3.39) that $$\lim_{n\to\infty} \sup {1 \over \sqrt[n]{n!}} = 0?$$
Can we state for the power series $\sum c_n z^n$ a result analogous to Theorem 3.39, using the ratio test (i.e. Theorem 3.34)?