Exponential power series - why is the limit like that? I am looking at the exponential power series.
I have to calculate this:
$$R=\frac{1}{{\lim\limits_{n \to \infty }} \sup \frac{1}{\sqrt[n]{n!}}}.$$
But why is it like that:
$$\sqrt[n]{n!} \to +\infty?$$
I thought that it would be $\lim\limits_{n \to \infty} (n!)^{\frac{1}{n}}=(n!)^{0}=1$.
Why isn't it like that?
 A: One way of seeing it is to consider $\log ((n!)^{\frac{1}{n}})$ and then use Stirling approximation for $n!$, so you get $\log n -1 +o(1)$, which of course diverges, hence the original expression $(n!)^{\frac{1}{n}}$ diverges too.  
EDIT: another way of seeing it is once again to consider the log and then do the following:
$$
\log(n!)^\frac{1}{n}=\frac{ \log n!}{n} =  \frac{\sum_{k=1}^{n} \log k}{n} \sim \frac{n \log n}{n} = \log n
$$
Now exponentiate back and you can see the series diverges.
A: The calculation $\lim_n (n!)^{1/n} = (n!)^0 = 1$ is flawed because you cant take the limit only in the exponent but not in the base!
That would be like calculating $\lim_n 1 = \lim_n \frac{n^{-1}}{n^{-1}} = \frac{0}{n^{-1}} = 0$ which is also wrong.
Showing $\sqrt[n]{n!} \rightarrow \infty$ is not that straightforward; it is more easy in this case to show by different means (e.g. ratio test) that $\sum_{n=1}^{\infty} \frac{x^n}{n!}$ converges and then to conclude that $\limsup_n \frac{1}{\sqrt[n]{n!}} = 0$.
