How to calculate the limit of a sequence having a factorial $ \lim_{n\to \infty}\frac{\ln \left ( n \right )}{\left ( n! \right )^{\frac{1}{n}}} $ The limit of 

$$ \lim_{n\to \infty}\frac{\ln \left ( n \right )}{\left ( n! \right )^{\frac{1}{n}}} $$

Since the sequence is positive , can we find a sequence bigger having as a limit 0 ? 
 A: Use Stirling's Formula,
\begin{align}
n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n
\end{align}
then it follows
\begin{align}
\lim_{n\rightarrow \infty} \frac{e \ln n}{(2\pi n)^{1/2n} n} =0.
\end{align}
A: Assume first $n$ is even. Then there are exactly $n/2$ integers among $1,2,\dots , n$ that are $>n/2.$ Thus $n!> (n/2)^{n/2},$ giving $(n!)^{1/n}> (n/2)^{1/2}.$ So
$$\tag 1 \frac{\ln n}{(n!)^{1/n}}< \frac{\ln n}{(n/2)^{1/2}}.$$
Since $(\ln n)/n^{1/2} \to 0,$ the limit of $(1)$ is $0$ as $n\to \infty$ through even values.
The case of odd $n$ is basically the same.
A: Yes, you can.  First, take the log of the factorial:
$$\ln(n!)=\ln(1\times2\times\dots\times n)=\sum_{k=1}^n\ln(k)$$
Which is derived from log rules.  Then note that
$$\sum_{k=1}^n\ln(k)>\int_1^n\ln(x)\ dx=n\ln(n)-n+1>n\ln(n)-n$$
Thus,
$$\log(n!)>n\ln(n)-n$$
$$n!>\left(\frac ne\right)^n$$
$$(n!)^{1/n}>\frac ne$$
Which reduces your limit to the following:
$$0<\frac{\ln(n)}{(n!)^{1/n}}<e\frac{\ln(n)}{n}$$
which can be proven to go to zero in one application of L'Hospital's rule.
