Why is $\lim\limits_{n \to \infty} \ln \left(\frac{n}{(n!)^{\frac{1}{n}}}\right)=1$? Why is $$\lim_{n \to \infty} \ln \left(\frac{n}{(n!)^{\frac{1}{n}}}\right)=1?$$
I see from looking at the graph that it goes to $1$ but I am not too sure how to prove this algebraically.
The only way I can see this function going to $1$ is if $(n!)^{\frac{1}{n}}>n$ but I am not too sure if that is true.
 A: It is
$$\lim_{n\to\infty}\Bigl(\ln (n)-\frac {1}{n}\sum_{k=1}^{n}\ln (k) \Bigr)$$
But
$$\ln (n)-\frac {1}{n}\sum_{k=1}^n\ln (k)=$$
$$-\frac {1}{n}\sum_{k=1}^n\ln \left(\frac {k}{n}\right) $$
It is a Riemann sum, its limit is
$$-\int_0^1\ln (x)dx=-\Big[x\ln (x)-x\Big]_0^1=1$$
A: It is equivalent to proving that $\left(\frac{n^n}{n!}\right)^{1/n}\to e$.
Use that if $a_n=n^n/n!$ and $a_{n+1}/a_n\to L$ then $a_n^{1/n}\to L$.
The limit is easier to compute using that, after simplification, $\frac{a_{n+1}}{a_n}=(1+1/n)^n\to e$.
A: Note that
$$\ln\left(\frac{n}{(n!)^{\frac{1}{n}}}\right)=\frac1n\ln\left(\frac{n^n}{n!}\right)=\frac{a_n}n$$
and by Stolz-Cesaro
$$\frac{a_{n+1}-a_n}{n+1-n}=\ln\left(\frac{(n+1)^{n+1}}{(n+1)!}\right)-\ln\left(\frac{n^n}{n!}\right)=\ln\left(1+\frac1n\right)^n\to 1$$
A: Use the properties of the logarithm: 
$\ln \frac{a}{b} = \ln a - \ln b$, 
and 
$\ln a^b = b \ln a$. 
So we get: 
$\lim_{n\to\infty} \ln(\frac{n}{(n!)^{1/n}})= \ln n - \ln (n!^{1/n}) = \ln n - \frac{1}{n} \ln n! = \ln n - \frac{1}{n} n \ln n + \frac{1}{n} n = 1$
where we used Stirlings approximation: $n! \approx n \ln n  - n$ for large $n$, see http://mathworld.wolfram.com/StirlingsApproximation.html.
