Show that this limit is related to Euler number I am calculating the limit $\lim_{n \rightarrow \infty} \left( \frac{n!^{\frac{1}{n}}}{n} \right)= \frac{1}{e}.$ I got this limit from wolframalpha, but don't know how to show this.wolframalpha
 A: You do not need Stirling's approximation or logarithms. Since:
$$ n=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right) $$
we have:
$$ N! = \prod_{n=2}^{N}\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right) = \prod_{k=1}^{N-1}\left(1+\frac{1}{k}\right)^{N-k}=\frac{N^N}{\prod_{k=1}^{N-1}\left(1+\frac{1}{k}\right)^k}$$
so, if we define:
$$ a_N = \frac{N!}{N^N} $$
we have:
$$ \frac{a_{N+1}}{a_N}=\left(1+\frac{1}{N}\right)^{-N}$$
so:
$$ \lim_{N\to +\infty}\frac{a_{N+1}}{a_N}=\frac{1}{e}$$
implies:
$$ \lim_{N\to +\infty}\sqrt[N]{a_N}=\frac{1}{e}$$
as wanted.
A: Hint :
$$log(\frac{(n!)^{\frac{1}{n}}}{n})=\frac{1}{n}\sum_{i=1}^nlog(i)-log(n) $$
... and compare with a good primitive of $log(x)$.
A: Hint
Say that $$A=\frac{n!^{\frac{1}{n}}}{n} $$ Take logarithms of both sides $$\log(A)=\frac{1}{n}\log(n!)-\log(n)$$ Now, use Stirling approximation $$\log(n!)\approx n\log(n)-n+\frac 12 \log(2\pi n)+\cdots$$ I am sure that you can take from here.
A: Start with Stirling
$$n!\sim \sqrt{2\pi n}\left({n\over e}\right)^n$$
So our sequence say $u_n$ is such that:
$$u_n\sim \left(2\pi n\right)^{1\over n}\cdot {1\over e}$$
And therefore $u_n\to {1\over e}$
A: Hint: try evaluating the limit of $e^{ln} $ of your function.
Answer: $e^{(1/n)(ln(n!))-ln(n)}$, whose limit as $n$ approaches infinity is $e^{-1}$ 
