Compute $\lim_{n\rightarrow\infty}\sqrt[{n+1}]{(n+1)!}-\sqrt[n]{n!}$ 
Evaluate $L=\lim_{n\rightarrow\infty}\sqrt[{n+1}]{(n+1)!}-\sqrt[n]{n!}$

How I approached it and where I get stuck:
$$\lim_{n\rightarrow\infty}\sqrt[{n+1}]{(n+1)!}-\sqrt[n]{n!}=\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}nn(\frac{\sqrt[{n+1}]{(n+1)!}}{\sqrt[n]{n!}}-1)$$
Now:
$$\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}n=\lim_{n\rightarrow\infty}\sqrt[n]{\frac{n!}{n^n}}=\lim_{n\rightarrow\infty}e^{\frac1n\sum_{k=1}^n\ln(\frac kn)}=e^{\int_0^1\ln(x)dx}=e^{-1}=\frac1e$$
So $L=\lim_{n\to\infty}\frac 1e\times n(\frac{\sqrt[{n+1}]{(n+1)!}}{\sqrt[n]{n!}}-1)$. Now this is where I get stuck. What should I do?
 A: I replaced the factorials with Sterling's approximation: $n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n.$  The expression became
$$\left(\frac{n}{e}\right)\left[\sqrt[n+1]{2\pi (n+1)}-\sqrt[n]{2\pi n}\right]+\frac{1}{e}\sqrt[n+1]{2\pi (n+1)}.$$
The second term is easily seen to go to $1/e$.  The first term goes to $0$.  I argue like this:
Let $f(x) = \sqrt[n]{2\pi n}$ and note that the bracketed part of the first term is $f(n+1)-f(n)$.  Invoke the Mean Value Theorem.  
$$f'(c) = \frac{ \sqrt[c]{2\pi c}(1-\ln 2\pi c)}{c^2} = f(n+1)-f(n)$$
for some $c$ between $n$ and $n+1$.  Note that $f'(x)$ is increasing, so the bracketed bit is greater than $f'(n)$.  Also note that $f'(x)$ is negative, so the $f(x)$ is decreasing and so the bracketed bit is negative.  We have
$$\left(\frac{n}{e}\right)f'(n)<  \left(\frac{n}{e}\right)\left[\sqrt[n+1]{2\pi (n+1)}-\sqrt[n]{2\pi n}\right]<0.$$
Then see that 
$$\lim_{n\to \infty} \left(\frac{n}{e} \right)\frac{ \sqrt[n]{2\pi n}(1-\ln 2\pi n)}{n^2} = 0$$
so final answer is $1/e$.
A: You may invoke Stirling's double inequality
$$ \left(\frac{n}{e}\right)^n\sqrt{2\pi n}\,e^{\frac{1}{12n+1}}\leq n! \leq \left(\frac{n}{e}\right)^n\sqrt{2\pi n}\,e^{\frac{1}{12n}} $$
to get that
$$ \sqrt[n]{n!} = \frac{n}{e}+\frac{\log n}{2e}+\frac{\log(2\pi)}{2e}+O\left(\frac{\log^2 n}{n}\right) $$
and
$$ \sqrt[n+1]{(n+1)!}-\sqrt[n]{n!} = \frac{1}{e}+O\left(\frac{\log^2 n}{n}\right).$$
In other terms, we just have to show that $\sqrt[n]{n!}=\text{GM}(1,2,\ldots,n)$ does not deviate much from $\frac{n}{e}$. Since $n=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)$ holds for any $n\geq 2$ we have
$$ n! = \prod_{m=2}^{n}\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{n-k}=\frac{n^{n}}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k} $$
$$ \frac{\sqrt[n]{n!}}{n}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{-k/n}=\exp\sum_{k=1}^{n-1}-\frac{k}{n}\log\left(1+\frac{1}{k}\right). $$
Over the interval $[0,1]$ the function $x-\log(1+x)$ is non-negative and $O(x^2)$, hence
$$ \frac{\sqrt[n]{n!}}{n}=\exp\left[-\frac{n-1}{n}+O\left(\frac{H_{n-1}}{n} \right)\right]=\exp\left[-1+O\left(\frac{\log n}{n}\right)\right]$$
and this proves (in a elementary way) the sharper
$$ \sqrt[n+1]{(n+1)!}-\sqrt[n]{n!} = \frac{1}{e}+O\left(\frac{\log n}{n}\right).$$
A: I have an idea in my mind which is to long for a comment and provides another way of computing this limit. However I am not sure whether my attempt is legit or not. Therefore I would just add it now and remove it again in the case that it turns out to be absolute nonsense.
Hence we are dealing with the limit $n\to\infty$ one can use Stirlings formula as a valid approximation for $n!$. Therefore the limit becomes
$$\begin{align}
\lim_{n\to\infty}\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}&=\lim_{n\to\infty}\sqrt[n+1]{\sqrt{2\pi(n+1)}\left(\frac{n+1}{e}\right)^{n+1}}-\sqrt[n]{\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}}\\
&=\lim_{n\to\infty}\sqrt[2n+2]{2\pi(n+1)}\left(\frac{n+1}{e}\right)-\sqrt[2n]{2\pi n}\left(\frac{n}{e}\right)\\
&\stackrel{?}{=}\lim_{n\to\infty}\left(\frac{n+1}{e}\right)-\left(\frac{n}{e}\right)\\
&=\lim_{n\to\infty}\frac1{e}=\frac1e
\end{align}$$
I marked the weak point of my argumentation with $\stackrel{?}{=}$ since I am not sure whether I can do so in this situation or not. The given roots are clearly approaching to one while $n$ is going up to infinity therefore the two fractions as remaining part of the limit are the only thing of relevance. Since my evaluation ends up in the same result as WolframAlphas I wanted to add my attempt at least.
