TITLE: Compute $\lim_{n\rightarrow\infty}\sqrt[{n+1}]{(n+1)!}-\sqrt[n]{n!}$
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: 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
$$ \sqrt[n+1]{(n+1)!}-\sqrt[n]{n!} = \frac{1}{e}+O\left(\frac{\log n}{n}\right).$$