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

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 $1/e$.

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