Computation of a limit involving factorial $\lim_{n \to \infty} \sqrt[n+1] {(n+1)!} - \sqrt[n] {(n)!} = \frac{1}{e}$ I want to prove the following limit:
$$\lim_{n \to \infty} \sqrt[n+1\;] {(n+1)!} - \sqrt[n] {(n)!} = \frac{1}{e}.$$
I searched the forum & found the link here: If $\frac{p_{n+1}}{np_n} \to p > 0 $, then $\sqrt[n+1]{p_{n+1}}-\sqrt[n]{p_{n}} \to \frac{p}{e}$ .
But still, there is no way out of the problem. So, please solve it.

One way is to write the expression $a(n) = e^{\log a_n}$, then you can use the bonds on $\int_{1}^{n} \log x dx < \sum_{k=1}^{n} \log k < 1+ \int_{1}^{n} \log x dx$ to get the asymptotic result. 
