Computation of a limit involving factorial $\lim_{n \to \infty} \sqrt[n+1] {(n+1)!} - \sqrt[n] {(n)!} = \frac{1}{e}$

Question

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.

Answer 1

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.

Answer 2

Hint: Set $p_n = n!$.

Then what is $$\lim_{n \rightarrow \infty}\frac{p_{n+1}}{np_n}?$$

Can you take it from here?

Answer 3

By Stolz-Cesaro theorem , we have

$$\lim_{n\to\infty}\frac{\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}}1=\lim_{n\to\infty}\frac{\sqrt[n]{n!}}n=\lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^n}}$$

and by the root-to-ratio limit, we also have

$$\lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^n}}=\lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\times\frac{n^n}{n!}=\lim_{n\to\infty}\frac1{\left(1+\frac1n\right)^n}$$

which, by the limit definition of $e$, is given by

$$\lim_{n\to\infty}\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}=\frac1e$$