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.

Solution 1:

Hint: Set $p_n = n!$.

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

Can you take it from here?

Solution 2:

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.

Solution 3:

It appears (after looking at several other answers posted here) that it is difficult to avoid the use of Stirling's approximation. Stirling's approximation is given by $$\log n! = n\log\left(\frac{n}{e}\right) + \frac{1}{2}\log (2\pi n) + \frac{1}{12n} + O(n^{-2})$$ and hence $$\log (n!)^{1/n} = \log\left(\frac{n}{e}\right) + \frac{\log (2\pi n)}{2n} + O(n^{-2})$$ and therefore $$(n!)^{1/n} = \exp\left(\log\left(\frac{n}{e}\right) + \frac{\log (2\pi n)}{2n} + O(n^{-2})\right) = \frac{n}{e}\exp\left(\frac{\log (2\pi n)}{2n} + O(n^{-2})\right)$$ and thus we get $$(n!)^{1/n} = \frac{n}{e}\left(1 + \frac{\log (2\pi n)}{2n} + O(n^{-2}(\log n)^{2})\right)$$ or $$(n!)^{1/n} = \frac{n}{e} + \frac{\log(2\pi n)}{2e} + O(n^{-1}(\log n)^{2})$$ and then using the above formula for $((n + 1)!)^{1/(n + 1)}$ and on subtracting we get $$\sqrt[n + 1]{(n + 1)!} - \sqrt[n]{n!} = \frac{1}{e} + \frac{1}{2e}\log\left(\frac{n + 1}{n}\right) + O(n^{-1}(\log n)^{2}) $$ and the result follows by taking limit as $n \to \infty$.

I later checked the linked question and found that the result in that question crucially depends on the limit in this question and hence the problem in current question is the more fundamental one and its genuine answer can not be based on the answer of the linked question. Also the linked question has an excellent answer by Daniel Fischer which uses the Stirling's approximation to obtain a limit which is same as the one in current question.