Find the limit of: $\lim_{n\to\infty} \frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}}$ Could be the following limit computed without using Stirling's approximation formula?
$$\lim_{n\to\infty} \frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}}$$
I know that the limit is $e$, but I'm looking for some alternative ways that doesn't require to resort
 to the use of Stirling's approximation. I really appreciate any support at this limit. Thanks. 
 A: The Stolz–Cesàro theorem implies that if the limit exists, then it is equal to $\lim\limits_{n\to\infty}\dfrac{n}{\sqrt[n]{n!}}$.  Some ways to evaluate the latter limit, including a method that uses the Stolz–Cesàro theorem again, are included in the answers to the question Finding the limit of $\frac {n}{\sqrt[n]{n!}}$.
This leaves existence of the original limit to be proved.  
A: This completes Jonas's answer, here is an idea. This is too long for a comment.
To prove that the limit exists, we can prove that $a_n$ is decreasing and positive:
$$\frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}} \geq 0 \Leftrightarrow $$
$$\sqrt[n+1]{(n+1)!} \geq \sqrt[n]{(n)!} \Leftrightarrow $$
$$(n+1)!^n\geq (n)!^{n+1} \Leftrightarrow $$
$$(n+1)^n\geq (n)! \Leftrightarrow $$
$$(n+1)\cdot(n+1)...\cdot(n+1)\geq 1\cdot 2..\cot n \checkmark $$
Now for decreasing
$$\frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}} \geq \frac{1}{\sqrt[n+2]{(n+2)!} - \sqrt[n+1]{(n+1)!}} \Leftrightarrow $$
$$\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!} \leq \sqrt[n+2]{(n+2)!} - \sqrt[n+1]{(n+1)!} \Leftrightarrow $$
$$2\sqrt[n+1]{(n+1)!}  \leq \sqrt[n]{(n)!}+ \sqrt[n+2]{(n+2)!}$$
Now, by AM-GM inequality
$$\frac{\sqrt[n]{(n)!}+ \sqrt[n+2]{(n+2)!}}{2} \geq (n!)^\frac{1}{2n}[(n+2)!]^\frac{1}{2n+1}$$
So if we can prove that 
$$(n!)^\frac{1}{2n}[(n+2)!]^\frac{1}{2n+4} \geq \sqrt[n+1]{(n+1)!}$$
we are done.
Now
$$(n!)^\frac{1}{2n}[(n+2)!]^\frac{1}{2n+4} \geq \sqrt[n+1]{(n+1)!} \Leftrightarrow $$
$$(n!)^\frac{1}{2n}[(n+2)]^\frac{1}{2n+4} \geq (n+1)!^{\frac{1}{n+1}-\frac{1}{2n+4}} \Leftrightarrow $$
$$(n!)^{\frac{1}{2n}+\frac{1}{2n+4}-\frac{1}{n+1}}[(n+2)]^\frac{1}{2n+4} \geq (n+1)^{\frac{1}{n+1}-\frac{1}{2n+4}}\,. $$
To keep it simple:
The power of $n!$ is
$$\frac{(2n^2+6n+4)+(2n^2+2n)-(4n^2+8n)}{(n+1)2n(2n+4)}=\frac{2}{n(n+1)(2n+4)}$$
The power of $n+1$ is
$$\frac{1}{n+1}-\frac{1}{2n+4}=\frac{n+3}{(n+1)(2n+4)}$$
Thus, after bringing the inequality to the $n(n+1)(2n+4)$, it becomes:
$$(n!)^\frac{1}{2n}[(n+2)!]^\frac{1}{2n+4} \geq \sqrt[n+1]{(n+1)!} \Leftrightarrow $$
$$(n!)^2(n+2)^{n(n+1)} \geq (n+1)^{n(n+3)} $$
Now, I ma not sure that this is true, but might work....
