# 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.

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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.

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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....

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There's another question about showing that the sequence is increasing: math.stackexchange.com/questions/238245/… – Jonas Meyer Feb 6 '13 at 21:13