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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.

  • 0
    See also: [Limit of the sequence $a_n=\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}$](http://math.stackexchange.com/q/2020153) and [Limit of the sequence $a_n=\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}$](http://math.stackexchange.com/q/2020153)2017-01-15

2 Answers 2

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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.

2

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....

  • 0
    There's another question about showing that the sequence is increasing: http://math.stackexchange.com/questions/238245/prove-elementarily-that-sqrtn1-n1-sqrtn-n-is-strictly-decrea?lq=12013-02-06