Finding $n$ such that $\frac{n+1}{n}$ < $\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}$

Question

I have been thinking about this limit:

$$\lim\limits_{n \rightarrow \infty}\frac{n}{\sqrt[n]{n!}} = e$$

Using a spreadsheet, I noticed that for $0 < n \le 150, \frac{n+1}{n} > \frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}$.

The difference $\frac{n+1}{n} - \frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}$ is strictly decreasing as $n$ increases.

I wondered if there exists an integer $k$ such that if $n \ge k$, then $\frac{n+1}{n} < \frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}$

What would be a standard way to determine if $k$ exists? And if $k$ exists, what would be a standard way to determine $k$?

I suspect that there is a simple way to approach this question without using the gamma function. Am I right? Or does the determination of $k$ require using the gamma function?

FYI, using an Approach0 search, I found the AoPS thread a limit. In message #2, it says "Proving this is less than one is equivalent to proving $\frac{n+1}{n}>\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}$", but then later "I don't know how to proceed from here". ... — John Omielan, Jul 13 '22 at 06:48
(cont.) I also found several somewhat related questions here, but they dealt with somewhat other issues like $\lim_{n\to\infty}\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}} = 1$, and the difference between the numerator & denominator, e.g., Prove elementarily that $\sqrt[n+1] {(n+1)!} - \sqrt[n] {n!}$ is strictly decreasing and What is $\lim_{n\to \infty }\left(\sqrt[\leftroot{-2}\uproot{2}n+1]{(n+1)!}-\sqrt[\leftroot{-2}\uproot{2}n]{n!}\right)$?. — John Omielan, Jul 13 '22 at 06:51

score 3 · Accepted Answer · answered Jul 13 '22 at 06:29

The answer is no. By Stirling's formula $$ \log n! = n\log n - n + \frac{1}{2}\log (2\pi n) + \mathcal{O}\!\left( {\frac{1}{n}} \right), $$ i.e., $$ \log \sqrt[n]{{n!}} = \frac{1}{n}\log n! = \log n - 1 + \frac{1}{{2n}}\log (2\pi n) + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right). $$ Consequently, \begin{align*} \log \frac{{\sqrt[{n + 1}]{{(n + 1)!}}}}{{\sqrt[n]{{n!}}}} & = \log \left( {\frac{{n + 1}}{n}} \right) + \frac{1}{{2n + 2}}\log (2\pi (n + 1)) - \frac{1}{{2n}}\log (2\pi n) + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right) \\ & = \log \left( {\frac{{n + 1}}{n}} \right) - \frac{1}{{2n^2 }}\log (2\pi n) + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right) < \log \left( {\frac{{n + 1}}{n}} \right), \end{align*} for all sufficiently large values of $n$. Using explicit error estimates, it may be shown that this inequality holds for all $n\geq 1$.

Thanks very much. This is a great example of applying Stirling's formula! :-) — Larry Freeman, Jul 13 '22 at 06:41

score 3 · Answer 2 · answered Jul 13 '22 at 23:50

We may use Mathematical Induction.

Problem: Prove that, for any positive integer $n$, $$\frac{n+1}{n} > \frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}.$$

Proof.

Taking logarithm on both sides, it suffices to prove that $$\ln n! > n(n + 1) \ln n - n^2 \ln (n + 1).$$

We use Mathematical Induction.

It is easy to verify the case $n = 1$.

Assume that it is true for $n$.

For $n+1$, by the inductive hypothesis, we have \begin{align*} \ln (n+1)! &= \ln n! + \ln(n + 1)\\ &> n(n + 1) \ln n - n^2 \ln (n + 1) + \ln(n + 1). \end{align*} It suffices to prove that $$n(n + 1) \ln n - n^2 \ln (n + 1) + \ln(n + 1) > (n + 1)(n + 2)\ln (n + 1) - (n + 1)^2\ln (n + 2)$$ or $$n\ln n + (n + 1)\ln(n + 2) > (2n + 1)\ln(n + 1)$$ or $$(n + 1)\ln\left(1 + \frac{1}{n+1}\right) > n \ln\left(1 + \frac1n\right)$$ which is true (easy).

We are done.

Finding $n$ such that $\frac{n+1}{n}$ < $\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}$

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