Finding $n$ such that $\frac{n+1}{n}$ < $\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}$ 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?
 A: 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 $n\geq 1$.
A: 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.
