Another formula for $n$-th root of $n$ factorial, approximate or exact?

Question

I was playing around with graphing formulas on Desmos, and I stumbled upon $\sqrt[n]{n!}$, which when graphed outputted a line that seemed to be perfectly linear, but wasn't. Is there another exact (or approximate) formula that could be used to fit the values outputted by $\sqrt[n]{n!}$, since it resembles very closely a linear equation? And if so, out of curiosity, is there one that doesn't use the factorial?

Answer 1

From the known asymptotic expansion of $\log n!$, we obtain $$ \frac{1}{n}\log n! \sim \log n - 1 + \frac{{\log (2\pi n)}}{{2n}} + \frac{1}{{12n^2 }} - \frac{1}{{360n^4 }} + \ldots\, . $$ Taking the exponential of each side and expanding the exponential using its power series gives \begin{align*} \sqrt[n]{{n!}} & \sim \frac{n}{\mathrm{e}}\exp \left( {\frac{{\log (2\pi n)}}{{2n}} + \frac{1}{{12n^2 }} - \frac{1}{{360n^4 }} + \ldots } \right) \\ & = \frac{n}{\mathrm{e}}\left( {1 + \frac{{\log (2\pi n)}}{{2n}} + \frac{{3\log ^2 (2\pi n) + 2}}{{24n^2 }} + \frac{{\log ^3 (2\pi n) + 2\log (2\pi n)}}{{48n^3 }} + \ldots } \right), \end{align*} as $n\to +\infty$. At leading order $$ \sqrt[n]{{n!}} \sim \frac{n}{\mathrm{e}}, $$ confirming your observation.