Is $x(x!)^{1/x}$ an increasing function of $x$, for $x > 0$?

Question

Is $x(x!)^{1/x}$ an increasing function of $x$, for $x > 0$?

Here $x!$ is the factorial of $x$.

Sure, I do know differential calculus, but my problem is that I do not know how to compute for the derivative of $x!$.

Lastly, I asked WolframAlpha to plot the function $$f(x)=x(x!)^{1/x}$$ and it does appear to be increasing.

Is it possible to prove this fact rigorously? Thanks!

Answer 1

$$\log f(x) = \log x + \frac{1}{x}\log\Gamma(x+1) $$ and $\log x$ is increasing, hence it is enough to show that $\frac{1}{x}\log\Gamma(x+1)$ is increasing. However:

$$\frac{1}{x}\log\Gamma(x+1) = \frac{\log\Gamma(x+1)-\log\Gamma(0+1)}{x-0} $$ is increasing since $\log\Gamma$ is convex by the Bohr-Mollerup theorem.

Answer 2

If $x \in \mathbb Z^+$, the proof is simpler. Let $f(x) = (x!)^\frac{1}{x}$, for $x \in \mathbb Z^+$. Then, $f(x)>0$ and $\displaystyle \Big(\frac{f(x+1)}{f(x)}\Big)^{x(x+1)}= \frac{((x+1)!)^x}{(x!)^{x+1}}= \frac{(x!)^{x}(x+1)^x}{(x!)^{x+1}} = \frac{(x+1)^x}{x!} =\prod_{i = 1}^{x} \frac{x+1}{i} > 1$, since it's the non-empty product of reals > 1. Thus, $\displaystyle \frac{f(x+1)}{f(x)}> 1$, since $x(x+1) > 0$. Hence, $f(x)$ is strictly increasing. Therefore, $xf(x)$ is strictly increasing.