Smells Like Something's Stirling

The function $\large f(x)$ is defined as $\large\sqrt[x]{x!}$ .

Now, using Stirling's approximation that $\large n! \sim \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n$ , we rewrite the above function as $\large\begin{aligned}f(x) &\sim \sqrt[x]{\sqrt{2\pi x}{\left(\frac{x}{e}\right)}^x}\\ &=\sqrt[2x]{2\pi x}\frac{x}{e}\end{aligned}$

Then, $\large\frac{f(x)}{x} \sim \frac{\sqrt[2x]{2\pi x}}{e}$

As $\large x\to \infty$ , $\large\frac{f(x)}{x}=\frac{1}{e}$

I used Riemann's sum. $\frac{f(x)}{x}=(\frac1x\times\frac2x\times\frac3x\times...\times\frac xx)^{\frac1x}$ $ln(\frac{f(x)}{x})=\frac1x[ln(\frac1x)+ln(\frac2x)+ln(\frac3x)+...+\ln(\frac xx)]$ $\lim_{x\to\infty}ln(\frac{f(x)}{x})=\int_{0}^{1}lnx=1$ $\lim_{x\to\infty}\frac{f(x)}{x}=e$

Suppose one looked at the natural log of this limit. One could rewrite the limit as: $\lim_{x \to \infty} (ln(x!)-xln(x))/x$ Using L’Hopital’s rule and the fact that the derivative of the natural logarithm of the gamma function gives you the digamma function. $\lim_{x \to \infty} (ψ(x+1)-ln(x)-1)/1$ Since $ψ(x+1)=-\gamma+H_x$ , where $H_x$ is the xth Harmonic number and $\gamma$ is the Euler Mascheroni constant, making the substitution yields: $\lim_{x \to \infty} (-\gamma+H_x-ln(x)-1)/1=lim_{x \to \infty} (-\gamma+\gamma-1)/1=-1$ Exponentiating both sides will give us our answer: $exp(-1)$ .