I'm looking for the limit
$$\lim_{x \to \infty} \left[[(x+1)!]^\frac{1}{1+x} - (x!)^\frac{1}{x}\right].$$
I've put the above in a computer program, and evaluated it at very high values of $x$ (at $x = 100\text{ }000$, it is approximately $0.367881$). The value seems to be caving in to $1/e$, which is $0.3678794412\ldots$
That makes sense, as $e$ has an expansion related to factorial. However I'm stuck trying to figure out a proof if there is any. Thanks for any help.
By Stirling's Approximation, we have $$x! \approx \sqrt{2 \pi x} \left(\frac{x}{e}\right)^x,$$ and so for large $n$ $$(x!)^{1 / x} \approx (2 \pi)^{1 / 2x} x^{1 / x} \frac{x}{e}.$$ Thus, $$[(x + 1)!]^{1 / (x + 1)} - (x!)^{1 / x} \approx (2 \pi)^{1 / 2(x + 1)} (x + 1)^{1 / (x + 1)} \frac{(x + 1)}{e} - (2 \pi)^{1 / 2x} x^{1 / x} \frac{x}{e}.$$ Now, $(2 \pi)^{1 / 2x} x^{1 / x} \to 1$ as $x \to \infty$, so we have that $$[(x + 1)!]^{1 / (x + 1)} - (x!)^{1 / x} \approx \frac{x + 1}{e} - \frac{x}{e} = \frac{1}{e}.$$
Stirling's approximation has error $O\left(\frac{1}{x}\right)$ (and in fact is very good even for modestly small $x$), and hence we can translate the above asymptotic approximations to prove rigorously the desired limit $$\lim_{n \to \infty} \left[ [(x + 1)!]^{1 / (x + 1)} - (x!)^{1 / x} \right] = \frac{1}{e}.$$
From DeMoivre's formula, we have $$x! \sim C \sqrt{x}\left(\dfrac{x}e\right)^x + \mathcal{O}(1/x)$$ This means we have \begin{align} ((x+1)!)^{1/(x+1)} - (x!)^{1/x} & \sim C^{1/(x+1)} \sqrt{x}^{1/(x+1)} \left(\dfrac{x+1}e \right) - C^{1/x} \sqrt{x}^{1/x} \left(\dfrac{x}e\right) + \mathcal{O}(1/x)\\ & \sim \dfrac{x+1-x}e = \dfrac1e \end{align}
If $x$ is an integer, we can use Sterling's formula
$$x! \approx. \sqrt{2\pi x} \left(\frac{x}{e}\right)^x$$
Thus,
$$((x+1)!)^{1/(x+1)}\approx. (\sqrt{2\pi (x+1)})^{1/(x+1)}\left(\frac{x+1}{e}\right)$$
and
$$(x!)^{1/x}\approx. (\sqrt{2\pi x})^{1/x}\left(\frac{x}{e}\right)$$
The term $\sqrt{2\pi x}^{1/x}$ and $\sqrt{2\pi (x+1)}^{1/(x+1)}$ can easily be shown to approach $1$ as $x \to \infty$. Thus,
$$((x+1)!)^{1/(x+1)}-(x!)^{1/x}\approx. \frac{x+1}{e}-\frac{x}{e} = 1/e$$
as $x\to \infty$
Aside, we do not need to assume that $x$ is an integer. Indeed, the Gamma function provides an extension of the factorial for complex values. Using the large argument asymptotic expansion of the Gamma function with positive real arguments, we have
$$x!=\Gamma(x+1)=\sqrt{2\pi x}\left(\frac{x}{e}\right)^x \left(1+O\left(\frac{1}{x}\right)\right)$$
Analysis proceeds identically and the result of the post is unaffected (i.e., the limit of the function of this post is $1/e$).
