# Evaluating the factorial-related limit $\lim_{x \to \infty} (x + 1)!^{1 / (x + 1)} - x!^{1/x}$

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.

• Yeah, Dr. MV has a point. We can use that to approximate $n!$ at infinity. – SalmonKiller May 3 '15 at 3:58
• @Dr.MV Comment I was referring to is gone. – apnorton May 3 '15 at 4:15

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 $$\lim_{n \to \infty} \left[ [(x + 1)!]^{1 / (x + 1)} - (x!)^{1 / x} \right] = \frac{1}{e}.$$

• Same as mine, though DeMoivre is sufficient than Stirling. – Adhvaitha May 3 '15 at 4:08
• I'm not sure what you mean by "sufficient than". Anyway, the approximation $C \sqrt{x} \left(\frac{x}{e}\right)^x + O\left(\frac{1}{x}\right)$ you mention is exactly the approximation used in this solution (except that it doesn't use the particular value of $C$, which turns out to be irrelevant), so I don't see how "DeMoivre" is any different. – Travis Willse May 3 '15 at 4:16
• How do you make this rigorous? Isn't $(c n^{-1})^\frac{1}{n}\rightarrow 1$ ? – simonzack May 3 '15 at 4:24
• @simonzack What's important is the behavior of $\left(\sqrt{2 \pi x} \left(\frac{x}{e}\right)^x + c\frac{1}{x}\right)^{1 / x}$, not the limit of the $x$th root of the error term alone. – Travis Willse May 3 '15 at 4:33

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}

• I've never seen that credited to DeMoivre before; typically, I've just seen $C$ taken to be $\sqrt{2\pi}$ and then had it called "Stirling's formula." – apnorton May 3 '15 at 4:15

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 $1/e$).

• Same as mine, though DeMoivre is sufficient than Stirling. – Adhvaitha May 3 '15 at 4:08