How can I improve my proof of Stirling's Theorem?

Question

I'm trying to prove Robbin's inequality: $$ n! \le \sqrt{2 \pi n}(n/e)^n e^{1/(12n)}. $$ Step 1: I start from the integral formulation \begin{align} n! = \int_0^\infty x^n e^{-x} dx &= (n/e)^n\int_0^\infty (x/n)^n e^{n-x} dx \\&= (n/e)^n\int_{-n}^\infty (1+x/n)^n e^{-x} dx \\&= (n/e)^n\int_{-n}^\infty \exp[-x + n\log(1+x/n)] dx \\&= (n/e)^n n \int_{-1}^\infty \exp[-x n + n\log(1+x)] dx. \end{align} Step 2: I want to use Laplace's method, so I pull out Taylor's theorem $$ \log(1+x)=x - x^2/2 + \frac{2}{(1 + \xi_x)^3}x^3/6, $$ where $\xi_x \in [-x,x]$. (Here we used $\frac{d^3\log(1+x)}{dx^3}=\frac{2}{(1+x)^3}$.)

Step 3: I split the integral into segments: $[-1, 0]$, $[0, L]$ and $[L, \infty)$. In the first interval I guess I should use $\xi_x=-1$, but actually we know $\log(1+x)\le x - x^2/2$, so we don't really need Taylor's theorem. (Good, since we would have had a zero divisor.) In the interval $[0,L]$ we set $\xi_x=0$ since $x^3>0$. Finally in $[L,\infty)$ in use a simple linear bound: $$\log(1+x) \le \log(1+L) + \frac{x-L}{1+L}$$ since $\log(1+x)$ is a concave function. (We have $f(x) \le f(L) + (x-L)f'(L)$ for any concave function and constant $L$.)

Step 4: I now do the integrals:

1. Integral one from $-1$ to $0$: \begin{align} \int_{-1}^0 \exp[-x n + n\log(1+x)] dx &\le \int_{-1}^0 \exp[-nx^2/2] dx \end{align}

2. Integral two from $0$ to $L$: \begin{align} \int_{0}^L \exp[-x n + n\log(1+x)] dx &\le \int_{0}^L \exp[-nx^2/2 + nL^3/3] dx \\&= e^{nL^3/3} \int_{0}^L \exp[-nx^2/2] dx \end{align}

3. Integral from $L$ to $\infty$: \begin{align} \int_{L}^\infty \exp[-x n + n\log(1+x)] dx &\le\int_{L}^\infty \exp[n \log(1+L) - n(1+x)L/(1+L)] dx \\&\le(1+L)^n \int_{L}^\infty \exp[- n(1+x)L/(1+L)] dx \\&=(1+L)^n e^{-L n} \frac{1+L}{L n} \\&\le \exp(-n L^2/2 + n L^3/3) \frac{1+L}{L n}. \end{align}

Step 5: Combine the integrals. First combining (1) and (2): \begin{align} \int_{-1}^0 \exp[-nx^2/2] dx +e^{nL^3/3} \int_{0}^L \exp[-nx^2/2] dx &\le e^{nL^3/3} \int_{-1}^L \exp[-nx^2/2] dx \\&\le e^{nL^3/3} \int_{-\infty}^\infty \exp[-nx^2/2] dx \\&= e^{nL^3/3} \sqrt{2\pi/n}. \end{align}

Then combining (1) and (2) with (3): \begin{align} \int_{-1}^\infty \exp[-x n + n\log(1+x)] dx \le e^{nL^3/3}\left(\sqrt{2\pi/n} + \exp(-n L^2/2) \frac{1+L}{L n}\right) \end{align}

Step 6: Choose $L$.

I choose $L=n^{-2/3}$. Putting it all together we have shown $$ n! \le (n/e)^n \left( \sqrt{2\pi n} + (1+n^{2/3})e^{-n^{1/3}/2}\right) e^{1/(3n)}. $$

That is nearly the $(n/e)^n \sqrt{2\pi n} e^{1/(12n)}$ bound we wanted, but it is off by a factor $1/4$ in the error exponent, and we have that extra annoying $o(1)$ term added onto $\sqrt{2\pi n}$.

Question: How can I improve my proof? I would like to get the full strength of Robbins's inequality, ideally with similar methods. If there are any tricks I can use to make the derivation easier, I'm also very interested.

Answer 1

The approach that you take is discussed on Wikipedia and yields $n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n(1+\frac{1}{12n}+o(\frac{1}{n}))$, which is not the desired bound by Robbins. While Laplace's method is very useful, I think that Robbins' bound is too sharp for this approach.

I thought about presenting the proof, which follows the other approach discussed on Wikipedia. But since the original paper is freely accessible and excellently written, there is no need to repeat the argument. Also, there are tons of additional resources, for example a very thorough discussion can be found here.

Answer 2

I tried to redo the computations with as few approximations as possible, and then optimize the constants numerically.

In particular, I took the $e^{-n x^2/2}$ integral from $-1$ to $1$, the $e^{-n x^2/2 + n L^3/3}$ from $0$ to $L$, and I generalized the linear bound in the tail to use an arbitrary constant, $a$.

I then plotted the error, $r_n$, when $n!\approx (n/e)^n \sqrt{2\pi n} e^{r_n}$:

It seems like, while I'm definitely getting $r_n=o(1)$, even tuning every parameter as much as I can, there's no way to get to $r_n\sim 1/(12n)$, at least for small $n$.

I wrote this update as an answer, since that's what it may be: "There's no way to get Robbin's inequality with Laplace's method."

However, I'm definitely very interested in being told otherwise.