TITLE: Stirling's Series Derivation
QUESTION [8 upvotes]: I was reading this paper and at the top of page 9 it says that as $n\to\infty$, $$\left(1+\frac{1}{n}\right)^{n+1/2}e^{-1}\left(1+\frac{a_1}{(n+1)}+\frac{a_2}{(n+1)^2}+\cdots \right)=1+\frac{a_1}{n}+\frac {a_2-a_1+1/12}{n^2}+\frac{(13/12) a_1-2a_2+a_3-1/12}{n^3}...$$ I just do not understand where the $1/12,~13/12,~\text{etc.}$ come from, so can anybody enlighten me?
And something else: 
If I have shown the exact relation $n!= \sqrt{n}(n/e)^n e^{1-E(n)}$ where $$E(n)=\sum\limits_{k=1}^{n-1} \biggl[\left(\frac {2k+1}{2}\right)\ln \left(\frac{k+1}{k}\right)-1\biggl]$$ that after some working gets to $$\sum\limits_{k=1}^{n-1} \biggl[\left(\frac{1}{3(2k+1)^2}+\frac{1}{5(2k+1)^4}+
\cdots\right)\biggl]<\frac{n-1}{12n},$$ can I use this in any way to derive Stirling's series (the series, not the first term)?
I know that I can derive the series using the Euler-Maclaurin formula but I want this for an essay and I am well off the word limit to introduce a new result. 
Thank you.

REPLY [3 votes]: Stirling formula can be derived using steepest descent.
\[  n! = \int_0^\infty t^n e^{-t} \, dt = \int_0^\infty e^{-t + n \ln t} \, dt\]
The exponent $f(t) = -t + n \ln t$ has critical point at $t = n$ since $f'(t) = -1 + n/t$
\[ f(t) = -n + n \ln n - \frac{1}{2n} (t-n)^2 + O(t-n)^3 \]
This second order expansion means we can approximate the factorial function using a Gaussian integral:
\[ n! \simeq \int_0^\infty e^{-n + n \ln n - \frac{1}{2n} (t-n)^2 } \, dt = e^{-n + n \ln n} \sqrt{2\pi n}\]
If you're more careful, you can get all the terms this way. Nobody really has a good explanation for it, though.

In general, if $x_0$ is a critical point of $f$ then Laplace's method gives
\[ \int_a^b e^{M f(x)} \, dx \approx \sqrt{\frac{ 2\pi  }{ M | f''(x_0)|}}e^{M f(x_0)}  \]