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.