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\begin{document}
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%      Since this involves a lot of formulas it's written using TeX
%
%         Charles Blair [BITNET:CEBLAIR@UIUCVMD]
%         (submitted to the Monthly)
%
\par This derivation of Stirling's approximation (including 
upper and lower bound) uses infinite series for logarithms
instead of integrals.
\def\otn{\sum_1^{n-1}}
%%\def\frac#1#2{{{#1}\over{#2}}}
\def\nin{\sum_n^\infty}
\def\mess{\frac{k-9}{120k^4(k+1)}}
\def\omess{\left(\frac1{L+1}-\frac1{2L}\right)}
\begin{eqnarray*}
     \ln  n! & = & n\ln\ n-\sum_{k=1}^{n-1} k\ln\left(1+\frac1k\right)\\
        & = &
          n\ln n+\sum_{L=1}^\infty\left(\sum_{k=1}^{n-1}k(-1)^L\frac{k^{-L}}L\right)\\
       & = & n\ln\ n-(n-1)+\frac12\otn k^{-1}-\frac13\otn k^{-2}+\dots\\
\end{eqnarray*}

\par Approximate $\sum k^{-1}$ using
$$\ln\ n = \otn \ln\left(1+\frac1k\right)=\otn k^{-1}-\frac12\otn k^{-2}+\dots$$
\pagedone
 When we group according to powers of $k$ we get:
$$\ln\ n! = n\ln\ n-(n-1)+\frac12\ln\ n +\left(\frac14-\frac13\right)
       \otn k^{-2}$$
       $$+\left(-\frac16+\frac14\right)\otn k^{-3}+\dots$$
       
       
$$\hbox{Let:}\quad S=\left(n+\frac12\right)\ln n-(n-1)$$
$$M_L=\sum_{k=1}^\infty k^{-L}$$
$$M=\sum_{L=2}^\infty(-1)^L\omess M_L$$
\begin{eqnarray*}
     \ln n! & = & S-\frac1{12}\left(M_2-\nin k^{-2}\right) \\
          & &  -\sum_{L=3}^\infty(-1)^L\omess\left(M_L-\nin k^{-L}\right)\\	& = & S-M+\frac1{12}\nin k^{-2}-\frac1{12}\nin k^{-3} \\
         & &     +\frac3{40}\nin k^{-4}-\frac1{15}\nin k^{-5}+\dots \\
\end{eqnarray*}

Since:
$$ \frac1{12k^2}-\frac1{12k^3}+\frac3{40k^4}=
\frac1{12}\left(\frac1k-\frac1{k+1}\right)-\mess ,$$

we have
$$S-M+\frac1{12n}-\nin\mess $$
$$> \ln\ n! >$$
$$ S-M+\frac1{12n}-\nin\mess
-\nin\frac1{15k^5}$$
For $n\ge 9$, $\ln n!<S-M+1/12n$ is immediate.  For a lower bound,
we can use $[k\ge9]$: $$\mess+\frac1{15k^5}=\frac{k^2-k+8}{120k^5(k+1)}$$
$$ < \frac1{120k^3(k+1)}<\frac1{360}\left(\frac1{(k-1)^3}-%
\frac1{k^3}\right)$$
to obtain $\ln n!>S-M+1/(12n)-1/(360(n-1)^3)$.
\par To determine $M$, the usual argument involving Wallis' product
can be used:
\begin{eqnarray*}
\lim_{n\to\infty}\frac{4^n(n!)^2}{\sqrt{2n}(2n!)} & = & \lim\frac{2\cdot
4\cdot6\dots(2n-2)\sqrt{2n}}{3\cdot5\dots(2n-1)} \\
 & = & \sqrt{\frac\pi2} \\
 & = &  e^{1-\ln 2-M} \\
\end{eqnarray*}
 So: $e^{-M}=\frac{\sqrt{2\pi}}e$


\end{document}