# 一个形式较精细的 Strling 公式的证明

Strling公式$$n!=\sqrt{2\pi n}\left(\dfrac{n}{\mathrm{e}}\right)^n\mathrm{e}^{\frac{\theta_n}{12n}},$$其中$$\theta_n\in\left(\dfrac{n}{n+1},1\right)$$是一个与$$n$$有关的变量。

$f'(t)=\dfrac1{1+t}-\dfrac1{1-t}-2-2t^3=\dfrac{2t^4}{1-t^2}>0\quad(0<|t|<1)$

$$f(t)$$单增，而又有$$f(t)=0$$，故可见$$f(t)>0$$$$0<|t|<1$$时恒成立，而这正是要证的不等式的左半部分。同样的道理，设$$g(t)=\ln\dfrac{1+t}{1-t}-2t-\dfrac23\cdot\dfrac{t^3}{1-t^2}$$，可验证得到

$g'(t)=\dfrac1{1+t}-\dfrac1{1-t}-2-\dfrac23\cdot\dfrac{3t^2(1-t^2)-t^3\cdot(-2t)}{(1-t^2)^2}=-\dfrac43\dfrac{t^4}{(1-t^2)^2}<0\quad(0<|t|<1)$

$\dfrac1{12}\left(\dfrac1{n+1}-\dfrac1{n+2}\right)<\alpha_n-\alpha_{n+1}<\dfrac1{12}\left(\dfrac1n-\dfrac1{n+1}\right).$

$\alpha_n-\alpha_{n+1}=\left(n+\dfrac12\right)\ln\dfrac{n+1}n-1=\left.\left(\dfrac1{2x}\ln\dfrac{1+x}{1-x}-1\right)\right|_{x=\frac1{2n+1}}$

$\alpha_n-\alpha_{n+1}>\left.{\dfrac{x^2}3}\right|_{x=\frac1{2n+1}}=\dfrac12\left(\dfrac1{2n+1}\right)^2=\dfrac13\dfrac1{4n^2+4n+1}>\dfrac13\dfrac1{4(n^2+3n+2)}=\dfrac1{12}\left(\dfrac1{n+1}-\dfrac1{n+2}\right).$

$\alpha_n-\alpha_{n+1}<\left.{\dfrac13\dfrac{x^2}{1-x^2}}\right|_{x=\frac1{2n+1}}=\dfrac13\dfrac1{(2n+1)^2-1}=\dfrac1{12}\left(\dfrac1n-\dfrac1{n+1}\right)$

$\begin{cases} a_n-a_{n+1}=(\alpha_n-\alpha_{n+1})-\dfrac1{12}\left(\dfrac1n-\dfrac1{n+1}\right)<0\\ b_n-b_{n+1}=(\alpha_n-\alpha_{n+1})-\dfrac1{12}\left(\dfrac1{n+1}-\dfrac1{n+2}\right)>0 \end{cases}$

Strling公式$$n!=\sqrt{2\pi n}\left(\dfrac{n}{\mathrm{e}}\right)^n\mathrm{e}^{\frac{\theta_n}{12n}},$$其中$$\theta_n\in\left(\dfrac{n}{n+1},1\right)$$是一个与$$n$$有关的变量。

$\begin{cases} A_n=\mathrm{e}^{a_n}=\dfrac{n!}{\sqrt n}\left(\dfrac{\mathrm{e}}n\right)^n\mathrm{e}^{-\frac1{12n}}\\ B_n=\mathrm{e}^{b_n}=\dfrac{n!}{\sqrt n}\left(\dfrac{\mathrm{e}}n\right)^n\mathrm{e}^{-\frac1{12(n+1)}} \end{cases}$

$n!=A\sqrt{n!}\left(\dfrac{n}{\mathrm{e}}\right)^n\mathrm{e}^{\frac{\theta_n}{12n}}.$

$\dfrac\pi2=\lim\limits_{n\to\infty}\left(\dfrac{(2n)!!}{(2n-1)!!}\right)^2\dfrac1{2n+1}=\lim\limits_{n\to\infty}\left(\dfrac{2^{2n}(n!)^2}{(2n)!}\right)^2\dfrac1{2n+1}$

$\lim\limits_{n\to\infty}\left[\dfrac{\left(2^{2n}\cdot A\sqrt n\left(\dfrac{n}{\mathrm{e}}\right)^n\mathrm{e}^{\frac{\theta_n}{12n}}\right)^2}{A\sqrt{2n}\left(\dfrac{2n}{\mathrm{e}}\right)^2n\cdot\mathrm{e}^{\frac{\theta_{2n}}{24n}}}\right]^2\dfrac1{2n+1}=\dfrac\pi2.$

$\lim\limits_{n\to\infty}\dfrac{nA^2}{2}\dfrac{\mathrm{e}^{\frac{\theta_n}{6n}}}{\mathrm{e}^{\frac{\theta_{2n}}{12n}}}\cdot\dfrac1{2n+1}=\dfrac{A^2}4=\dfrac\pi2$

$n!=\sqrt{2\pi n}\left(\dfrac{n}{\mathrm{e}}\right)^n\mathrm{e}^{\frac{\theta_n}{12n}}$

posted @ 2018-07-30 15:45  黑山雁  阅读(2042)  评论(0编辑  收藏  举报