Stirling 逼近, 2023 天津压轴

Czhang271828

题: 求证
\[
\dfrac{5}{6}<\ln (n!)-\left(n+\dfrac{1}{2}\right)\ln n+n\leq 1.
\]

---------------------------原题---------------------------

$\mathrm{(20)}\quad $ 已知函数 $f(x)=\left(\dfrac{1}{x}+\dfrac{1}{2}\right)\ln (x+1)$.

• 求曲线 $y=f(x)$ 在 $x=2$ 处切线的斜率;
• 当 $x>0$ 时, 证明: $f(x)>1$;
• 证明: $\dfrac{5}{6}<\ln (n!)-\left(n+\dfrac{1}{2}\right)\ln n+n\leq 1$.

---------------------------Stirling 逼近---------------------------

常用的 Stirling 逼近:
\[
\sqrt{2\pi n}\cdot \left(\frac ne\right)^{n}e^{\frac {1}{12n+1}}< n!< {\sqrt{2\pi n}} \left(\frac ne\right)^{n}e^{\frac{1}{12n}}.
\]

　

天津卷给出的较为宽泛的逼近:
\[
e^{5/6}\sqrt n\cdot \left(\dfrac{n}{e}\right)^n< n!< e\sqrt n\cdot \left(\dfrac{n}{e}\right)^n.
\]

又: 数学分析原理 (Rudin) 中类似题目(P200-T20)

20. The following simple computation yields a good approximation to Stirling's formula.

For $m=1,2,3, \ldots$, define
$$
f(x)=(m+1-x) \log m+(x-m) \log (m+1)
$$
if $m \leq x \leq m+1$, and define
$$
g(x)=\frac{x}{m}-1+\log m
$$
if $m-\frac{1}{2} \leq x<m+\frac{1}{2}$. Draw the graphs of $f$ and $g$. Note that $f(x) \leq \log x \leq g(x)$ if $x \geq 1$ and that
$$
\int_1^n f(x) d x=\log (n!)-\frac{1}{2} \log n>-\frac{1}{8}+\int_1^n g(x) d x
$$
Integrate $\log x$ over $[1, n]$. Conclude that
$$
\frac{7}{8}<\log (n!)-\left(n+\frac{1}{2}\right) \log n+n<1
$$
for $n=2,3,4, \ldots $(Note: $\log \sqrt{2 \pi} \sim 0.918 \ldots$) Thus
$$
e^{7 / 8}<\frac{n!}{(n / e)^n \sqrt{n}}<e
$$

isee

天津卷快封魔了（如果只是尾巴翘，可以接受

kuing

hbghlyj 发表于 2025-3-19 21:13
链接失效了

估计是这个 PDF：
david92jackson.neocities.org/images/Principle … l_Analysis-Rudin.pdf