求极限：$\displaystyle\lim_{n\to +\infty}{\dfrac{\sqrt[n]{n!}}n}$

其妙

求极限：$\displaystyle\lim_{n\to +\infty}{\dfrac{\sqrt[n]{n!}}n}$

tommywong

$\displaystyle \frac{\sqrt[n]{n!}}{n}=\frac{\sqrt[n]{e^{lnn!}}}{n}=\frac{\sqrt[n]{e^{\sum_{x=1}^n lnx}}}{n}$

$\displaystyle \int_1^n lnxdx=nlnn-n+1$

$\displaystyle nlnn-n+1 \le \sum_{x=1}^n lnx \le (n+1)lnn -n+1$

$\displaystyle \frac{\sqrt[n]{e^{nlnn-n+1}}}{n} \le \frac{\sqrt[n]{n!}}{n} \le \frac{\sqrt[n]{e^{(n+1)lnn-n+1}}}{n}$

$\displaystyle \frac{e^{lnn-1+\frac{1}{n}}}{n} \le \frac{\sqrt[n]{n!}}{n} \le \frac{e^{(1+\frac{1}{n})lnn-1+\frac{1}{n}}}{n}$

$\displaystyle e^{-1+\frac{1}{n}} \le \frac{\sqrt[n]{n!}}{n} \le e^{\frac{1}{n}lnn-1+\frac{1}{n}}$

$\displaystyle e^{-1} \le \frac{\sqrt[n]{n!}}{n} \le e^{-1}$

kuing

FAQ哇这是

\begin{align*}
\lim_{n\to +\infty }\frac{\sqrt[n]{n!}}n&=\exp \left( \lim_{n\to +\infty }\ln \sqrt[n]{\frac{n!}{n^n}} \right) \\
& =\exp \left( \lim_{n\to +\infty }\frac1n\left( \ln \frac nn+\ln \frac{n-1}n+\cdots +\ln \frac1n \right) \right) \\
& =\exp \left( \int_0^1{\ln x\rmd x} \right) \\
& =\exp \left( -1 \right).
\end{align*}

战巡

回复 1# 其妙

斯特灵公式瞬秒.......
\[n!\sim \sqrt{2\pi n}(\frac{n}{e})^n\]

其妙

，夹逼，定积分，Stirling，齐登场！
Stirling：baike.baidu.com/view/4113061.htm?fr=aladdin