有关$n^n/n!$

hbghlyj

維基百科說

`{\displaystyle {\sqrt {|D|}}\geq \left({\frac {\pi }{4}}\right)^{r_{2}}{\frac {n^{n}}{n!}}\geq \left({\frac {\pi }{4}}\right)^{n/2}{\frac {n^{n}}{n!}}\ .}`
For n at least 2, it is easy to show that the lower bound is greater than 1

如何證明它說的$$\forall n\geqslant2,\ \left({\frac {\pi }{4}}\right)^{n/2}{\frac {n^{n}}{n!}}>1$$呢？

hbghlyj

注意到
\[
\frac{(n+1)^{n+1}/(n+1)!}{n^n/n!}=\left(1+\frac{1}{n}\right)^n>\sqrt\frac4\pi,
\]
递归即可.

hbghlyj

D = Δ_K = discriminant of a field K over $\mathbb {Q}$.
从上面知道|D|>1
|D|最小是多少？是3，惟一$\mathbb{Q}(\sqrt{-3})$，见math.stackexchange.com/questions/820990/is-ev … gebraic-number-field

hbghlyj

平方得，$\left({\frac {\pi }{4}}\right)^{n}(\frac {n^n}{n!})^2>1$
$\iff\left({\frac {\pi }{4}}\right)^{n}\prod_{j=1}^{n}\frac {n}{j}\prod_{j=1}^{n}\frac {n}{j}>1$
$\iff\left({\frac {\pi }{4}}\right)^{n}\frac{n^2}{n^2}\prod_{j=1}^{n-1}\frac {n}{j}\prod_{j=1}^{n-1}\frac {n}{n-j}>1$
$\iff\left({\frac {\pi }{4}}\right)^{n}\prod_{j=1}^{n-1}\frac {n}{j}\frac {n}{n-j}>1$
$\iff\frac{\pi ^{n}}4\prod_{j=1}^{n-1}\frac {n/2}{j}\frac {n/2}{n-j}>1$
因为$\frac n2\frac n2\ge j(n-j)$，所以每项都>1，证毕