# Quod Erat Demonstrandum

## 2014/02/11

### 二項極限

Filed under: NSS,Pure Mathematics — johnmayhk @ 3:16 下午
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（圖片來源：Mathematics）

“…is a recently discovered beauty, probably in 2012…"

ＯＫ，做數時間：

$\frac{s_{n-1}s_{n+1}}{s_n^2}$

$=\frac{C_0^{n-1}C_1^{n-1}C_2^{n-1}\dots C_{n-1}^{n-1}C_0^{n+1}C_1^{n+1}C_2^{n+1}\dots C_{n+1}^{n+1}}{C_0^nC_1^nC_2^n\dots C_n^nC_0^nC_1^nC_2^n\dots C_n^n}$

$=(\frac{C_0^{n-1}}{C_0^n})(\frac{C_1^{n-1}}{C_1^n})(\frac{C_2^{n-1}}{C_2^n})\dots (\frac{C_{n-1}^{n-1}}{C_{n-1}^n})\frac{1}{C_n^n}(\frac{C_0^{n+1}}{C_0^n})(\frac{C_1^{n+1}}{C_1^n})(\frac{C_2^{n+1}}{C_2^n})\dots (\frac{C_n^{n+1}}{C_n^n})C_{n+1}^{n+1}$

$(\frac{n-1}{n})(\frac{n-2}{n})\dots (\frac{2}{n})(\frac{1}{n})\times(\frac{n+1}{n})(\frac{n+1}{n-1})(\frac{n+1}{n-2})\dots (\frac{n+1}{2})(\frac{n+1}{1})$

$=\frac{(n-1)!}{n^{n-1}}\times \frac{(n+1)^n}{n!}$

$=(1+\frac{1}{n})^n$

$\displaystyle \lim_{n\rightarrow \infty}\frac{s_{n-1}s_{n+1}}{s_n^2}$

$=\displaystyle \lim_{n\rightarrow \infty}(1+\frac{1}{n})^n$

$=e$

Evaluate

$\displaystyle \lim_{n\rightarrow \infty}\sqrt[n]{\frac{C_n^{3n}}{C_n^{2n}}}$