連乘不等式

分數連乘積不等式

A = $\Large\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{97}{98} \cdot\frac{99}{100}$

B = $\Large\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\ldots\cdot\frac{96}{97} \cdot\frac{98}{99}$

AB = $\Large\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\ldots\cdot\frac{96}{97} \cdot\frac{97}{98}\cdot\frac{98}{99} \cdot\frac{99}{100}$ = $\Large\frac{1}{100}$

因為 AA < AB，所以 AA < $\Large\frac{1}{100}$，因此 A < $\Large\frac{1}{10}$，即

$\Large\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{97}{98} \cdot\frac{99}{100}$ < $\Large\frac{1}{10}$

其他想法

A = $\Large\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{97}{98} \cdot\frac{99}{100}$

AA = $\Large(\frac{1}{2})^2\cdot(\frac{3}{4})^2\cdot(\frac{5}{6})^2\cdot\ldots\cdot(\frac{97}{98})^2 \cdot(\frac{99}{100})^2$

< $\Large(\frac{1^2}{2^2-1})\cdot(\frac{3^2}{4^2-1})\cdot(\frac{5^2}{6^2-1})\cdot\ldots\cdot(\frac{97^2}{98^2-1}) \cdot(\frac{99^2}{100^2-1})$

=$\Large\frac{1}{3}\cdot\frac{3}{5}\cdot\frac{5}{7}\cdot\ldots\cdot\frac{97}{99} \cdot\frac{99}{101}$ = $\Large\frac{1}{101} $< $\Large\frac{1}{100}$

因此 A < $\Large\frac{1}{10}$，即

$\Large\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{97}{98} \cdot\frac{99}{100}$ < $\Large\frac{1}{10}$

可以推廣得

$\Large\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\ldots\cdot\frac{2n-1}{2n}$ < $\Large\frac{1}{\sqrt{2n}}$，n是自然數。