ongelijkheid

Zij $A=\frac12\cdot\frac34\cdots\frac{99}{100}$ en $B=\frac23\cdot\frac45\cdots\frac{98}{99}$, zodat $AB=\frac{1}{100}$.

• Enerzijds is $\frac23<\frac34, \frac45<\frac56, \cdots , \frac{98}{99}<\frac{99}{100}$ dus $2A>B$.
• Anderzijds is $\frac12<\frac23,\frac34<\frac45, \cdots, \frac{99}{100}<1$ dus $A

Dus is $\frac{1}{200}