Bài giải :

Tổng $S=\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}$ có 20 số hạng.

$S=\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}=\left( \frac{1}{21}+\frac{1}{22}+...+\frac{1}{30} \right)+\left( \frac{1}{31}+\frac{1}{32}+...+\frac{1}{40} \right)$

$S<\left( \frac{1}{20}+\frac{1}{20}+...+\frac{1}{20} \right)+\left( \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30} \right)=\frac{10}{20}+\frac{10}{30}=\frac{5}{6}$

Vậy $S<\frac{5}{6}$.