Bài giải :

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

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

$S>\left( \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30} \right)+\left( \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40} \right)=\frac{10}{30}+\frac{10}{40}=\frac{7}{12}$

Vậy $S>\frac{7}{12}$.