Cho

- Hướng dẫn giải

$\begin{array}{l}A=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+\mathrm{...}+\frac{1}{200}\\ =\left(\frac{1}{101}+\frac{1}{102}+\mathrm{...}+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+\mathrm{...}+\frac{1}{200}\right)\\ >\frac{1}{150}.50+\frac{1}{200}.50=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\end{array}$