Cho 

- Hướng dẫn giải

$$\begin{gathered} A=\frac{1}{6}+\frac{1}{7}+\mathrm{..}+\frac{1}{18}+\frac{1}{19} \\ =\left(\frac{1}{6}+\frac{1}{7}+\mathrm{...}\frac{1}{14}\right)+\left(\frac{1}{15}+\frac{1}{16}+\mathrm{...}+\frac{1}{19}\right) \end{gathered}$$

Ta có

$\frac{1}{6}+\frac{1}{7}+\mathrm{...}\frac{1}{14}<\frac{1}{6}+\frac{1}{6}+\mathrm{...}+\frac{1}{6}=\frac{9}{6}=\frac{3}{2}\left(1\right)$

$\frac{1}{15}+\frac{1}{16}+\mathrm{...}+\frac{1}{19}<\frac{1}{15}+\frac{1}{15}+\mathrm{...}+\frac{1}{15}=\frac{5}{15}=\frac{1}{3}<\frac{1}{2}\left(2\right)$

Từ (1) và (2) ta có: A<2