Ta có: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{1}{5}.4=\frac{4}{5}\)
\(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}< \frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}=\frac{1}{10}.4=\frac{2}{5}\)
\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}< \frac{1}{15}+\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\frac{1}{15}.4=\frac{4}{15}\)
\(\frac{1}{17}+\frac{1}{18}< \frac{1}{17}.2=\frac{2}{17}\)
\(\Rightarrow B< \frac{4}{5}+\frac{2}{5}+\frac{4}{15}+\frac{2}{17}=\frac{404}{255}=1\frac{149}{255}< 2\)