Đặt \(N=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
ta có: \(M.N=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}=\frac{1}{101}\)
ta có: \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow M=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< N=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
\(\Rightarrow M.M< M.N\)
\(\Rightarrow M^2< \frac{1}{101}< \frac{1}{100}=\left(\frac{1}{10}\right)^2\)
\(\Leftrightarrow M^2< \left(\frac{1}{10}\right)^2\)
\(\Rightarrow M< \frac{1}{10}\left(đpcm\right)\)
