Đặt \(Q=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{400}{401}\)
Áp dụng tính chất \(\frac{a}{b}< \frac{a+m}{b+m}\left(a,b,m\inℕ^∗\right)\)ta có
\(\frac{1}{2}< \frac{1+1}{2+1}=\frac{2}{3}\)
\(\frac{2}{3}< \frac{2+1}{3+1}=\frac{3}{4}\)
...
\(\frac{399}{400}< \frac{399+1}{400+1}=\frac{400}{401}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{399}{400}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{400}{401}\)
hay P < Q
=> \(P^2< P.Q\)
\(P^2< \frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{399}{400}.\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{400}{401}\)
\(P^2< \frac{1.2.3.4.....400}{2.3.4.5.....401}\)
\(P^2< \frac{1}{401}< \frac{1}{400}< \left(\frac{1}{20}\right)^2\)
Vì P và 1/20 có cùng dấu
\(\Rightarrow P< \frac{1}{20}\)
