Đặt \(B=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{10000}{10001}\)
Ta có: \(S=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{9999}{10000}< \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{10000}{10001}=B\)
Do đó: \(S\cdot S^2< S\cdot B\)
hay \(S^2< \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{9999}{10000}\cdot\frac{10000}{10001}=\frac{1}{10001}\)
⇔\(S^2< \frac{1}{10001}< \frac{1}{10000}=\frac{1}{100}\cdot\frac{1}{100}=\left(0,01\right)^2\)
hay S<0,01(đpcm)