Xét với mọi n > 2 , ta có \(\frac{n}{n+2}< \frac{n-1}{n}\) (vì \(n^2< n^2+n-2\))
Áp dụng : \(A=\frac{1}{3}.\frac{4}{6}.\frac{7}{9}.\frac{10}{12}...\frac{208}{210}< \frac{1}{3}.\frac{3}{4}.\frac{6}{7}.\frac{9}{10}...\frac{207}{208}\)
Suy ra : \(A^2< \frac{1.4.7.10...208}{3.6.9.12...210}.\frac{1.3.6.9...207}{3.4.7.10...208}=\frac{1}{210}.\frac{1}{3}=\frac{1}{630}< \frac{1}{625}=\left(\frac{1}{25}\right)^2\)
Do đó \(A< \frac{1}{25}\)