Đặt A = \(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{9998}{9999}.\frac{10000}{10000}\)

Rõ ràng A < A'

=> A² < A . A' \(=\frac{1}{10000}=\frac{1}{100^2}\)

Nên A < 0,01

So sánh

ta có:

A > 0,01 

Vì 1/2<2/3;3/4<4/5;5/6<6/7;....;9999/10000<10000/10001

-->S<2/3.4/5.5/6.....10000/10001

Gọi 2/3.4/5.5/6.....10000/10001 là D và D>S

Có D=2/3.4/5.5/6.....10000/10001

-->S.D=(1/2 . 3/4 . 5/6.....9999/10000).(2/3.4/5.5/6.....10000/10001)

S.D=1/2.2/3.3/4.4/5.5/6.6.7.....9999/10000.10000/10001

S.D=1/10001

Vì S<D nên S.S<S.D hay S.S<1/10001

-->S<1/10001

mà 1/10001<1/100=0.01

-->S<1/100=0.01

-->S<0.01

Vậy

Có \(\frac{1}{2}< \frac{2}{3};\frac{2}{3}< \frac{3}{4};.....;\frac{9999}{10000}< \frac{10000}{10001}\)

\(\Rightarrow S^2< \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)

\(\Rightarrow S^2< \frac{1}{10001}< \frac{1}{10000}=\left(\frac{1}{100}\right)^2\)

\(\Rightarrow S< \frac{1}{1000}< 0,01\)

Vậy S<0,01