A là tích của 99 số âm.Do đó :
\(-A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{100^2}\right)\)
\(=\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{9999}{100^2}=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\frac{3\cdot5}{4^2}\cdot...\cdot\frac{99\cdot101}{100^2}\)
\(-A=\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot99\cdot100}\cdot\frac{3\cdot4\cdot5\cdot....\cdot101}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\cdot\frac{101}{2}=\frac{101}{200}>\frac{1}{2}\)
Do đó : \(A< B\)
