\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)...\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{-1}{2}\right).\dfrac{3}{2}.\left(-\dfrac{2}{3}\right).\dfrac{4}{3}...\left(\dfrac{-99}{100}\right).\dfrac{101}{100}\)
\(=\dfrac{\left(-1\right).\left(-2\right)...\left(-99\right)}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{101}{2}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}>\dfrac{100}{200}=\dfrac{1}{2}\)
Vậy...
Nhận thấy A có 99 hạng tử mà mỗi hạng tử chứa dấu âm nên viết gọn\(A=-\dfrac{3}{4}.\dfrac{8}{9}.....\dfrac{9999}{10000}=-\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}....\dfrac{99.101}{100^2}=-\dfrac{\left(1.2...99\right).\left(3.4...101\right)}{\left(2.3..100\right).\left(2.3...100\right)}=-\dfrac{101}{2.100}=-\dfrac{101}{200}< -\dfrac{1}{2}\)
