A=\(\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.....\frac{2016^2-1}{2016^2}\)
A=\(\frac{\left(2+1\right)\left(2-1\right)}{2^2}.\frac{\left(3+1\right)\left(3-1\right)}{3^2}......\frac{\left(2016+1\right)\left(2016-1\right)}{2016^2}\)
A=\(\frac{3.4......2017}{2.3....2016}.\frac{1.2...2015}{2.3...2016}\)
A=\(\frac{2017}{2}.\frac{1}{2016}\)
A=\(\frac{2017}{2.2106}>\frac{1}{2}\)
Vậy A\(>\frac{1}{2}\)