B = \(1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}\right)\)
Xét \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}=1-\frac{1}{2016}=\frac{2015}{2016}\)
Do đó B > 1 - \(\frac{2015}{2016}=\frac{1}{2016}\)