Đặt \(S=\frac{1}{2^2}+\frac{1}{3^2}+........+\frac{1}{2014^2}\)
Đặt A=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{2014.2015}\)
\(A=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+.....\left(\frac{1}{2014}-\frac{1}{2015}\right)\\ =>A=\frac{1}{2}-\frac{1}{2015}\\ =>A=\frac{2013}{4030}\)
Mà S>A =>S>\(\frac{2013}{4030}\)