\(\frac{2}{3^2}+\frac{2}{4^2}+\frac{2}{5^2}+......+\frac{2}{2014^2}\)
=\(2.\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2014^2}\right)<2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.........+\frac{1}{2013.2014}\right)\)
<\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.......+\frac{1}{2013}-\frac{1}{2014}\right)\)
< \(2.\left(\frac{1}{2}-\frac{1}{2014}\right)\)
<\(\frac{2012}{2014}\)
Ma \(\frac{2012}{2014}<1\)
\(\Rightarrow\frac{2}{3^2}+\frac{2}{4^2}+\frac{2}{5^2}+........+\frac{2}{2014^2}<1\)
Có: \(A=\frac{2}{3^2}+\frac{2}{4^2}+......+\frac{2}{2014^2}\)
Chuyển đổi: \(A=2.\left(\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{2014}\right)=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{2013.2014}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.......+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2014}\right)=2.\frac{503}{1007}=\frac{1006}{1007}\)
Mà: \(\frac{1006}{1007}=\)\(\frac{2}{3^2}+\frac{2}{4^2}+......+\frac{2}{2014^2}\)
Mà \(\frac{1006}{1007}<1\) nên \(A<1\)
Vậy \(\frac{2}{3^2}+\frac{2}{4^2}+\frac{2}{5^2}+...........+\frac{2}{2014^2}<1\)
