\(A=2\cdot\left(\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2017^2}\right)< 2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}\right)\)
Đặt \(M=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}=\left(1+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
\(\Rightarrow M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(\Rightarrow M=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}< \frac{1}{1009}+\frac{1}{1009}+...+\frac{1}{1009}\)(1008 số hạng )
hay\(M< \frac{1008}{1009}\Rightarrow A< 2\cdot\frac{1008}{1009}=\frac{504}{1009}\left(ĐPCM\right)\)
