\(\dfrac{2}{3^2}+\dfrac{2}{4^2}+\dfrac{2}{5^2}+....\dfrac{2}{2016^2}\)
Ta thấy: \(\dfrac{2}{3^2}< \dfrac{2}{2.3}\)
\(\dfrac{2}{4^2}< \dfrac{2}{3.4}\)
...\(\dfrac{2}{2016^2}< \dfrac{2}{2015.2016}\)
Đặt:A=\(\dfrac{2}{3^2}+\dfrac{2}{4^2}+\dfrac{2}{5^2}+...+\dfrac{2}{2016^2}\)
=>\(A< \dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{2015.2016}\)
=>\(A< \dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{5}+...+\dfrac{2}{2015}-\dfrac{2}{2016}\)
=>A<\(\dfrac{2}{2}-\dfrac{2}{2016}\)
=>A<\(\dfrac{1007}{1008}\) mà \(\dfrac{1007}{1008}\) < 1
=>A<1
Vậy \(\dfrac{2}{3^2}+\dfrac{2}{4^2}+\dfrac{2}{5^2}+...+\dfrac{2}{2016^2}\)<1 (\(đpcm\))
\(\dfrac{2}{3^2}+\dfrac{2}{4^2}+...+\dfrac{2}{2016^2}=2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)\)
Ta có: \(\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{2016^2}< \dfrac{1}{2015.2016}\)
\(\Rightarrow2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)< 2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}\right)\)
\(\Rightarrow2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)< 2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(\Rightarrow2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)< 2\left(\dfrac{1}{2}-\dfrac{1}{2017}\right)=1-\dfrac{2}{2017}< 1\)
=> đpcm
