Đặt $A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{1009^2}$
Ta có: $A<\dfrac{1}{4}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{1008.1009}$
`=>` $A<\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1008}-\dfrac{1}{1009}$
`=>`$A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{1009}=\dfrac{1}{4}+\dfrac{2}{4}-\dfrac{1}{1009}=\dfrac{3}{4}-\dfrac{1}{1009}$
`=>`$A<\dfrac{3}{4}$
Vậy $\dfrac{1}{1+3}+\dfrac{1}{1+3+5}++...+\dfrac{1}{1+3+...+2017}<\dfrac{3}{4}$
