Giải:
Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2017^2}\)
Dễ thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(....\)
\(\dfrac{1}{2017^2}=\dfrac{1}{2017.2017}< \dfrac{1}{2016.2017}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2016.2017}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2016}-\dfrac{1}{2017}\)
\(\Rightarrow A< 1-\dfrac{1}{2017}< 2\)
\(\Rightarrow A< 2\left(dpcm\right)\)