Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{2021^2}< \dfrac{1}{2020.2021}\)
Cộng vế với vế ta đc
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2021^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2020}-\dfrac{1}{2021}=1-\dfrac{1}{2021}=\dfrac{2020}{2021}< 1\)
Vậy ta có đpcm