Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
...
\(\frac{1}{2020^2}<\frac{1}{2019\cdot2020}=\frac{1}{2019}-\frac{1}{2020}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2020^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{2019}-\frac{1}{2020}\)
=>\(A<1-\frac{1}{2020}\)
=>A<1