Đặt \(A=1-\frac{1}{2^2}-\frac{1}{3^2}-.........-\frac{1}{2020^2}\)
Ta có: \(2^2=2.2< 2.3\)\(\Rightarrow\frac{1}{2.2}>\frac{1}{2.3}\)\(\Rightarrow\frac{1}{2^2}>\frac{1}{2.3}\)
Tương tự, ta có: \(\frac{1}{3^2}>\frac{1}{3.4}\), ........... , \(\frac{1}{2020^2}>\frac{1}{2020.2021}\)
\(\Rightarrow A>1-\frac{1}{2.3}-\frac{1}{3.4}-...........-\frac{1}{2020.2021}\)
\(=1-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)-.......-\left(\frac{1}{2020}-\frac{1}{2021}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-..........-\frac{1}{2020}+\frac{1}{2021}\)
\(=1-\frac{1}{2}+\frac{1}{2021}\)\(=\frac{1}{2}+\frac{1}{2021}=\frac{2023}{4042}>\frac{1}{2020}\)
\(\Rightarrow A>\frac{1}{2020}\)
