Đặt A = $1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+...+\frac{1}{2021^{2}}$
A < $1+\frac{1}{2^{2}-1}+\frac{1}{3^{2}-1}+...+\frac{1}{2021^{2}-1}$ = $1+\frac{1}{(2 - 1)(2 + 1)}+\frac{1}{(3 - 1)(3 + 1)}+...+\frac{1}{(2021 - 1)(2021 + 1)}$ = $1+\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{2020.2022}$
Đặt B = $\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2019.2021}$, C = $\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2020.2022}$
B = $\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2019.2021}$
B = $\frac{1}{2}(\frac{1}{1}-\frac{1}{3})+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+...+\frac{1}{2}(\frac{1}{2019}-\frac{1}{2021})$
B = $\frac{1}{2}(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5})+...+\frac{1}{2019}-\frac{1}{2021})$
B = $\frac{1}{2}(\frac{1}{1}-\frac{1}{2021})$ = $\frac{1010}{2021}$
Tương tự, ta có: C = $\frac{505}{2022}$
Vậy A < 1 + B + C = $1+\frac{1010}{2021}+\frac{505}{2022}$ = $1+\frac{1515}{2021}$ < 1 + 1 = 2
Vậy A < 2
