Chứng minh 1/2^2 + 1/3^2+ 1/4^2 +... + 1/2022^2 < 1

Ta có: $\frac{1}{2^2}<\frac{1}{1.2}; \frac{1}{3^2}<\frac{1}{2.3}; \frac{1}{4^2}<\frac{1}{3.4};...; \frac{1}{{2021}^2}<\frac{1}{2020.2021}$

Đặt $\mathrm{A}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\mathrm{...}+\frac{1}{{2022}^2}$

$$\begin{gathered} \Rightarrow \mathrm{A}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\mathrm{...}+\frac{1}{{2022}^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\mathrm{...}+\frac{1}{2020.2021} \\ \Rightarrow \mathrm{A}<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\mathrm{...}+\frac{1}{2020}-\frac{1}{2021} \\ \Rightarrow \mathrm{A}<1-\frac{1}{2021} \\ \Rightarrow \mathrm{A}<\frac{2020}{2021} \end{gathered}$$

Vì 2020 < 2021 nên $\frac{2020}{2021}<1$

Do đó $\mathrm{A}<\frac{2020}{2021}<1$ .

Vậy $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\mathrm{...}+\frac{1}{{2022}^2}<1$ (đpcm).