-Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{2023^2}< \dfrac{1}{2022.2023}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2022.2023}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2022}-\dfrac{1}{2023}=1-\dfrac{1}{2023}=\dfrac{2022}{2023}\)\(\Rightarrow0< \dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1\) nên \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}\) không là một số tự nhiên.
\(A=\dfrac{3}{2^2}+\dfrac{8}{3^2}+\dfrac{15}{4^2}+...+\dfrac{2023^2-1}{2023^2}=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{2023^2}=\left(1+1+1+...+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2023^2}\right)\)
\(\Rightarrow\)A không là một số tự nhiên.
