ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1\)
mà \(\frac{1}{1^2};\frac{1}{2^2};\frac{1}{3^2};...;\frac{1}{n^2}>0\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}>0\)
\(\Rightarrow0< \frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\)không phải số tự nhiên
\(\Rightarrow A=\frac{1}{1^2}+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\) là hỗn số
\(\Rightarrow A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\) không phải số tự nhiên ( đ p c m)
