D2:
\(a=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}>1\)
\(a=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+..+\dfrac{1}{n^2}< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)
=> \(a< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
=> \(a< 1+1-\dfrac{1}{n}=2-\dfrac{1}{n}< 2\)
=> 1 < a < 2
=> a không phải số tự nhiên
-Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
\(\Rightarrow1< 1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< `+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=2-\dfrac{1}{n}< 2\)\(\Rightarrow1< a< 2\) nên a không phải là số tự nhiên.
