Ta có: \(S=\frac34+\frac89+\frac{15}{16}+\ldots+\frac{n^2-1}{n^2}\)
\(=\left(1-\frac14\right)+\left(1-\frac19\right)+\cdots+\left(1-\frac{1}{n^2}\right)\)
\(=\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+\cdots+\left(1-\frac{1}{n^2}\right)\)
\(=\left(n-1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\right)\)
Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
...
\(\frac{1}{n^2}<\frac{1}{\left(n-1\right)\cdot n}=\frac{1}{n-1}-\frac{1}{n}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{n-1}-\frac{1}{n}<1\)
=>\(0<\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1\)
=>\(-1<-\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\right)<0\)
=>\(-1+\left(n-1\right)<-\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\right)+\left(n-1\right)<0+\left(n-1\right)\)
=>n-2<S<n-1
=>S không là số tự nhiên
