$\begin{array}{l}3)\\S=\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+\dots+\dfrac1{(n-1)^2}+\dfrac1{n^2}\\\quad<\dfrac1{1.2}+\dfrac1{2.3}+\dfrac1{3.4}+\dots+\dfrac1{(n-2)(n-1)}+\dfrac1{(n-1)n}\\\quad=1-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+\dots+\dfrac1{n-1}-\dfrac1n\\\quad=1-\dfrac1n<1\\\text{- Vậy $S<1$}\\\,\\4)\\S=\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+\dots+\dfrac1{9^2}\\\quad>\dfrac1{2.3}+\dfrac1{3.4}+\dfrac1{4.5}+\dots+\dfrac1{9.10}\\\quad=\dfrac12-\dfrac13+\dfrac13-\dfrac14+\dfrac14-\dfrac15+\dots+\dfrac19-\dfrac1{10}\\\quad=\dfrac12-\dfrac1{10}\\\quad=\dfrac25\\\to S>\dfrac25\quad(1)\\S=\dfrac1{2^2}+\dfrac1{3^2}+\dfrac1{4^2}+\dots+\dfrac1{9^2}\\\quad<\dfrac1{1.2}+\dfrac1{2.3}+\dfrac1{3.4}+\dots+\dfrac1{8.9}\\\quad=1-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+\dots+\dfrac18-\dfrac19\\\quad=1-\dfrac19\\\quad=\dfrac89\\\to S<\dfrac89\quad(2)\\\text{- Từ (1) và (2) $\to \dfrac25<S<\dfrac89$} \end{array}$
