Ta có
$S = \dfrac{5}{2^2} + \dfrac{5}{3^2} + \cdots + \dfrac{5}{100^2}$
$> \dfrac{5}{2.3} + \dfrac{5}{3.4} + \cdots + \dfrac{ 5}{100.101}$
$ = \dfrac{5}{2} - \dfrac{5}{3} + \dfrac{5}{3} - \dfrac{5}{4} + \cdots + \dfrac{5}{100} - \dfrac{5}{101}$
$= \dfrac{5}{2} - \dfrac{5}{101}$
$> \dfrac{5}{2} - \dfrac{1}{2} = 2$
Vậy $S > 2$.
Mặt khác, ta lại có
$S = \dfrac{5}{2^2} + \dfrac{5}{3^2} + \cdots + \dfrac{5}{100^2}$
$< \dfrac{5}{1.2} + \dfrac{5}{2.3} + \cdots + \dfrac{5}{99.100}$
$= \dfrac{5}{1} - \dfrac{5}{2} + \dfrac{5}{2} - \dfrac{5}{3} + \cdots + \dfrac{5}{99} - \dfrac{5}{100}$
$= 5 - \dfrac{5}{100} < 5$
Vậy $S < 5$
Suy ra $2 < S < 5$.
