Bài 3
Ta có
$D = \dfrac{1}{2^2} + \dfrac{1}{3^2} + \cdots + \dfrac{1}{10^2}$
$< \dfrac{1}{1.2} + \dfrac{1}{2.3} + \cdots + \dfrac{1}{9.10}$
$= 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \cdots + \dfrac{1}{9} - \dfrac{1}{10}$
$= 1 - \dfrac{1}{10} < 1$
Vậy $D < 1$.