Ta xét vế trái
$\dfrac{1}{1.2} + \dfrac{1}{3.4} + \cdots + \dfrac{1}{99.100} = \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \cdots + \dfrac{1}{99} - \dfrac{1}{100}$
$= \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{99} + \dfrac{1}{100} - 2.\dfrac{1}{2} - 2. \dfrac{1}{4} - 2.\dfrac{1}{6} - \cdots - 2.\dfrac{1}{100}$
$= \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{99} + \dfrac{1}{100} - \dfrac{1}{1} - \dfrac{1}{2} - \dfrac{1}{3} - \cdots - \dfrac{1}{50}$
$= (\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{49} + \dfrac{1}{50} + \dfrac{1}{51} + \cdots + \dfrac{1}{100}) - (1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{50})$
$ = \dfrac{1}{51} + \cdots + \dfrac{1}{100}$
Biểu thức cuối cùng là vế phải.
Vậy ta đã chứng minh xong.
