Ta để ý rằng
$\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$
Áp dụng vào ta có
$\dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} + \cdots + \dfrac{1}{A.(A+1)}$
$= \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \cdots + \dfrac{1}{A} - \dfrac{1}{A+1}$
$= 1-\dfrac{1}{A+1} = \dfrac{A}{A+1}$