Ta có
$\dfrac{1}{1.3} + \dfrac{1}{3.5} + \cdots + \dfrac{1}{(2n-1)(2n+1)} = \dfrac{1}{2} \left( 1 - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} +\cdots + \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right)$
$= \dfrac{1}{2} \left( 1 - \dfrac{1}{2n+1} \right)$
Do đó
$\lim \left( \dfrac{1}{1.3} + \dfrac{1}{3.5} + \cdots + \dfrac{1}{(2n-1)(2n+1)} \right) = \dfrac{1}{2} \lim \left( 1 - \dfrac{1}{2n+1} \right)= \dfrac{1}{2} . 1 = \dfrac{1}{2}$.
