Dễ dàng chứng minh được:
$\dfrac{1}{1.2.3} +\dfrac{1}{2.3.4} +\dfrac{1}{3.4.5} +\dots +\dfrac{1}{(n-1)n(n+1)}=\dfrac{1}{4} -\dfrac{1}{2n(n+1)}$
Đặt $S =\dfrac{1}{2.3^2} +\dfrac{1}{3.4^2} +\dfrac{1}{4.5^2} +\dots +\dfrac{1}{n(n+1)^2}$
Ta có:
$\quad \begin{cases}\dfrac{1}{2.3^2} <\dfrac{1}{1.2.3}\\\dfrac{1}{3.4^2}<\dfrac{1}{2.3.4}\\\cdots\\\dfrac{1}{n(n+1)^2}<\dfrac{1}{(n-1)n(n+1)}\end{cases}\quad \forall n\geq 2$
$\to S < \dfrac{1}{1.2.3} +\dfrac{1}{2.3.4} +\dfrac{1}{3.4.5} +\dots +\dfrac{1}{(n-1)n(n+1)}$
$\to S < \dfrac{1}{4} -\dfrac{1}{2n(n+1)}$
$\to S < \dfrac14\quad \forall n \geq 2$
