$u_n=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{n(n+2)}$
$2u_n=1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n}-\dfrac{1}{n+2}$
$=1+\dfrac{1}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}$
$=\dfrac{3}{2}+\dfrac{-n-2-n-1}{(n+1)(n+2)}$
$=\dfrac{3}{2}+\dfrac{-2n-3}{(n+1)(n+2)}$
$=\dfrac{3(n^2+3n+2)-2n-3}{2(n+1)(n+2)}$
$=\dfrac{3n^2+7n+3}{2(n+1)(n+2)}$
$\Leftrightarrow u_n=\dfrac{3n^2+7n+3}{4(n+1)(n+2)}$
$=\dfrac{3n^2+7n+3}{4n^2+12n+8}>0$
$\Rightarrow (u_n)$ bị chặn dưới
$\lim u_n=\lim\dfrac{3n^2+7n+3}{4n^2+12n+8}$
$=\lim\dfrac{3+\dfrac{7}{n}+\dfrac{3}{n^2}}{4+\dfrac{12}{n}+\dfrac{8}{n^2}}$
$=\dfrac{3}{4}$
$\Rightarrow u_n<\dfrac{3}{4}$
$\Rightarrow (u_n)$ bị chặn trên
Vậy $(u_n)$ là dãy bị chặn
