Ta có: $\frac{1}{k(k+1)}=\frac{k+1-k}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$.
Như vậy: $\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{(n-1)n}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}$
$=1-\frac{1}{n}$.
$lim(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{(n-1)n})$
$=lim(1-\frac{1}{n})$
$=1-0=1$