Answer
$\dfrac{1}{2}$
Work Step by Step
Here $a_n=\dfrac{1}{(n+1)(n+2)}$
or, it can be re-written as: $a_n=\dfrac{1}{(n+1)}-\dfrac{1}{(n+2)}$
The Nth partial sum of a geometric series can be found as:
$s_n=\dfrac{a(1-r^n)}{1-r}$
The sum of a geometric series can be found as:
$S=\dfrac{a}{1-r}$
Thus, $s_n= (\dfrac{1}{2}-\dfrac{1}{3})+(\dfrac{1}{3}-\dfrac{1}{4})+....(\dfrac{1}{(n+1)}-\dfrac{1}{(n+2)})$
and $\lim\limits_{n \to \infty} s_n=\dfrac{1}{2}$
