## University Calculus: Early Transcendentals (3rd Edition)

$5$
Here $a_n=\dfrac{5}{n(n+1)}$ or, it can be re-written as: $a_n=\dfrac{5}{n}-\dfrac{5}{n+1}$ 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, $s_n= (5-\dfrac{5}{2})-(\dfrac{5}{2}-\dfrac{5}{3})+....(\dfrac{5}{n}-\dfrac{5}{n+1})=5-\dfrac{5}{n+1}$ and $\lim\limits_{n \to \infty} s_n=5$