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

$s_n=3(1-\dfrac{1}{3^n})$ and series sum $=3$
Consider the series, $2+\dfrac{2}{3}+\dfrac{2}{9}+\dfrac{2}{27}+...$ The given sequence shows a geometric series with common ratio, $r=\dfrac{1}{3}$ The Nth partial sum of a geometric series can be found as: $s_n=\dfrac{a(1-r^n)}{1-r}=\dfrac{2(1-(\dfrac{1}{3})^n)}{1-\dfrac{1}{3}}=3(1-\dfrac{1}{3^n})$ The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}=\dfrac{2}{1-\dfrac{1}{3}}=3$ Hence, $s_n=3(1-\dfrac{1}{3^n})$ and series sum $=3$