Answer
$s_n=3(1-\dfrac{1}{3^n})$ and series sum $=3$
Work Step by Step
Consider the series, $ 2+\dfrac{2}{3}+\dfrac{2}{9}+\dfrac{2}{27}+...$
This shows a geometric series.
Here, common ratio, $r=\dfrac{1}{3}$
Formula for Nth partial sum of a geometric series is
$s_n=\dfrac{a(1-r^n)}{1-r}$
or, $s_n=\dfrac{2(1-(\dfrac{1}{3})^n)}{1-\dfrac{1}{3}}$
or, $s_n=3(1-\dfrac{1}{3^n})$
Formula for sum of a geometric series is
$S=\dfrac{a}{1-r}=\dfrac{2}{1-(\dfrac{1}{3})}=3$
Hence, $s_n=3(1-\dfrac{1}{3^n})$ and series sum $=3$
