Answer
$S_{1}=\frac{2}{3}$
$S_{2}=\frac{8}{9}$
$S_{3}=\frac{26}{27}$
$S_{4}=\frac{80}{81}$
$S_n=\frac{3^n-1}{3^n}$
Work Step by Step
We are given:
$a_{n}= \frac{2}{3^{n}}$
We find the partial sums:
$S_{1}= \frac{2}{3^1}$
$S_{2}= \frac{2}{3^1}+\frac{2}{3^{2}}=\frac{8}{9}$
$S_{3}= \frac{2}{3^1}+\frac{2}{3^{2}}+\frac{2}{3^{3}}=\frac{26}{27}$
$S_{4}= \frac{2}{3^1}+\frac{2}{3^{2}}+\frac{2}{3^{3}}+\frac{2}{3^{4}}=\frac{80}{81}$
We notice that the partial sums have powers of $3$ in the denominator and $1$ value less in the numerator. Therefore:
$S_n=\frac{3^n-1}{3^n}$
