Answer
Convergent with sum $\frac{5}{8}$
Work Step by Step
$\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{2}{81}+\frac{1}{243}+\frac{2}{729}+...$
$= \Sigma^{\infty}_{n=1} \frac{1}{3^{2n-1}} + 2\Sigma^{\infty}_{n=1} \frac{1}{9^{n}}$
The series $\Sigma^{\infty}_{n=1} \frac{1}{3^{2n-1}}$ is a geometric series with $a=\frac{1}{3}$ and $r=\frac{1}{9}$.
$\Sigma^{\infty}_{n=1} \frac{1}{3^{2n-1}}= \frac{\frac{1}{3}}{1-\frac{1}{9}}$
$= \frac{3}{8}$
The series $\Sigma^{\infty}_{n=1} \frac{1}{9^{n}}$ is a geometric series with $a=\frac{1}{9}$ and $r=\frac{1}{9}$
$\Sigma^{\infty}_{n=1} \frac{1}{9^{n}}= \frac{\frac{1}{9}}{1-\frac{1}{9}}$
$=\frac{1}{8}$
Hence the series is convergent
$\Sigma^{\infty}_{n=1} \frac{1}{3^{2n-1}} + 2\Sigma^{\infty}_{n=1} \frac{1}{9^{n}}=\frac{3}{8}+2(\frac{1}{8})$
$=\frac{5}{8}$
