What is the sum of the following infinite series?

It seems you have

$$\begin{equation}\begin{aligned} & \frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729} + \ldots \\ & = \left(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \frac{1}{243} + \frac{1}{729} + \ldots\right) + \left(\frac{1}{9} + \frac{1}{81} + \frac{1}{729} + \ldots\right) \\ & = \sum_{i=1}^{\infty}\left(\frac{1}{3}\right)^i + \sum_{i=1}^{\infty}\left(\frac{1}{9}\right)^i \\ & = \frac{\frac{1}{3}}{1 - \frac{1}{3}} + \frac{\frac{1}{9}}{1 - \frac{1}{9}} \\ & = \frac{\frac{1}{3}}{\frac{2}{3}} + \frac{\frac{1}{9}}{\frac{8}{9}} \\ & = \frac{1}{2} + \frac{1}{8} \\ & = \frac{5}{8} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Note I was able to split the sum into $2$ parts in the second line due to the series being absolutely convergent, with details about this in the Rearrangements and unconditional convergence section. Also note I used, such as described in Geometric series, that for $|r| \lt 1$, you have

$$\sum_{i=0}^{\infty}ar^i = \frac{a}{1 - r} \tag{2}\label{eq2A}$$

$$9\left(\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729}+\cdots\right)=3+2+\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \cdots$$

so that

$$9S=5+S.$$

Denote by $S$ the value of the infinite sum:

$$S=\frac13+\frac29+\frac1{27}+\frac2{81}+\frac1{243}+\frac2{729}+\cdots$$

Some rearranging of terms lets us write

$$S=\frac23-\frac19+\frac2{27}-\frac1{81}+\frac2{243}-\frac1{729}+\cdots$$

That is, $\frac29=\frac39-\frac19=\frac13-\frac19$, and so on.

Adding these sums together gives

$$\begin{align*} 2S&=1+\frac29+\frac2{81}+\frac2{729}+\cdots\\[1ex] S&=\frac12+\sum_{n\ge1}\frac1{9^n}\\[1ex] &=\frac58 \end{align*}$$