<s>[INST] What is the sum of the following infinite series? $$
\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729} + \cdots
$$
So basically I separated it into two series
where:
one of them is $\left(\frac{1}{3}\right)^n$
And I use geometric series formula to find that this series equals $\frac{1}{2}$.
But I can't figure out the series of the other one.
Apparently the answer for the series combined is: $\frac{5}{8}$
What is the other series?
 [/INST] 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
$$\sum_{i=0}^{\infty}ar^i = \frac{a}{1 - r} \tag{2}\label{eq2A}$$
</s>