$$
2\frac{1}{3} + \frac{1}{3^2} + 2\frac{1}{3^3} + \frac{1}{3^4} + 2\frac{1}{3^5} + \cdots = \frac{1}{3}\sum_{n = 0}^{\infty}\left(\frac{1}{3}\right)^n
$$
I am stuck on what to add in to account for the 2 at every other term.
The first parenthesis is a geometric series with ratio $\frac{1}{3}$ and the second with ratio $\frac{1}{9}$.
This doesn't seem right though, because the sum would be $\frac{7}{8}$, whereas on the right side you'd get $\frac{1}{2}$. Perhaps this change of terms in the series I did is wrong, but I doubt that because the series on the right, which it is supposed to be, is absolutely convergent, being a geometric series with ratio $\frac{1}{3}$.
I hope this helps spring the discussion into the right direction.