TITLE: What is the sum of the following infinite series?
QUESTION [6 upvotes]: $$
\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729} + \cdots
$$
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?

REPLY [3 votes]: Note that when you group adjacent terms, $\frac{1}{3} + \frac{2}{9} = \frac{5}{9}$. Therefore, the sum of the series becomes:
$$\frac{5}{9} + \frac{1}{9} \cdot \frac{5}{9} + \left(\frac{1}{9} \right)^2 \cdot \frac{5}{9} + \cdots$$
$$= \frac{\frac{5}{9}}{1 - \frac{1}{9}}$$
$$= \frac{5/9}{8/9} = \frac{5}{8}$$