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?
 A: $$9\left(\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729}+\cdots\right)=3+2+\frac{1}{3} + \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \cdots$$
so that
$$9S=5+S.$$
A: 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
$$\begin{align*}
2S&=1+\frac29+\frac2{81}+\frac2{729}+\cdots\\[1ex]
S&=\frac12+\sum_{n\ge1}\frac1{9^n}\\[1ex]
&=\frac58
\end{align*}$$
A: Here is a trick that I like to use:
Recall that if you wanted to find the value of the infinite repeating decimal $0.\overline{12}$, you would get $\frac{12}{99}$. The $99$ in the denominator comes from the fact that we use a base $10$ number system, and $10^2-1=99$.
Create the decimal in base $3$, and you get $\frac{12_3}{3^2-1}=\frac{5}{8}$.
A: Note that when you group adjacent terms, $\frac{1}{3} + \frac{2}{9} = \frac{5}{9}$. Therefore, the sum of the series becomes:
$$\frac{5}{9} + \frac{1}{9} \cdot \frac{5}{9} + \left(\frac{1}{9} \right)^2 \cdot \frac{5}{9} + \cdots$$
$$= \frac{\frac{5}{9}}{1 - \frac{1}{9}}$$
$$= \frac{5/9}{8/9} = \frac{5}{8}$$
A: 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
$$\sum_{i=0}^{\infty}ar^i = \frac{a}{1 - r} \tag{2}\label{eq2A}$$
