I was asked to determine whether $\frac{1}{3}+ \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729} +...$ converges or diverges and, should it converge, to show what it converges to. To do this, I took the approach of partitioning the series into the sum of two geometric sequences. Since I am very new to infinite series (started learning this today), I was wondering if my approach is correct.
$\alpha )$ Let $a_n = \frac{1}{3^{2n-1}}= \{\frac{1}{3}, \frac{1}{27}, \frac{1}{243},... \}$ and $b_n = \frac{2}{3^{2n}} = \{\frac{2}{9}, \frac{2}{81}, \frac{2}{729}, ... \}$. Then, due to the commutative property of addition,
$$\frac{1}{3}+ \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729} +... = \sum_{n=1}^\infty a_n + \sum_{n=1}^\infty b_n$$
$\beta)$ Notice that $a_n= \frac{1}{3^{2n-1}}=\frac{1}{3}(\frac{1}{9})^{n-1}$ and $b_n=\frac{2}{3^{2n}} = \frac{2}{9}(\frac{1}{9})^{n-1}$.
$\gamma)$ Then,
$$\frac{1}{3}+ \frac{2}{9} + \frac{1}{27} + \frac{2}{81} + \frac{1}{243} + \frac{2}{729} +... \overset{\text{due to }\alpha, \beta}{=} \sum_{n=1}^\infty \frac{1}{3}(\frac{1}{9})^{n-1} + \sum_{n=1}^\infty \frac{2}{9}(\frac{1}{9})^{n-1}$$
$$\overset{\text{due to g.s.}}{=} \frac{\frac{1}{3}}{1-\frac{1}{9}} + \frac{\frac{2}{9}}{1-\frac{1}{9}} = \frac{5}{8}$$
where $\text{g.s.}$ is the following theorem regarding geometric series: $\sum_{n=1}^\infty a r^{n-1}=\frac{a}{1-r}$ for $r \in (-1, 1)$.
Is this solution correct? Thank you.