How do I find the sum of the infinite geometric series? $$2/3-2/9+2/27-2/81+\cdots$$
The formula is $$\mathrm{sum}= \frac{A_g}{1-r}\,.$$
To find the ratio, I did the following:
$$r=\frac29\Big/\frac23$$
Then got:
$$\frac29 \cdot \frac32= \frac13=r$$
and $$A_g= \frac23$$
Then I plug it all in and get:
$$\begin{align*}
\mathrm{sum} &= \frac23 \Big/ \left(1-\frac13\right)\\
             &= \frac23 \Big/ \left(\frac33-\frac13\right)\\
             &= \frac23 \Big/ \frac23\\
             &= \frac23 \cdot \frac32\\
             &= 1\,.
\end{align*}$$
But the real answer is $\frac12$.
What did I do wrong?

$$\frac{2}{3}-\frac{2}{9}+\frac{2}{27}-\frac{2}{81}+...$$
$$\frac{2}{3}+\frac{2}{27}+\cdots - (\frac{2}{9}+\frac{2}{81}+\cdots)$$
$$2(\dfrac{\frac{1}{3}}{1-\frac{1}{9}}) - 2(\dfrac{\frac{1}{9}}{1-\frac{1}{9}})$$
$$2.(\frac{3}{8}) - 2(\frac{1}{8})$$
$$\frac{1}{2}$$
