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?

$$
\frac23 - \frac29 + \frac2{27} - \frac2{81} + \dots
= 
\frac23\left(1 + (-\frac13) + (-\frac13)^2 + (-\frac13)^3 + ...\right)
$$
Now just use the formula:
$$1 + x + x^2 + x^3 + \dots = \frac1{1-x}$$
