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?

If we are willing to be a little too casual (OK, much too casual), let the sum be $S$. Then 
$$S=\frac{2}{3}-\frac{2}{9}+\frac{2}{27}-\frac{2}{81}+\cdots.\tag{1}$$
Multiply through by $3$. We get 
$$3S=2-\frac{2}{3}+\frac{2}{9}-\frac{2}{27}+\cdots.\tag{2}$$
Add, and note the almost total cancellation. We get
$$4S=2.$$
Remark: Note that for the series $1-3+9-27+\cdots$ the procedure yields the nonsensical (?) sum $\frac{3}{4}$. But the procedure yields the correct answer whenever the original series converges. 
