$$2/3-2/9+2/27-2/81+\cdots$$
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?
Best Answer
Hint: the terms are alternating in sign. $\displaystyle r = \frac{\frac{2}{9}}{-\frac{2}{3}} = -\frac{1}{3}$.
Note the minus sign.
Hence the sum is $\displaystyle \frac{\frac{2}{3}}{1 - (-\frac{1}{3})} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{1}{2}$.