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?
 A: 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 $\displaystyle \frac{\frac{2}{3}}{1 - (-\frac{1}{3})} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{1}{2}$.
A: Your common ratio should be 
$$\frac{a_{n+1}}{a_n}=\frac{-2/9}{2/3}=\frac{-1}{3}.$$
(You missed the sign of the successive term).
Now compute your answer.
A: $$
\frac23 - \frac29 + \frac2{27} - \frac2{81} + \dots
= 
\frac23\left(1 + (-\frac13) + (-\frac13)^2 + (-\frac13)^3 + ...\right)
$$
Now just use the formula:
$$1 + x + x^2 + x^3 + \dots = \frac1{1-x}$$
A: $$\frac{2}{3}-\frac{2}{9}+\frac{2}{27}-\frac{2}{81}+...$$
$$\frac{2}{3}+\frac{2}{27}+\cdots - (\frac{2}{9}+\frac{2}{81}+\cdots)$$
$$2(\dfrac{\frac{1}{3}}{1-\frac{1}{9}}) - 2(\dfrac{\frac{1}{9}}{1-\frac{1}{9}})$$
$$2.(\frac{3}{8}) - 2(\frac{1}{8})$$
$$\frac{1}{2}$$
A: 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. 
A: Let $S=2/3-2/9+2/27-2/81+\cdots$
Let this be equation 1
Multiply, both sides by $\large\frac{-1}{3}$
$\large \frac{-S}{3}=-2/9+2/27-2/81+\cdots$
Let this be equation 2
Subtract 2 from 1
$\large {S+\frac{S}{3}=\frac {2}{3}}$
$\large  \frac {4S}{3}=\large \frac{2}{3}$
Or, $S=\large \frac{1}{2}$
