<s>[INST] using a base 3 decimal to express as a base 10 fraction using geometric series 
Express $0.\overline{21}_3$ as a base 10 fraction in reduced form.

So I was able to solve it by setting $x=\overline{.21}$, but the solution also briefly mentioned another way using the geometric series:

A quick way to get the answer by using the geometric series is: $(0.212121 \ldots)_3 = \frac{7}{9} + \frac{7}{81} + \frac{7}{729} + \dots = \frac{7}{8}.$

However, I'm having a hard time understanding how to actually use the geometric series (the above answer is not clear to me). 
 [/INST] If you are not sure on how to compute $(0.2121\ldots)_{3}$, we have: $$\begin{align} E = (0.\color{red}{2}\color{green}{1}\color{red}{2}\color{green}{1}\ldots)_3 = \frac{1}{3}\times \color{red}{2} + \frac{1}{3^2}\times \color{green}{1} + \frac{1}{3^3}\times \color{red}{2} + \frac{1}{3^4}\times \color{green}{1} + \ldots \\ = \frac{2}{3} + \frac{1}{9} + \frac{2}{27} + \frac{1}{81} + \ldots \\ = \frac{7}{9}+ \frac{7}{81}+\ldots \end{align}$$

Now note that: $$\begin{align} E = \frac{7}{9}+\frac{7}{81}+\frac{7}{729}+\ldots \\ =\frac{7}{9}+\frac{7}{9}\times \frac{1}{9} + \frac{7}{9}\times [\frac{1}{9}]^2\ldots \\ =\frac79(1+\frac19 + [\frac19]^2+\ldots)\end{align}$$
Now, this is an infinite geometric progression. Can you compute the required sum?
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