Base 3

Let $x = 0.\overline{2}_3$ . Then $3x = 2.\overline{2}_3$ . This means that $2x = 3x - x = 2.\overline{2}_3 - 0.\overline{2}_3 = 2$ , and solving $2x = 2$ gives $x = \boxed{1}$ .

$.\overline{2}_3 = \frac{2}{3} + \frac{2}{9} + ... = 2\sum_{n=1}^\infty\frac{1}{3^n} = 1$