\(0.\overline{1}_b = \\ \sum \limits_{k=1}^\infty \dfrac{1}{b^k} \\ \text{let $u = \dfrac 1 b$}\\ \sum \limits_{k=1}^\infty \dfrac{1}{b^k} = \sum \limits_{k=1}^\infty u^k = \\ \dfrac{u}{1-u} = \\ \dfrac{\frac 1 b}{1-\frac 1 b} = \\ \dfrac{1}{b-1} \)
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+12 
