True or False ? - Base 2

In general, in base $b$ , let $k=b-1$ . Then $\begin{aligned} 0.\overline{k}_b &= \sum_{i=1}^\infty \frac{k}{b^i} \\ &= \sum_{i=0}^\infty \frac{b-1}{b^i} \\ &=(b-1)\sum_{i=1}^\infty \frac{1}{b^i} \\ &= (b-1)\frac{^1\!/\!_b}{1-{}^1\!/\!_b} \\ &= \frac{b(b-1)}{b(b-1)} \\ &= \boxed{1} \end{aligned}$

We know that : $1 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} +...$ (base 10)

Or, $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} +...= 1 \times 2^{-1} + 1 \times 2^{-2} + 1 \times 2^{-3} + 1 \times 2^{-4} +...$ (base 10)

And $(1 \times 2^{-1} + 1 \times 2^{-2} + 1 \times 2^{-3} + 1 \times 2^{-4} +...)_{10} = (0.1 +0.01 +0.001 +0.0001 +...)_2$

Thus, $\boxed{1_2 = 0.111111111111..._2}$