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Find the sum of the base-2 geometric series
$0.1_2-0.01_2+0.001_2-0.0001_2+0.00001_2\ldots$;
give your answer as a fraction in which the numerator and denominator are both expressed in base 10.

$\begin{array}{|rcll|} \hline 0.1_{2} = 1\cdot 2^{-1} &=& \frac{1}{2} \\ 0.01_{2} = 1\cdot 2^{-2} &=& \frac{1}{4} \\ 0.001_{2} = 1\cdot 2^{-3} &=& \frac{1}{8} \\ 0.0001_{2} = 1\cdot 2^{-4} &=& \frac{1}{16} \\ 0.00001_{2} = 1\cdot 2^{-5} &=& \frac{1}{32} \\ \ldots \\ \hline \end{array}$

Infinite Geometric Series:

$\begin{array}{|rcll|} \hline s &=& \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{8} - \dfrac{1}{16} + \dfrac{1}{32} \pm \ldots \quad | \quad a=\frac{1}{2},~ r = -\frac{1}{2} \\\\ s &=& \dfrac{a}{1-r} \quad | \quad |r| < 1. \\\\ s &=& \dfrac{\frac{1}{2}}{1- \left(-\frac{1}{2} \right)} \\\\ s &=& \dfrac{\frac{1}{2}}{1+\frac{1}{2}} \\\\ s &=& \dfrac{\frac{1}{2}}{\frac{3}{2}} \\\\ s &=& \dfrac{1}{2} \cdot \dfrac{2}{3} \\\\ \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{1}{3}} \\ \hline \end{array}$

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