jmc

Places to the right of the decimal point represent negative powers of the base, so we notice that the series in base 10 is $2^{-1}-2^{-2}+2^{-3}\dots =\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\dots$. We recognize this as a geometric series with a common ratio of $-\frac{1}{2}$ and apply the formula $\frac{a}{1-r}$ for the sum of a geometric series with first term $a$ and common ratio $r$. We get $$\frac{\frac{1}{2}}{1-\left(-\frac{1}{2}\right)}=\frac{\frac{1}{2}}{\frac{3}{2}}=\frac{1}{3}.$$So, the sum of the geometric series is $\boxed{\frac{1}{3}}$.