Since b_{i}has to be either 1 or 0, that sum is less than or equal to which is a geometric series converging to 1.

To find the binary decimal for 1/3, note that 1/3< 1/2 so the first decimal place is 0. It is larger than 1/4 so the first two places are 0.01. Now, 1/3- 1/4= (4- 3)/12= 1/12. That is less than 1/8 but larger than 1/16 so the first four places are 0.0101. Now, 1/12- 1/16= 4/48- 3/48= 1/48. That is less than 1/32 but larger than 1/64 so the first six places are 0.010101. You might be able to guess, now, how that continues!

Thanks in advance