Suppose x is a positive number less than 1. What is the sum of the positive powers of x?

For example, suppose x=12.

∞∑i=1xi=∞∑i=112i=121+122+123+…=12+14+18+…=1

The case where x=12 is intuitively obvious in base-10, but it's even more intuitively obvious in base-2. I will use a subscript to indicate base e.g.0.5=0.12, 4=1002, 16=100002 and 1017=17.

It does work for negative bases. Representation of a number in any base is in essence a sum of base powers multiplied by coefficients. The geometric series just has all coefficients equal to 1 after the radix point (and a 1 before it, if we start addition from the 0th power).

Suppose x is a positive number less than 1. What is the sum of the positive powers of x?

For example, suppose x=12.

∞∑i=1xi=∞∑i=112i=121+122+123+…=12+14+18+…=1

The case where x=12 is intuitively obvious in base-10, but it's even more intuitively obvious in base-2. I will use a subscript to indicate base

e.g.0.5=0.12, 4=1002, 16=100002 and 1017=17.∞∑i=1xi=∞∑i=1110i2=∞∑i=10.1i2=0.12+0.012+0.0012+0.00012+…=0.1111111112…=0.¯12=1

The above trick works for the inverse of any positive integer. Suppose x=117.

∞∑i=1xi=∞∑i=1110i17=∞∑i=10.1i17=0.117+0.0117+0.00117+0.000117+…=0.11111111117…=0.¯117=116

We can generalize to any denominator d.

∞∑i=11di=1d−1∞∑i=01di=1+1d−1=d−1d−1+1d−1=dd−1=11−1d

The relationship holds even when d is not an integer. Let r=1d.

∞∑i=0ri=11−r

This is the equation for a geometric series.