The Geometric Series of 1/​(d+1) is a Fraction in Base-d

[lsusr]

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

For example, suppose $x=\frac{1}{2}$.

$$\begin{array}{ccc}\sum _{i=1}^{\infty }{x}^i& =& \sum _{i=1}^{\infty }\frac{1}{2^i}\\ & =& \frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\dots \\ & =& \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots \\ & =& 1\end{array}$$

The case where $x=\frac{1}{2}$ 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. $0.5={0.1}_2$, $4={100}_2$, $16={10000}_2$ and ${10}_{17}=17$.

$$\begin{array}{ccc}\sum _{i=1}^{\infty }{x}^i& =& \sum _{i=1}^{\infty }\frac{1}{{10}_2^i}\\ & =& \sum _{i=1}^{\infty }{0.1}_2^i\\ & =& {0.1}_2+{0.01}_2+{0.001}_2+{0.0001}_2+\dots \\ & =& {0.111111111}_2\dots \\ & =& 0.{\overline{1}}_2\\ & =& 1\end{array}$$

The above trick works for the inverse of any positive integer. Suppose $x=\frac{1}{17}$.

$$\begin{array}{ccc}\sum _{i=1}^{\infty }{x}^i& =& \sum _{i=1}^{\infty }\frac{1}{{10}_{17}^i}\\ & =& \sum _{i=1}^{\infty }{0.1}_{17}^i\\ & =& {0.1}_{17}+{0.01}_{17}+{0.001}_{17}+{0.0001}_{17}+\dots \\ & =& {0.111111111}_{17}\dots \\ & =& 0.{\overline{1}}_{17}\\ & =& \frac{1}{16}\end{array}$$

We can generalize to any denominator $d$.

$$\begin{array}{ccc}\sum _{i=1}^{\infty }\frac{1}{d^i}& =& \frac{1}{d-1}\\ \sum _{i=0}^{\infty }\frac{1}{d^i}& =& 1+\frac{1}{d-1}\\ & =& \frac{d-1}{d-1}+\frac{1}{d-1}\\ & =& \frac{d}{d-1}\\ & =& \frac{1}{1-\frac{1}{d}}\end{array}$$

The relationship holds even when $d$ is not an integer. Let $r=\frac{1}{d}$.

$$\sum _{i=0}^{\infty }{r}^i=\frac{1}{1-r}$$

This is the equation for a geometric series.