Trivium: 02nov2008

Let’s have some fun with divergent geometric series. It can be shown (or rather, argued) that $1+2+4+8+\cdots{}=-1$

In fact, Euler already knew this. Intuitively one can simply try the well-known closed form for geometric series, that is

$x^0 + x^1 + x^2 + x^3 + \cdots{} = {1\over 1-x}$

for $x>1$ (at your own risk!), to find the solution:

$2^0 + 2^1 + 2^2 + 2^3 + \cdots{} = {1\over 1-2} = -1$

Alternatively, you can use Riemann’s Zeta function

$\zeta(s) = \sum_{n=1}^{\infty} {1\over n^s}$

which, with $s=-2$ , results after heavy rewriting in $-1$ again.

For a different perspective, see Bill Gosper in HAKMEM 154, where he figures:

By this strategy, consider the universe, or, more precisely, algebra:

let $X = \text{the sum of many powers of two} = \cdots{}111111$

now add X to itself; $X + X = \cdots{}111110$

thus, $2X = X - 1$ so $X = -1$

therefore algebra is run on a machine (the universe) which is twos-complement.