# Divisibility with sums going to infinity

I can't quite wrap my head around this.

Given the formula

$(1-x)(1+x+x^2+...) = 1$

It seems clear to me why this is true. All the x terms cancel out and we are left with one. And this is clearly true for all values of x.

However what I can't figure out is

$\displaystyle\sum_{i=0}^\infty x^i = \frac{1}{1-x}$

If x is something like 2, then

$\displaystyle\frac{1}{1-2} = -1$

But $\displaystyle\sum_{i=0}^\infty 2^i$ is the sum of infinitely many positive numbers. How is it possible that they are equal?

-
That be true 2-adically! – jspecter Oct 18 '11 at 1:40
@Olives: The sum $a_1+a_2+\cdots+a_n$ of a finite set of reals has a reasonably clear meaning. However, before working with sums $a_1+a_2+\cdots +a_n+\cdots$ (adding forever), we must assign meaning to such "sums," and make sure that under our definition, such "sums" behave well under the usual operations of arithmetic. That is why it is necessary to define what one means by a convergent series. If we are going to behave purely "formally," there is no problem. But when we substitute a real number for the formal symbol $x$, the issue of convergence arises. – André Nicolas Oct 18 '11 at 2:13
You'll want to see this... – J. M. Oct 18 '11 at 2:55

The series $\sum_{i=0}^\infty x^i$ converges if $-1<x<1$ and otherwise diverges. If it converges, then it is $1/(1-x)$.
One might add that OP's first formula is not true for all values of $x$. – Gerry Myerson Oct 18 '11 at 2:21