Prove $ 1 + 2 + 4 + 8 + \dots = -1$ 
Possible Duplicate:
Infinity = -1 paradox 

I was told by a friend that $1 + 2 + 4 + 8 + \dots$ equaled negative one. When I asked for an explanation, he said:

Do I have to?
Okay so, Let $x = 1+2+4+8+\dots$
$2x-x=x$
$2(1+2+4+8+\dots) - (1+2+4+8+\dots) = (1+2+4+8+\dots)$
Therefore, $(2+4+8+16+\dots) + (-1-2-4-8+\dots) = (1+2+4+8+\dots)$. Now $-2$ and $2$, $-4$ and $4$, $-8$ and $8$ and so on, cancel out, and the only thing left is $-1$.
Therefore, $1+2+4+8+\dots = -1$.

I feel that this conclusion is not right, but I cannot express it. Can anyone tell if this proof is wrong, and if it is, how it is wrong?

Look at Calculus textbook, undergraduate level.  When we treat infinite sum, we cannot change the order to compute.
Example. $1-1+1-\cdots$
$$(1-1)+(1-1)+\cdots=0+0+\cdots=0$$
$$1+(-1+1)+(-1+1)+\cdots=1.$$
