Prove $ 1 + 2 + 4 + 8 + \dots = -1$ [duplicate]

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?

The first mistake is right at the beginning, in writing "Let $x=1+2+4+\cdots$." This builds in the assumption that there is such an object as $1+2+4+\cdots$. The second mistake lies in treating this supposed object as if it were a finite but maybe very long sum, to which the sensible rules for manipulating finite sums apply.

Remark: Think about the "sum" $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$. If we call this $x$, and use a manipulation analogous to the one that you made, we end up with the conclusion that the "sum" is $2$. When infinite series are formally defined, it turns out that the answer is indeed $2$. So some "natural" manipulations yield nonsense, and some yield correct results. Of course that is not a tolerable state of affairs: we cannot use manipulational techniques that sometimes yield a correct result, and sometimes don't. This sort of issue, at a more sophisticated level, led mathematicians in the second half of the $19$th century to look for very careful definitions of the fundamental objects of mathematics, and rigorous proofs of their basic properties.

The infinite series $\displaystyle 1 + 2 + 4 + 8 + \ldots$ diverges. However, the sum $f(z) = 1 + z + z^2 + z^3 + \ldots$, which converges to $1/(1-z)$ for $|z| < 1$, has an analytic continuation to the complex plane with the point $1$ removed, and indeed $f(2) = -1$. So in that sense you could regard $-1$ as the value of the divergent series.

For more on this, see http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%C2%B7_%C2%B7_%C2%B7 and references there.

Look at Calculus textbook, undergraduate level. When we treat infinite sum, we cannot change the order to compute.

Example. $1-1+1-\cdots$

$$(1-1)+(1-1)+\cdots=0+0+\cdots=0$$

$$1+(-1+1)+(-1+1)+\cdots=1.$$