Dear Uncle Colin,

A friend of mine told me that $1 + 2 + 4 + 8 + … = -1$. Is he crazy, or is there something going on here?

-- Somehow Enumerating Ridiculous Infinitely Extended Sum

Dear SERIES,

There are a couple of ‘proofs’ of this non-fact that fall down basically because your friend is wrong. I’m no psychology-ologist, so I can’t tell whether your friend is also crazy.

### Erroneous proof #1

Let $S = 1 + 2 + 4 + … + 2n + …$

Then $2S = 2 + 4 + 8 + … + 4n + …$

So $2S - S = -1$, because all of the terms apart from the first one cancel out. That means $S=-1$.

### Erroneous proof #2

Consider the binomial expansion of $\left(1-x\right)^{-1}$. That gives $1 + x + x^2 + x^3 + …$. Putting in $x=2$, the bracket is $-1$ and the expansion is $1 + 2 + 4 + 8 + … = S$. That means $S = -1$, as before.

### So what’s wrong?

Both of the proofs are wrong, for slightly different but related reasons.

The first proof isn’t valid because it implicitly assumes that $S$ is a number. $S$ is not a number – it’s larger than any number you care to think of, and as the number of terms increases, it goes to infinity. Once you start trying to do arithmetic with infinitely large numbers, you get into all sorts of trouble.

A better way to examine this is to consider the first $n$ terms of $2S$, and take away the first $n$ terms of $S$. You get $4n - 1$, which is clearly larger than $-1$, unless you’re some sort of freaky physicist who thinks that ignoring infinities is ok.

The second proof is invalid because the binomial series is only valid for $|x| < 1$ – sticking 2 in really isn’t allowed, because it leads to just this sort of nonsense.

-- Uncle Colin