# Sum of All Natural Numbers II

Let us treat the divergent series $s=1+2+3+4+\cdots$ as if it were absolutely covergent. This means that we can manipulate it as if it were a sum of a finite number of terms. So let us asssume that $s = 1 + 2 + 3 + 4 + 5 + \ldots$ There are several ways to show that the result is $s=\displaystyle\ -\frac{1}{12}.$

Grégoire Nicollier's proof has been placed on a separate page.

### Proof

A more conventional proof is based on another (questionable) identity

$1-1+1-1+1-1+\ldots=\displaystyle\frac{1}{2}.$

See, e.g., Divergent Series: why $1 + 2 + 3 + · · · = -1/12$ by Bryden Cais or an article and a video Infinity or -1/12? at +Plus Magazine.

\begin{align} -3s = (1-4)s =& (1+2+3+\ldots )-2(2+4+6+\ldots )\\ =& 1-2+3-4+5-6+\ldots \\ =& 1-(2-3+4-5+6-\ldots) \\ -&\frac{1}{2}+(1-1+1-1+1-1\ldots) \\ =&\frac{1}{2}-(1-2+3-4+5-\ldots)\\ =& \frac{1}{2}+3s, \end{align}

implying $-6s=\displaystyle\frac{1}{2},$ i.e., $s=-\displaystyle\frac{1}{12}.$