Sum of All Natural Numbers

Grégoire Nicollier
University of Applied Sciences of Western Switzerland
December 3, 2014.

Let us treat the divergent series $s=1+2+3+4+\cdots$ abusively like an ordinary number. By grouping the corresponding terms in all parentheses of the following sum, one gets

\begin{equation*}\begin{split} -12s=&\;s-4s+2s-8s+s-4s\\ =&\;(1+2+3+4+5+\cdots)-4\cdot(0+1+0+2+0+3+\cdots)\\ &+2\cdot(0+1+2+3+4+\cdots)-8\cdot(0+0+1+0+2+0+\cdots)\\ &+(0+0+1+2+3+\cdots)-4\cdot(0+0+0+1+0+2+0+\cdots)\\ =&\;1. \end{split}\end{equation*}

This wrong result is "correct" in another context. The Riemann zeta function given by $\zeta(p)=\displaystyle\sum_{n=1}^\infty n^{-p}$ for every real number $p>1$ can be extended in a natural way to the negative integers $-m$ by defining $$\zeta(-m) = -2^{-m}\pi^{-m-1}\ \sin\left(\frac{m\pi}{2}\right)\ m!\ \zeta(m+1).$$ For $m=1$ one has then $\zeta(-1)=-1/12$ because of $\zeta(2)=\pi^2/6$, but $\zeta(-1)$ is no longer the sum $\displaystyle\sum_{n=1}^\infty n^{-p}$ for $p=-1$, which would effectively be $1+2+3+4+\cdots$.

(A more conventional proof can be found on a separate page.)

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