The “Sum” of the Positive Integers

Featured in the Science section of The New York Times on February 4, 2014: $$1+2+3+4+5+\dots =\zeta \left(-1\right)=-\frac{1}{12}$$

You find this puzzling? It goes back to what is meant by the value of the sum for an infinite series. There is a common understanding of that when the infinite series is convergent, but this one, on the contrary, is divergent.

It is commonly known that with $$\zeta \left(s\right)=\sum _{n=1}^{\infty }\frac{1}{n^s}\text{,}$$ which is the definition of $\zeta \left(s\right)$ as the sum of a convergent infinite series when $s>1$, one has $$\zeta \left(2\right)=\frac{{\pi }^2}{6}\text{.}$$ In fact, a computer algebra system with sufficient symbolic power, such as Sage, will produce this precise result without showing a decimal approximation.

$\zeta \left(-1\right)$ can be derived from the value of $\zeta \left(2\right)$ using the functional equation for Riemann's zeta function, which relates $\zeta \left(s\right)$ with $\zeta \left(1-s\right)$.

The definition of $\zeta$ by an infinite series gives rise to an analytic function of the variable $s$ for all real values of $s>1$ and, moreover, for all complex values of $s=u+iv$ with $u>1$. The fact that it makes sense for complex values is significant because, although $\zeta$ “blows up” at $s=1$, one can nonetheless investigate the possibility of analytic continuation of $\zeta$, which, due to the nature of analyticity, is unique whenever it is possible. Riemann did that, and so we know that $\zeta \left(s\right)$ thereby has a unique meaning for all complex values of $s$ other than $s=1$ and $s=0$. In particular, it is meaningful for $s=-1$.

For a clean statement of the functional equation it is desirable to “complete” the zeta function in the following way.

Because every integer $n\ge 1$ has a unique factorization $$n=\prod _{p\text{prime}}{p}^{{\mathrm{ord}}_p\left(n\right)}$$ where ${\mathrm{ord}}_p\left(n\right)\ge 0$ denotes the highest power of the prime $p$ dividing $n$, the infinite series defining $\zeta \left(s\right)$ can be re-written, using the formula for summing a geometric series, as an infinite product, which is due to Euler: $$\zeta \left(s\right)=\prod _{p\text{prime}}\frac{1}{1-\frac{1}{p^s}}\text{.}$$ This product is taken over all of the ordinary primes. Each of these primes $p$ gives rise to a “valuative metric” — a notion of distance $d_p$ – defined by $$d_p\left(m,n\right)={\left|m-n\right|}_p$$ where $${\left|n\right|}_p=p^{-{\mathrm{ord}}_p\left(n\right)}\text{.}$$ Of course, there is also the ordinary absolute value defined by $$\left|n\right|=\left\{\begin{array}{cc}\hfill n& \text{if}x\ge 0\hfill \\ \hfill -n& \text{if}n<0\hfill \end{array}\right.\text{.}$$ It turns out to be convenient to think of the ordinary absolute value as corresponding to a “prime at infinity” and work with an infinite Euler-style product that includes a factor for the prime at infinity.

Having for all “finite” primes $p$ the Euler factors $${\xi }_p\left(s\right)=\frac{1}{1-\frac{1}{p^s}}$$ one defines the Euler factor for the infinite prime to be $${\xi }_{\infty }\left(s\right)={\pi }^{-s⁄2}\Gamma \left(s⁄2\right)$$ where $\Gamma$ denotes the factorial interpolating “Gamma” function, which may be defined by $$\Gamma \left(s\right)={\int }_0^{\infty }{t}^s{e}^{-t}\frac{dt}{t}\text{.}$$ Basic facts about $\Gamma$ include:

1. $\Gamma \left(n+1\right)=n!$.

2. $\Gamma \left(1⁄2\right)=\sqrt{\pi }$.

3. $\Gamma \left(s+1\right)=s\Gamma \left(s\right)$.

These various Euler factors are tied together at a deep theoretical level by each being, relative to the prime $p$ at hand, the “Mellin transform” of a canonical “Gaussian density” in the world of the $p$-adic absolute value ${\left|\right|}_p$ that is equal to its own “Fourier transform” in that world. For $p=\infty$ the Gaussian density is the classical probability density function for the normal distribution: $$\Phi \left(x\right)=e^{-\pi x^2}\text{.}$$

Using all the Euler factors one defines the complete zeta function $$\begin{array}{cccc}& \hfill \xi \left(s\right)& \hfill =\hfill & \prod _{p\text{prime},p\le \infty }{\xi }_p\left(s\right)\hfill \\ & \hfill & \hfill =\hfill & {\xi }_{\infty }\left(s\right)·\zeta \left(s\right)\hfill \\ & \hfill & \hfill =\hfill & {\pi }^{-s⁄2}\Gamma \left(s⁄2\right)\zeta \left(s\right)\hfill \end{array}$$

Riemann proved that $\xi \left(1-s\right)=\xi \left(s\right)$. This equation may now be understood as a divergent model of the Poisson summation formula. Applying it with s = 2 one has $\xi \left(-1\right)=\xi \left(2\right)$. So $$\begin{array}{cccc}& \hfill \zeta \left(-1\right)& \hfill =\hfill & \frac{{\xi }_{\infty }\left(2\right)}{{\xi }_{\infty }\left(-1\right)}\zeta \left(2\right)\hfill \\ & \hfill & \hfill =\hfill & \frac{\frac{1}{\pi }}{\sqrt{\pi }·\Gamma \left(-1⁄2\right)}·\frac{{\pi }^2}{6}\hfill \\ & \hfill & \hfill =\hfill & \frac{1}{\pi ·\sqrt{\pi }·\left(-2\sqrt{\pi }\right)}·\frac{{\pi }^2}{6}\hfill \\ & \hfill & \hfill =\hfill & -\frac{1}{12}\text{.}\hfill \end{array}$$