Tuesday, July 12, 2011

Why is the sum of integers equal to -1/12

The answer is copied from Math Stackexchange




Exponential regulator method

Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $$$n$$$ is the base that is exponentiated to the $$$s$$$th power. Here, the regulator has $$$n$$$ in the exponent. Obviously, the original sum of integers is obtained in the $$$\epsilon\to 0$$$ limit of the formula for $$$S$$$. In physics, $$$\epsilon$$$ would be viewed as a kind of "minimum distance" that can be resolved.

We will only need some Laurent expansion around $$$\epsilon = 0$$$.
We have
You see that the $$$1/\epsilon^2$$$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command
and the result is
In the $$$\epsilon\to 0$$$ limit we were interested in, the $$$\epsilon^2/240$$$ term as well as the smaller ones go to zero and may be erased. The leading divergence $$$1/\epsilon^2$$$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $$$-1/12$$$, which is the correct result of the sum.

In physics applications, $$$\epsilon$$$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $$$\exp(-(\epsilon+\epsilon^2)n)$$$.


Euler's method

Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of integers is $$$-1/12$$$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too.

We will work with two sums,
$$ \begin{align} S &= 1+2+3+4+5+\dots, \\ \quad T &= 1-2+3-4+5-\dots \end{align} $$
Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $$$x$$$, the expression $$$T$$$ may be summed according to the Taylor expansion
Substitute $$$x=1$$$ to see that $$$T=+1/4$$$. The value of $$$S$$$ is easily calculated now:
$$ \begin{align}T &= (1+2+3+\dots) - 2\times (2+4+6+\dots)= \\ &= (1+2+3+\dots) (1 - 4) =\\ &= -3S \end{align} $$
so $$$S=-T/3=-1/12$$$.

A zeta-function calculation

A somewhat unusual calculation of $$$\zeta(-1)=-1/12$$$ of mine may be found here:
The comments are in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $$$s$$$ is added to the zeta function. The new function is the old zeta function for $$$s=0$$$ and for $$$s=1$$$, it only differs by one. We Taylor expand around $$$s=0$$$ to get to $$$s=1$$$ and we find out that only a finite number of terms survives if the main argument $$$x$$$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $$$1$$$, and prove that the function [vanishes][2] at negative even values of $$$x$$$.


  1. "actually confirmed experimentally whenever it is possible"
    Maybe true, but without references, this kind of claim is meaningless.

  2. That's one tough paper to read -- even when I tried to read the part about the sum of positive integers being -1/12.