# How to make sense of the sum: 1+2+3+ =-1/12 using regularization?

1. Aug 12, 2008

### arroy_0205

I read that for the divergent series:
$$1+2+3+...=-\frac{1}{12}$$
It was said that is obtained by using the so called regularization technique (zeta function regularization?). I would like to see an explicit proof for that. Can anybody suggest a suitable source where this can be found?

2. Aug 12, 2008

### blechman

consider the sum
$$\sum_{n=1}^{\infty}ne^{-n\epsilon}=-\frac{d}{d\epsilon}\sum_{n=1}^{\infty}e^{-n\epsilon}$$

This sum is just the geometric series, so it sums up to:

$$-\frac{d}{d\epsilon}(1-e^{-\epsilon})^{-1}$$

Now just do the differentiation and expand for small $\epsilon$ and you find:

$$\frac{A}{\epsilon^2}-\frac{1}{12}+\mathcal{O}(\epsilon)}$$

where A is some number I forget off the top of my head. Renormalizing away the first term, you get the right answer in the limit that $\epsilon\rightarrow 0$.

Another way to see what is going on is to realize that this sum looks like the Riemann zeta function $\zeta(-1)$ where

$$\zeta(z)=\sum_{n=1}^{\infty}n^{-z}$$

This expression for the zeta fcn is of course only valid for Re(z)>1. However, it can be analytically continued to the entire complex plane, analytic everywhere except for only a simple pole at z=1. This more general function indeed has $\zeta(-1)=-1/12$, so as long as you're willing to accept that this sum has a "natural definition" in terms of the well-defined zeta function, then that's another way to see this result come out.

3. Aug 12, 2008

### arroy_0205

Hi Belchman, thanks for such a detailed answer. However in the first method, do you not think you are deviating from usual mathematical rules when you just ignore the divergent quantity just because you are interested in the finite part in a particular limit? I am yet to check carefully but suppose I try to regularize by using some other identity (rather than the first one in your response) and then I end up with a different result. This seems possible though I have not found any example yet. Is there any result which tells that in all such regularization procedure the final results should be same otherwise we will have another problem, that of nonuniqueness of the regularized result of a particular sum.

4. Aug 13, 2008

### blechman

my renormalization follows the so-called "Minimal Subtraction Scheme" (MS) where you only cancel the divergent part. It is typically the case that there is an ambiguity in the definition of divergent sums (and integrals), and this ambiguity must be specified by a scheme. I've seen this sum regulated several different ways (the one I've shown you being the simplest in my opinion) and it's always given the result of -1/12, so I'm not sure where scheme-dependence comes in - it might not in this case, but I can't prove it.

However, even if it does come in, I don't care: the point is that the sum is divergent, and therefore, formally undefined! I can call it whatever I want! However, choosing this scheme makes the most amount of sense in the context of physical calculations. That is: if I choose a different scheme, I might have to find myself altering other terms in my expression, and then the FINAL result will be well-defined and scheme independent. But the sum, by itself, certainly won't be - after all, the thing's INFINITY!

Before you (or others) dismiss this, consider the often-used counterexample:

$$I=\int_{-\infty}^\infty x dx$$

This integral is not well-defined. But I can MAKE it well-defined by specifying a R-scheme. The most common (and most sensible) is the "Cauchy Principle Value":

$$I\rightarrow P\int_{-\infty}^\infty x dx\equiv \lim_{R\to\infty}\int_{-R}^R x dx = 0$$

This is certainly acceptable, but by no means unique! For example, why not regulate with:

$$\lim_{R\to\infty}\int_{-R-a/R}^{R+b/R} x dx = (b-a)$$

for ANY value of a and b - that is, I can make this integral ANYTHING I WANT!

So how do I CHOSE the value of the integral? That's where math ends and physics begins. There is usually a good choice, and this choice is not arbitrary, but set by boundary or initial conditions (for example: you find Cauchy's principle value appears as a unique regulator when solving Maxwell's equations (the so-called "Kramers-Kronig relations") in order to maintain causality). I like to think of the regulator that way: just like differential equations have an infinite number of solutions, you can pick the RIGHT one by fitting boundary/initial conditions. Same here.

So to answer your question: the sum is meaningless! But there is a "right" value to assign to it that comes not from the pure math, but from the physics. And the MS regulator (or zeta fcn analytic continuation argument) picks out that correct value.

Hope that helps!