The equation $$ 1 + 2 + 3 + \dots = -1/12 $$ is quite famous.

From the point of view of mathematics, I have no problem with it. My (probably naive) understanding is that there are certain "sums'' which can be assigned to an infinite series through zeta function regularisation, other types of regularisation, Ramanujan summation, and probably quite a few other methods I've never heard of. These "sums" have a lot of nice properties; in particular they agree with the usual sum when both are defined, and so we use the same notation as for the usual sums.

What I would like to understand is how this sum is useful in physics. Now, I have heard in somewhat vague terms that it's useful in string theory and in computation of Casimir force, and presumably in a couple of different places. To the extent that I understand the explanations, they are saying something along the lines:

A certain value with a physical meaning has, according to our computations, the value $\infty - 1/12$. However, we know that infinities do not physically exists, hence the $\infty$ we got has to cancel out with another $\infty$ (coming from some other consideration; something to do with symmetry?). Hence, what we are really left with is $-1/12$.

Is the above in any way an accurate understanding of how this works? If yes, then I have two questions.

  1. Why are we so quick to rule out the infinite value? For instance, if I were to imagine Earth as an infinite half-space with uniform density, then I could compute that the gravity on the surface is infinite (if I recall correctly). The conclusion is that this was not a good model (even though Earth looks quite flat and infinite from where I stand). Why cannot the conclusion in Casimir effect be: Hmm, we have all these contributions which together give a divergent series, we must have forgotten that there's a cutoff, or something.

  2. How do we know that the $\infty$ will cancel out so nicely? In other words, suppose I know that the answer to my problem is a difference between two divergent series, and one of them has the associated "sum" equal to $-1/12$. How do I know that the other one will automatically have the "sum" equal to $0$?

  • 1
    $\begingroup$ Are you sure you got the signs right in the title? $\endgroup$
    – zeta-band
    Nov 28, 2017 at 0:04
  • 2
    $\begingroup$ Quite sure... en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF $\endgroup$ Nov 28, 2017 at 0:12
  • 3
    $\begingroup$ This particular sum is also discussed here and here, and on Math.SE here. See also this Phys.SE post. Also related Phys.SE post here. $\endgroup$
    – Qmechanic
    Nov 28, 2017 at 0:13
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    $\begingroup$ Forget about string theory. If one tries to compute the partition function of the simple 1d harmonic oscillator using the path integral one encounters the need to regularize the functional determinant. Applying $\zeta$-function regularization gives the correct result. $\endgroup$ Nov 28, 2017 at 13:16
  • $\begingroup$ This calculation is of $\zeta(-1)$. The Casimir effect in $d$-dimensional space is proportional to $\zeta(-d)$. For example, in our $d=3$ universe we use $\sum_{n\ge 1}n^3=+\frac{1}{120}$. $\endgroup$
    – J.G.
    Dec 19, 2017 at 21:48

1 Answer 1


In QFT and String theory this divergences usually shows up when we take a continuum limit. As we have learned in the Wilson's Renormalization Group, the continuum limit in Statistical Mechanics and Quantum Mechanics is very subtle. The divergences shows up if we are naive about this limit.

There are various ways to proceed with this continuum limit, almost all characterized by the following steps:

  • Regularization: we do some kind of deformation in the theory controlled by some number $\varepsilon$ or $N$ in order to make everything finite for $\varepsilon\neq0$ or $N\neq\infty$.
  • Renormalization condition: we fix some conditions $A_0$ in a given (arbitrary) scale $\mu_0$, the renormalization scale. This is know as the parametrization of the renormalization flow, a initial point $(A_0,\mu_0)$ of the renormalization flow $(A(\mu),\mu)$ in the parameter space.
  • Limit to the continuum: at the and we do $\varepsilon\rightarrow0$ and obtain the physical quantities we want to compute.

Symmetries are very important here. When we do the first step, the regularization, there are two possibilities: preserve the symmetry or not. If we are able to preserve the symmetry, things are more straightforward. If we don't, we need to impose the symmetry by hand through the calculation and check by hand if this is really possible, i.e. if there is no anomaly.

In your particular case, you want to calculate: $$ S=\sum_{n=0}^{\infty}n $$ the obvious regularization will be a hard cut-off: $$ S(N)=\sum_{n=0}^{N}n $$ and then at the end of the calculation we make $N\rightarrow\infty$. This is obvious divergent. Turns out that this regularization does not preserve the symmetries of the problem. In order to preserve those symmetries you should add counter-terms to this quantity: $$ S(N)=\sum_{n=0}^{N}n + S_{ct}(N) $$ such that the symmetry is restored. This counter-terms should be compatible with the renormalization conditions of step 2.

There are others regularization like the Heat Kernel Regularization: $$ S(\varepsilon)=\sum_{n=0}^{\infty}ne^{-n\varepsilon} $$ this still does not preserve the symmetries of the problem but is more easy to impose the symmetries through the calculation and identify the counter-terms.

There is a regularization that in almost all the cases preserve all the symmetries of the problem, the Zeta regularization: $$ S=\zeta (-1) $$ and so, there is no need for counter-terms.

There is a very nice blog post about it here

  • $\begingroup$ But were in here do you use that the sum is -1/12? $\endgroup$ May 27, 2020 at 6:07
  • $\begingroup$ $S(\varepsilon)=\sum_{n=0}^{\infty}ne^{-n\varepsilon}=\frac{A}{\varepsilon}-\frac{1}{12}+O(\varepsilon)$. The counter-term cancel the $\frac{A}{\varepsilon}$. $\endgroup$
    – Nogueira
    May 27, 2020 at 6:40
  • $\begingroup$ I think I understand your arguement. But when you say that: " This is obvious divergent. Turns out that this regularization does not preserve the symmetries of the problem. In order to preserve those symmetries you should add counter-terms to this quantity" - you're now considering the limit of $S(N \rightarrow \infty) = \sum^N_{n=0}n + f(N)$. I don't see how you are using this notion that $\sum^\infty_{n=0}n = -1/12$. $\endgroup$ May 27, 2020 at 7:38
  • $\begingroup$ For example why can't I just make my "counterpropagating terms" = $\sum^N_n n - \sum^N_n(1/13)^n$. By doing so I show that my sum $S(N\rightarrow \infty) = -1/12$ without having to show that $\sum^\infty_n n = -1/12$ $\endgroup$ May 27, 2020 at 7:48
  • $\begingroup$ The choice of the counter-term is dictated by some renormalization condition. In this case it is the symmetries. The point is that the left-hand-side of $\sum^\infty_{n=0}n = -1/12$ is only well defined if you regularize and renormalize (add counter-terms). Under the same renormalization conditions we are always going to obtain $-\frac{1}{12}$, which allows us to identify $\sum^\infty_{n=0}n $ with $-1/12$. $\endgroup$
    – Nogueira
    May 27, 2020 at 19:02

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