# $\zeta(-1)=\frac{-1}{12}\nRightarrow \sum_\limits{i=1}^\infty i=\frac{-1}{12}$ so how do we justify its use in string theory? [duplicate]

I asked math.stackexchange about my concerns regarding the justification of the claim that $1+2+3+...=\frac{-1}{12}$

Once you have

1. defined $\zeta(s) =\sum_\limits{n=1}^\infty\frac{1}{n^s}$ where $|s|>1$
2. defined $\zeta^\prime(s)=...$ by analytic continuation for all $s$

then you can only claim

1. $\zeta(s)=\zeta^\prime(s)$ where $|s|>1$

Hence, $\zeta(-1)=\frac{-1}{12}\nRightarrow \sum_\limits{i=1}^\infty i=\frac{-1}{12}$

To my surprise, apparently I didn't drop the ball, the community there seems to agree with my objection.

I don't know string theory but I hear $1+2+3+...=\frac{-1}{12}$ is a standard lemma.

Is string theory's use of this "result" actually rigorous? If so, what did the math.stackexchange people miss?

• NB It is true using certain types of divergent "summing" techniques, like Ramanujan summation. – Argon Dec 15 '16 at 16:15
• I recommend the reading of the first 3 chapters of String Theory, by Polchinsk, specially at the final of each chapter. – Nogueira Dec 15 '16 at 16:18
• Possible duplicate of Critical Dimension of Bosonic Strings and Regularization of $\sum_{n=1}^\infty n$ – John Rennie Dec 15 '16 at 16:19
• You can show that IF there is a value that's usable for that sum, then it has to be -1/12. This is done via manipulation of sums. This of course doesn't justify it, but it makes a point for it. – Omry Dec 15 '16 at 16:28

## 1 Answer

It is true that $\zeta(-1)=-1/12$. But this doesn't imply that $$1+2+\cdots=\frac{-1}{12}$$

For one thing, the l.h.s. is undefined if we stick to the standard meaning of $+$. You could, if you want to, extend the definition of $+$ so that this is equation is true. But this is highly non-trivial: the standard rules of addition break down in this generalised notion of summation.

In QM (and string theory in particular) we sometimes have to consider improper sums, and they sometimes turn out to diverge. We don't like infinities for the obvious reason: a theory with divergent objects is ill-defined.

We therefore introduce a regularisation, which amounts to define a certain (context-dependent) deformation of the basic objects so that everything is finite. You perform some (rigorous) algebraic manipulations in terms of finite objects, calculate something measurable, and in the end you remove the regulator (by changing the deformation into the identity transformation). Sometimes this lengthy process is summarised as $$1+2+\cdots\to -\frac{1}{12}$$ which does not mean that these two objects agree; it means that you can replace your naïvely divergent sum by the r.h.s., and in some cases the result you get is right.

This is clearly non-rigorous if you don't go through all the details of regularisation and renormalisation, but it is a very solid mathematical procedure once you know what it actually means. I am sure that you'll remain skeptic until you calculate something yourself. You shouldn't believe people if they tell you "it can be justified"; you should check it yourself. But in order to do so, you'll have to learn the basics, for example, QFT.

Divergent sums and integrals appear everywhere in QM, but for some reason the popular choice to introduce the weirdness of QM to the general audience is $1+2+\cdots=-1/12$. There are thousands of more involved, more interesting and important examples, but this one here looks funny and so this is the one you'll find everywhere online.