I am a physicist interested in physics. In particular this question is related to quantum field theory.

I recently came across a derivation of the infinite sum $1+1+1+1+..... $ that produced the result -1/2, aka zeta regularization (from Terry Tao's blog)

This was quite surprising to me as I had previously known an infinite sum of 1s to be divergent - from taking a math physics course by the guy that wrote "the book" on asymptotic methods.

(Indeed I've known the sum of all positive integers to be finite for quite some time as well as many other "divergent-looking" sums and I get the whole idea behind summation methods like Padé, Shanks, Euler, etc.)

Anyways this prompted me to wonder;

  1. Are ALL infinite sums not divergent?
  2. If not then how can one determine whether a sum os divergent or not?
  3. What was all this business in undergrad calculus about learning tests of convergence and all this business in complex analysis about series if weird things like $1+1+1+1+.....$ are actually convergent??

I'm still confused by all this stuff. And I haven't found an answer to these questions that "click"

Any help understand this topic would be kindly appreciated.

  • $\begingroup$ arxiv.org/abs/hep-ph/0510142 $\endgroup$ Jul 10, 2016 at 0:36
  • 2
    $\begingroup$ You must distinguish between "the sum converges (or diverges)" and "we can assign a meaningful value to the sum via regularization". Those are two completely different things. $\endgroup$
    – Asaf Karagila
    Jul 10, 2016 at 13:35
  • $\begingroup$ A nice video by mathologer (youtube) about Ramanujan and weird divergent summations, explains that: if you change the convergence criteria you can sum some series. In particular: the Riemann z function for -1, is z(-1)=1+2+3+4+... which adds up to -1/12 $\endgroup$ May 26, 2021 at 20:13

2 Answers 2


No, many infinite sums converge. An infinite sum (or "series") $a_0 + a_1 + a_2 + \dots$ is defined to converge to a value $S$ if the limit $$ S = \lim_{n\rightarrow \infty} \sum_{i=1}^n a_i$$ exists. For example, if the value of $a_i$ falls off exponentially quickly or as a power law faster than $1/i$, then the series converges. The various convergence tests you learned in calculus can give you more precise criteria for convergence.

The infinite series $1 + 1 + 1 + \dots$ is not convergent - the above limit does not exist. However, it is regularizable - that is, you can play some tricks on it that "beat it into shape" well enough that you can assign some finite number to it. But this finite number is not actually the sum, which does not exist. Being regularizable is a much weaker criterion than being convergent.

So whenever you come across a divergent series in QFT and replace it with its regularized value, it's very important that you take into account that the two quantities aren't actually equal. Despite its being very important, there are approximately zero physicists who actually do it.

Edit: the OP asked a very good question in the comments that I though was worth addressing in my main answer: whether imposing different regulators on the same divergent series always yields the same result. If anyone has any thoughts, I've posed that question at Can different choices of regulator assign different values to the same divergent series?. Also, I once asked a related question at https://physics.stackexchange.com/questions/254051/how-can-dimensional-regularization-analytically-continue-from-a-discrete-set for which I never got a satisfactory answer.

  • 4
    $\begingroup$ @user122066 Ah, I see your confusion now. No, regularization does not give convergence. Some people might say something like "we can use $\zeta$-regularization to show that the series converges to ..." but they're being sloppy with their language, the series still doesn't converge no matter what clever regularization you use. Regularization is a way to assign a finite value to a divergent series that is somewhat useful, but it doesn't actually sum the series. $\endgroup$
    – tparker
    Jul 10, 2016 at 1:49
  • 1
    $\begingroup$ @user122066 The series $2 + 4 + 8 + 16 + \dots$ does not converge, as can be shown by any convergence test. (The terms have to decrease to zero for convergence to even by possible.) If you look at the formula you used to sum the series, you'll note that it only actually applies if the base of the exponent in the geometric sequence is less than 1, which isn't the case here. So strictly speaking, your formula does not apply. In order to get the value -2, you unknowingly used analytic continuation to assign a finite value to the divergent series. $\endgroup$
    – tparker
    Jul 10, 2016 at 1:56
  • 4
    $\begingroup$ This stuff youve said here makes so much sense it hurts! $\endgroup$
    – user122066
    Jul 10, 2016 at 2:00
  • 3
    $\begingroup$ @user122066 Yeah, that whole "formally replacing the series with S and then algebraically manipulating S" trick is actually just heuristic because the series doesn't actually converge so S isn't well-defined. And it can sometimes get you into trouble for more complicated series. Analytic continuation is the more rigorous way to do it. $\endgroup$
    – tparker
    Jul 10, 2016 at 2:06
  • 4
    $\begingroup$ @user122066 Regarding whether the sum is unique, in the sense that every choice of regulator gives you the same answer: that's a great question - I've wondered the exact same thing myself and I don't know the answer. $\endgroup$
    – tparker
    Jul 10, 2016 at 2:08

Read more carefully the first part of Terry Tao's post. He replaces the regular partial sums with a smoothed sum

$$\sum_{n=1}^N n^s \to \sum_{n=1}^\infty \eta(n/N) n^s$$

where $\eta$ is a cutoff function. The result he finds for the smoothed sum of $1+1+1+\dots$ (case $s=0$) is

$$\sum_{n=1}^\infty \eta(n/N) = - \frac 1 2 + C_{\eta,0} N + O(1/N)$$


$$C_{\eta,0} =\int_0^\infty \eta(x) dx$$

On the right side there is an asymptotic expansion of the smoothed sum. As you can see, there is indeed a constant term $-1/2$, but the following term is divergent in the limit $N \to \infty$. So it is misleading to state that

$$1+1+1+\dots = -1/2$$

because $-1/2$ is just the constant term of an asymptotic expansion which is divergent in the limit $N\to \infty$.

What is usually done in QFT (see Luboš Motl's answer here for example) is to cancel the leading divergence by means of a local counterterm. Basically, a "trick" is used to ged rid of the divergence and be able to write $1+1+1+\dots=-1/2$.

  • $\begingroup$ Why was that downvoted? $\endgroup$ Jul 10, 2016 at 10:47
  • $\begingroup$ I don't know, I voted it up. But: I have some feeling that something is missing here. Reading this 1+1+...=-1/2 thing it is unclear, how it comes out. Maybe this motivated the down (although I suspect the answer is very clearly available on some of the refered links). $\endgroup$
    – peterh
    Jul 25, 2016 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.