# Regulators and uniqueness

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Does the regularization of a divergent infinite sum yield a unique value?

I.e. do different regularization schemes acting on the same infinite sum produce the same exact value independent of the regulator?

What, exactly, do these values mean? Or what are they? My understanding is that they are not "convergent" values.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user user122066

asked Jul 10, 2016
retagged Jul 10, 2016
Dangit, you beat me to it! math.stackexchange.com/questions/1854642/…

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker
Relevant, as source/motivation of this question: physics.stackexchange.com/a/267243

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user Clement C.

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Different regularizations may lead to different regularized values. For instance $$\lim_{\lambda\to 0}\sum_{n\geq 0}n e^{-\lambda n} = +\infty$$ while the zeta regularization of $\sum_{n\geq 1} n$ gives the (in)famous value $\zeta(-1)=-\frac{1}{12}$.

If we take an hybrid between smoothed sums and the zeta regulatization we have: $$\sum_{n\geq 1}'' n = \sum_{N\geq 1}'\frac{N+1}{2} = \frac{\zeta(0)+\zeta(-1)}{2}=-\frac{7}{24}.$$

We also have a class of regularizations that depends on a positive parameter $\delta$: the Bochner-Riesz mean. There isn't a single regularization: a regularization is just a (somewhat arbitrary) way to extend the concept of convergence. About integrals, the Cauchy principal value can be interpreted as the Fourier transform of a distribution. About series, we may say that $$\sum_{n\geq 1}' a_n = L$$ à-la-Cesàro if $$\lim_{N\to +\infty}\frac{A_1+\ldots+A_N}{N}=L,$$ i.e. if the sequence of partial sums is converging on average. A convergent series is also a Cesàro-convergent series, but with such an extension $${\sum_{n\geq 0}}'(-1)^n = \frac{1}{2}=\lim_{\lambda\to 0}\sum_{n\geq 0}(-1)^n e^{-\lambda n}$$ where $\sum_{n\geq 0}(-1)^n$ is not convergent in the usual sense.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user Jack D'Aurizio
answered Jul 10, 2016 by (30 points)
I would say that your example is an instance where one regulator "works" and another "fails to work." But are there any examples where two different regulators give different, finite values?

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker
I see. I would suggest getting rid of the primes on the $\Sigma$ symbols because they don't seem to mean anything and I find them extremely confusing.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker
@tparker: without primes the chain of equality is simply wrong, so I prefer to leave the primes there.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user Jack D'Aurizio
What do the primes actually mean? What is the general distinction between $\Sigma'$ and $\Sigma''$?

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker
That's my question as well. Physicists never use the hybrid smoothing-and-regularization process you described, but they do often use the scheme I described. But if there exists a sum such that the scheme results in two different finite answers, that seems to pose serious questions about the validity of the physical theory, as you point out. I have revised my question math.stackexchange.com/questions/1854642/… to focus on this narrower question, and I believe it is no longer a duplicate of this one.

This post imported from StackExchange Mathematics at 2016-07-10 19:41 (UTC), posted by SE-user tparker
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