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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.

(Sorry in advance for being a physicist.

user122066
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  • Dangit, you beat me to it! http://math.stackexchange.com/questions/1854642/can-different-choices-of-regulator-assign-different-values-to-the-same-divergent – tparker Jul 10 '16 at 04:31
  • Relevant, as source/motivation of this question: http://physics.stackexchange.com/a/267243 – Clement C. Jul 10 '16 at 04:41
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    Also, no need to apologize for being a physicist. But you may need to apologize, however, for not matching your opening parentheses. :) – Clement C. Jul 10 '16 at 04:52

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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.

Jack D'Aurizio
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    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? – tparker Jul 10 '16 at 04:34
  • Could you explain your notation - what do the single and double primes on the sums mean? How did you go from summing over $n$ to summing over $(n+1)/2$? – tparker Jul 10 '16 at 05:05
  • @tparker: at first we perform a Cesàro regularization: the main term ($n$) is replaced by the averaged term $\frac{1}{N}\sum_{n=1}^{N}n = \frac{N+1}{2}$. A zeta regularization follows. – Jack D'Aurizio Jul 10 '16 at 05:07
  • 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. – tparker Jul 10 '16 at 05:17
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    @tparker: without primes the chain of equality is simply wrong, so I prefer to leave the primes there. – Jack D'Aurizio Jul 10 '16 at 05:18
  • What do the primes actually mean? What is the general distinction between $\Sigma'$ and $\Sigma''$? – tparker Jul 10 '16 at 05:18
  • $''$ means Cesàro+Zeta regularization, a single $'$ means Zeta regularization or Cesàro regularization, depending on the context. – Jack D'Aurizio Jul 10 '16 at 05:19
  • And there is no "general dinstinction" because we are just inventing regularizations for proving a non-uniqueness theorem. – Jack D'Aurizio Jul 10 '16 at 05:20
  • I'm not convinced that this is really in the spirit of Cesaro regularization, which regularizes a series, not individual terms in a series. Can you come up with a more straightforward regularization where you find a series of analytic functions $f_n(z)$ such that $f_n(0) = n$ and whose sum converges for some open set of $z$, then analytically continue the sum to $z = 0$? I checked your original attempt in the comment you deleted, and I think you made a mistake and it doesn't converge. – tparker Jul 10 '16 at 05:22
  • @tparker: all right, let me rephrase: we consider an hybrid between a particular smoothing (the main term is replaced by the average term) and the zeta regularization. Hope it is ok now. Out of curiosity, what is the purpose of this almost-non-math? – Jack D'Aurizio Jul 10 '16 at 05:25
  • In physics, we often use the analytic-continuation regularization prescription that I just described to assign values to divergent series, and I'm curious if the prescription always yields a unique answer. Unfortunately, the question appears to remain open. – tparker Jul 10 '16 at 05:30
  • @tparker: since the zeta regularition is not the only regularization, I think the question is pretty trivial, but I was interested to the problem in itself: if a series/integral is not converging in the usual sense, what is the physical meaning of taking any regularization of it? A regularization is a kind of arbitrary choice, like a choice of weights. The zeta regularization is perhaps God's weight choice? :D – Jack D'Aurizio Jul 10 '16 at 05:33
  • 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 http://math.stackexchange.com/questions/1854642/can-different-choices-of-regulator-assign-different-values-to-the-same-divergent to focus on this narrower question, and I believe it is no longer a duplicate of this one.

     – tparker Jul 10 '16 at 05:50
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    A physicist comment: both exponential and Zeta regularization give the same result for the finite part (1/12), which is the important part for example in the calculation of the Casimir effect, whereas the divergent parts (1/$\lambda^2$ for exponential, $0$ for Zeta) are different (and are absorbed into a renormalization of a coefficient (vacuum energy, etc.)). This universality of the finite part implies that there is no God regularization ;-) – Adam Jul 11 '16 at 07:00