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.


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.

  • 1
    $\begingroup$ 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? $\endgroup$ – tparker Jul 10 '16 at 4:34
  • $\begingroup$ 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 $(n+1)/2$? $\endgroup$ – tparker Jul 10 '16 at 5:05
  • $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio Jul 10 '16 at 5:07
  • 1
    $\begingroup$ @tparker: without primes the chain of equality is simply wrong, so I prefer to leave the primes there. $\endgroup$ – Jack D'Aurizio Jul 10 '16 at 5:18
  • 1
    $\begingroup$ 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 ;-) $\endgroup$ – Adam Jul 11 '16 at 7:00

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.