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.