When is an infinite product of natural numbers regularizable? I only recently heard about the concept of $\zeta$-regularization, which allows the evaluation of things like
$$\infty != \prod_{k=1}^\infty k = \sqrt{2\pi}$$
and
$$\infty \# = \prod_{k=1}^\infty p_k = 4\pi^2$$
where $n\#$ is a primorial, and $p_k$ is the $k$-th prime. (The expression for the infinite product of primes is proven here.) That got me wondering if, given a sequence of positive integers $m_k$ (e.g. the Fibonacci numbers or the central binomial coefficients), it is always possible to evaluate the infinite product
$$\prod_{k=1}^\infty m_k$$
in the $\zeta$-regularized sense. It would seem that this would require studying the convergence and the possibility of analytically continuing the corresponding Dirichlet series, but I am not too well-versed at these things. If such a regularization is not always possible, what restrictions should be imposed on the $m_k$ for a regularized product to exist?
I'd love to read up on references for this subject. Thank you!
 A: Given an increasing sequence $0<\lambda_1<\lambda_2<\lambda_3<\ldots$ one defines the regularized infinite product
$$
\prod^{\infty}_{n=1}\lambda_n=\exp\left(-\zeta'_{\lambda}(0)\right),
$$
where $\zeta_{\lambda}$ is the zeta function associated to the sequence $(\lambda_n)$,
$$
\zeta_{\lambda}(s)=\sum^{\infty}_{n=1}\lambda_n^{-s}.
$$
(See the paper: E.Munoz Garcia and R.Perez-Marco."The Product over all Primes is $4\pi^2$". http://cds.cern.ch/record/630829/files/sis-2003-264.pdf )
In the paper( https://arxiv.org/ftp/arxiv/papers/0903/0903.4883.pdf ) I have evaluated the
$$
\prod_{p-primes}p^{\log p}=\prod_{p-primes}e^{\log^2p}=\exp\left(24\zeta''(0)+12\log^2(2\pi)\right)
$$
where $\zeta''(0)$ is the second derivative of Riemann's Zeta function in $0$.
A: The claim that $\operatorname{reg} \prod_{k=1}^\infty k = \sqrt{2\pi}$ looks misleading and wrong to me.
Definitely, $\prod_{k=1}^\infty k=\exp \sum_{k=1}^\infty \ln k$.
Also, definitely $\operatorname{reg}\sum_{k=1}^\infty \ln k=\ln \sqrt{2\pi}$
But exponent of the regularized value is not the same as the regularized value of exponent!
More honestly, the value $\sqrt{2\pi}$ corresponds not to the regularized (finite) value of the product, but to its hypermodulus as defined here. Hypermodulus is the exponent of the scalar finite part of the logarithm of the object.
