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For a positive sequence $0\le\lambda_{1}\le\lambda_{2}\le\cdots$, consider an infinite product \begin{equation*} \prod_{i=1}^{\infty}\lambda_{i}:=\lim_{n\rightarrow\infty}\prod_{i=1}^{n}\lambda_{i}. \end{equation*} The product converges if and only if $\sum_{i=1}^{\infty}\log(\lambda_{i})$ converges.

Also, we consider the following zeta function given by a Dirichlet series \begin{equation*} \zeta(s)=\sum_{i=1}^{\infty}\lambda_{i}^{-s}. \end{equation*} We assume that the zeta function is absolutely convergent on some right-half plane, has an analytic continuation on $\mathrm{Re}(s)>-\epsilon$ for a positive number $\epsilon>0$, and is holomorphic at $s=0$. Then, for this case, we define a regularized product by \begin{equation*} \hat{\prod}_{i=1}^{\infty}\lambda_{i}:=\exp(-\zeta^{\prime}(0)). \end{equation*}

My question is that

If the ordinary infinite product converges, then does it coincide with the regularized product?, i.e. $$\prod_{i=1}^{\infty}\lambda_{i}=\hat{\prod}_{i=1}^{\infty}\lambda_{i}$$

The question is equivalent to the following

If $\sum_{i=1}^{\infty}\log(\lambda_{i})$ converges, then $$\sum_{i=1}^{\infty}\log(\lambda_{i})=-\zeta^{\prime}(0)?$$

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It doesn't make sense to compare the infinite product and the zeta-regularized determinant, because they can never be finite at the same time. The assumption that the infinite product converges is incompatible with the assumption that the series defining the zeta function converges.

If the infinite product $\Pi_i \lambda_i$ converges, then $\lambda_i \to 1$ (sufficiently fast).

Whereas if for some $s \in \mathbb{C}$, the series defining the zeta function converges, then $|\lambda_i^{-s}| = \lambda_i^{-\Re s} \to 0$, which implies either $\lambda_i \to \infty$ (if $\Re s > 0$) or $\lambda_i \to 0$ (if $\Re s < 0$) (sufficiently fast). (Usually, one has $\lambda_i \to \infty$ and one defines $\zeta(s)$ on a right-half plane, as you mentioned.)

So the two notions of determinant are useful in disjoint "realms".

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