# Relation between infinite product and regularized product

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

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.)