For a sequence of non-zero complex numbers $\lambda_{1}, \lambda_{2}, \cdots$, the infinite product is defined by $$ \prod_{n\in\mathbb{N}}\lambda_{n} := \lim_{i\rightarrow\infty}\prod_{n=1}^{i}\lambda_{n}$$ to be the limit of partial product.
For the same sequence $(\lambda_{n})_{n\in\mathbb{N}}$, we define the zeta-regularized product by $$ \widehat{\prod_{n\in\mathbb{N}}}\lambda_{n} := \exp(-\zeta_{\lambda}^{'}(0)), $$ where $\zeta_{\lambda}(s)$ is the zeta function which is given by $$ \zeta_{\lambda}(s)=\sum_{n\in\mathbb{N}}\lambda_{n}^{-s}. $$ We assume that the zeta function has an analytical continuation up to $0$.
My first question is:
What is the difference between the two products?
For more a precise question, I add an example.
For example, if we consider the sequence $(n)_{n\in\mathbb{N}}$, we have two different results: for the infinite product, we have $$ \prod_{n\in\mathbb{N}}n = \infty; $$ while for the zeta-regularized product we have $$ \widehat{\prod_{n\in\mathbb{N}}}n=\sqrt{2\pi}. $$ This results imply that, even for a same sequence, the zeta-regularized product could be convergent while the infinite product is not. (Under the condition that the zeta function for the sequence is well-defined.)
This example leads to my second question:
If an infinite product is convergent, then does it coincide with the zeta-regularized product?
Thank you for your time and effort.