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For a sequence of 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 $$ \hat{\prod_{n\in\mathbb{N}}}\lambda_{n} := \exp(-\zeta_{\lambda}^{'}(0)), $$ 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 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 $$ \hat{\prod_{n\in\mathbb{N}}}n=\sqrt{2\pi}. $$ 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.)

I got another question form the example that

If an infinite product is convergent, then does it coincide with the zeta-regularized product?

Thank you for your time and effort. Have a nice day.

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