Is anything known about the product of the Zeta function at odd numbers compared to at even functions?

Question

Is anything known about the product: $$\prod{\frac{\zeta(2n+1)}{\zeta(2n)}}$$ Over all n? I very much doubt there exists a closed form for it but what does it converge to?

Answer 1

For any $m\geq 2$ we have $$ \log\zeta(m) = -\sum_{p}\log\left(1-\frac{1}{p^m}\right) = \sum_{h\geq 1}\frac{1}{h p^{hm}}\tag{1} $$ $$ \frac{d}{dm}\log\zeta(m) = \frac{\zeta'(m)}{\zeta(m)} = -\sum_{n\geq 1}\frac{\Lambda(n)}{n^m}\tag{2} $$ with $\Lambda$ being the Von Mangoldt function. In particular: $$ \frac{d}{dm}\log\frac{\zeta(2m+1)}{\zeta(2m)}=\sum_{n\geq 1}\frac{\Lambda(n)}{n^{2m}}\left(1-\frac{1}{n}\right)\tag{3} $$ and: $$ \sum_{m\geq 1}\sum_{n\geq 1}\frac{\Lambda(n)}{n^{2m}}\left(1-\frac{1}{n}\right)=\sum_{n\geq 2}\frac{\Lambda(n)}{n(n+1)} \tag{4} $$ such that: $$ \sum_{m\geq 1}\log\frac{\zeta(2m+1)}{\zeta(2m)} = -\sum_{n\geq 2}\frac{\Lambda(n)}{n(n+1)\log n}\tag{5} $$ and:

$$ \prod_{m\geq 1}\frac{\zeta(2m+1)}{\zeta(2m)}=\color{red}{\exp\left(-\sum_{n\geq 2}\frac{\Lambda(n)}{n(n+1)\log n}\right)} \tag{6}$$

I guess the series appearing as the argument of the exponential function in the RHS of $(6)$ can be further accelerated by standard tricks, but I won't bet on a really simpler "closed form" for the given infinite product, apart from $$ \exp\left[-\sum_{k\geq 1}\sum_{p\in\mathcal{P}}\frac{1}{k}\left(\frac{1}{p^k}-\frac{1}{p^k+1}\right)\right].\tag{7}$$