TITLE: Can the exact value of the product over the Riemann zeta function at even arguments be evaluated?
QUESTION [6 upvotes]: According to wolframalpha, the product over the Riemann zeta function at even arguments converges : 
$$\prod_{n=1}^\infty \zeta(2n) \approx 1.82 $$ 
Q1: Can it be proved that this product actually converges? 
Furthermore, I am wondering whether we can determine to what exact value this product converges (once it has been established that this product actually converges). 
We know that the following formula holds: 
$$ \zeta(2n) = (-1)^{n+1} \frac{B_{2n} (2 \pi)^{2n} }{2 (2n)!}, \qquad (*) $$
where $B_{n}$ is the $n$'th Bernoulli number. So when we consider the product over these values from $n=1$ to infinity, we have have a term (in the numerator) $$(2 \pi)^{n(n+1)} ,  $$ where $n \to \infty $, and terms involving products over the even Bernoulli numbers and even factorial numbers, which I find harder to evaluate.  
Q2: Can we use the $(*)$-marked formula to evaluate the aforementioned product? Or some other formula?

REPLY [5 votes]: Proving the convergence of the product is easy, we have
$$\begin{align}
\log \prod_{n=1}^\infty \zeta(2n) &= \sum_{n=1}^\infty \log \zeta(2n)\\
&= \sum_{n=1}^\infty \log \left(1 + (\zeta(2n)-1)\right)\\
&\leqslant \sum_{n=1}^\infty (\zeta(2n)-1)\\
&= \sum_{n=1}^\infty \sum_{k=2}^\infty \frac{1}{k^{2n}}\\
&= \sum_{k=2}^\infty \sum_{n=1}^\infty \frac{1}{k^{2n}}\\
&= \sum_{k=2}^\infty \frac{1}{k^2-1}\\
&\leqslant \frac{4}{3}\sum_{k=2}^\infty \frac{1}{k^2}\\
&= \frac{4}{3}(\zeta(2)-1)\\
&= \frac{2\pi^2-12}{9}\\
&< +\infty,
\end{align}$$
where the change of order of summation is unproblematic since everything is non-negative.
However, determining the value of the infinite product is far more difficult. I have no idea how to do it.