# Can the exact value of the product over the Riemann zeta function at even arguments be evaluated?

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?

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.

• I love the honesty in your last sentence :) Dec 24, 2016 at 22:49