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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?

Michael Hardy
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Max Muller
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1 Answers1

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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.

Daniel Fischer
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