# What can be said about $\prod_{s=2}^{\infty} \zeta(s)$?

Another problem from quora.

What can be said about $$v =\prod_{s=2}^{\infty} \zeta(s)$$?

Wolfy says that $$v \approx 2.294856591673313794183$$.

The Inverse Symbolic Calculator (http://wayback.cecm.sfu.ca/projects/ISC/ISCmain.html) finds nothing.

Here are the close values:

$$2294856391585950 = (0064) sum((7/6*n^3-3*n^2+65/6*n-3)/(Fibo(n)+1),n=1..inf)\\ 2294856397653578 = (0001) GAM(5/6)*BesI(1,2)^{GAM(7/12)}\\ 2294856469349552 = (0006) 1/4357571\\ 2294856473388934 = (0324) 1/10*(10+14^{1/2}*10^{1/4})^{1/2}*10^{3/4}\\ 2294856488490537 = (0001) BesJ(0,1)^{GAM(1/3)/GAM(5/12)}\\ 2294856524215021 = (0314) sin(Pi*8/45)-sin(Pi*14/51)\\ 2294856591673313 \text{ would be here}\\ 2294856740350177 = (0011) sum((2*n^3-5*n^2+22*n-14)/(n!+2),n=1..inf)\\ 2294856991319600 = (0324) 23^{1/2}/(12^{3/4}-3^{2/3})^{1/2}\\ 2294857017919943 = (0001) Feig2*Li4(1/2)/GAM(5/6)\\ 2294857089879260 = (0248) F(11/26;27/50;1) \\ 2294857493303450 = (0404) Psi(1/21)+Psi(19/21)+Psi(13/14)\\$$

I tried using Euler's zeta product and partition sum, but this didn't seem to help. Here is what resulted:

$$\zeta(s) =\dfrac1{\prod_p (1-p^{-s})}$$.

$$\begin{array}\\ v &=\prod_{s=2}^{\infty} \zeta(s)\\ &=\prod_{s=2}^{\infty} \dfrac1{\prod_p (1-p^{-s})}\\ &=\prod_{s=2}^{\infty} \prod_p\dfrac1{ (1-p^{-s})}\\ &= \prod_p\prod_{s=2}^{\infty}\dfrac1{ (1-p^{-s})}\\ \end{array}$$

Euler's product identity states that $$\prod_{n=1}^{\infty}\dfrac1{1-zx^n} =\sum_{n=1}^{\infty} \dfrac{x^nz^n}{\prod_{k=1}^n(1-x^k)}$$. Putting $$x=z=\dfrac1{p}$$, $$\prod_{n=2}^{\infty}\dfrac1{1-p^{-n}} =\sum_{n=1}^{\infty} \dfrac{p^{-2n}}{\prod_{k=1}^n(1-p^{-k})}$$.

That's as far as I can go.

• $\sum_{m=2}^\infty \log\zeta(m) =\sum_{m=2}^\infty \sum_{p^k} \frac{p^{-mk}}{k} =\sum_{p^k} \sum_{m=2}^\infty \frac{p^{-mk}}{k}=\sum_{p^k} \frac{1}{k} \frac{p^{-2k}}{1-p^{-k}}$ which doesn't simplify May 24 '19 at 0:38

Marco's answer is about how to compute $$v$$ with desired precision.

My answer is to provide the meaning of the number. Let $$A_n$$ be the number of nonisomorphic abelian groups of order $$n$$. A result by Erdos and Szekeres shows that

$$\sum_{n\leq x} A_n = v x + O(\sqrt x),$$ where $$v=\prod_{k=2}^{\infty} \zeta(k)$$. Therefore, we can say that $$v$$ is the mean value of the number of nonisomorphic abelian groups of order $$n$$.

The proof can be done in two parts. The steps are outlined in Montgomery and Vaughan's Multiplicative Number Theory I. (2.1 #19).

(a) Show that $$\sum_{n=1}^{\infty} A_n n^{-s}=\prod_{k=1}^{\infty} \zeta(ks)$$.

(b) Show that $$\sum_{n\leq x} A_n = vx + O(\sqrt x)$$ where $$v=\prod_{k=2}^{\infty} \zeta(k)$$.

Proof of (a).

For any $$n$$, consider prime factorization of $$n$$. For each $$p^e||n$$, consider a partition $$e=\lambda_1+\lambda_2+\cdots+\lambda_r$$ with $$\lambda_i\leq \lambda_j$$ for $$i\leq j$$. Let this correspond to the $$p$$-primary part $$\mathbb{Z}/p^{\lambda_1}\mathbb{Z}\times\mathbb{Z}/p^{\lambda_2}\mathbb{Z}\times\cdots\times\mathbb{Z}/p^{\lambda_r}\mathbb{Z}$$ of an abelian group of order $$n$$.

Thus, $$\sum_{n=1}^{\infty}A_n n^{-s}$$ should be \begin{align} \prod_p \prod_{k=1}^{\infty} \left( 1- \frac1{p^{ks}}\right)^{-1}&=\prod_p\left(1-\frac1{p^s}\right)^{-1}\left(1-\frac1{p^{2s}}\right)^{-1}\cdots \left(1-\frac1{p^{ks}}\right)^{-1} \cdots \\ &=\prod_{k=1}^{\infty} \zeta(ks), \ \ \Re(s)>1.\end{align}

Proof of (b).

By (a), we have $$\sum_{n=1}^{\infty} A_n n^{-s} = \zeta(s) \zeta(2s) F(s)$$, where $$F(s) = \prod_{k=3}^{\infty} \zeta(ks)= \sum_{n=1}^{\infty} f(n)n^{-s}$$, $$\Re(s)>1/3$$. Then \begin{align} \sum_{n\leq x} A_n &= \sum_{mdk\leq x} \delta_2(d) f(m)\\ &=\sum_{m\leq x} f(m) \sum_{d\leq \frac x{m}}\delta_2(d)\sum_{k\leq \frac x{md}} 1\\ &=\sum_{m\leq x} f(m)\sum_{d\leq \frac x{m}}\delta_2(d)\left(\frac{ x}{md}+O(1)\right)\\ &=\sum_{m\leq x}f(m)\frac xm \left(\zeta(2)+O(\sqrt{\frac mx})\right)+O(\sqrt x)\\ &=xF(1)\zeta(2)+O(\sqrt x). \end{align} Here, $$\delta_2(d)=1$$ if $$d$$ is a square, and $$0$$ otherwise. The result follows by $$v=F(1)\zeta(2)$$.

• Very nice. This is an answer I could never have come up with. Have you checked the encyclopedia of integer sequences for the $A_n$ for other meanings? May 25 '19 at 14:07
• Yes there is. A000688. May 25 '19 at 16:25
• mathoverflow.net/questions/230960/… This posting has more information on the finer asymptotic's. May 25 '19 at 16:52
• I'll leave this for anyone else who didn't know this double bar notation. $a^b||c \iff a^b|c, a^{b+1} \nmid c$ May 29 '20 at 20:59

You can compute this product with arbitrary precision. From the well known inequality $$\zeta\left(s\right)-\sum_{n=1}^{N}\frac{1}{n^{s}}<\frac{N^{1-s}}{s-1},\,s>1$$ we have $$\sum_{n>N}\log\left(\zeta\left(n\right)\right)<\sum_{n>N}\left(\zeta\left(n\right)-1\right)<\sum_{n>N}\frac{1}{2^{n}}\left(1+\frac{2}{n-1}\right)<2^{-N}\left(1+\frac{2}{N}\right).$$