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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.

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

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  • $\begingroup$ 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? $\endgroup$ May 25 '19 at 14:07
  • $\begingroup$ Yes there is. A000688. $\endgroup$ May 25 '19 at 16:25
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    $\begingroup$ mathoverflow.net/questions/230960/… This posting has more information on the finer asymptotic's. $\endgroup$ May 25 '19 at 16:52
  • $\begingroup$ I'll leave this for anyone else who didn't know this double bar notation. $a^b||c \iff a^b|c, a^{b+1} \nmid c$ $\endgroup$
    – Mason
    May 29 '20 at 20:59
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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 $$\sum_{n>N}\log\left(\zeta\left(n\right)\right)<\sum_{n>N}\left(\zeta\left(n\right)-1\right)<\sum_{n>N}\frac{1}{2^{n}}\left(1+\frac{2}{n-1}\right)<2^{-N}\left(1+\frac{2}{N}\right).$$

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