TITLE: What can be said about $\prod_{s=2}^{\infty} \zeta(s) $?
QUESTION [11 upvotes]: 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.

REPLY [7 votes]: 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 $v=F(1)\zeta(2)$.