MoebiusMu product Some time back, I asked this question about 
$$\prod _{n=1}^{\infty } n^{\mu (n)}.$$
Recently, I found this numeric convergence: $0.6784223987077668596042536007\dots$, and when rationalized, found this:  $\frac{5040}{7429}$.
Factoring the fraction: $\lbrace\lbrace2, 4\rbrace, \lbrace3, 2\rbrace, \lbrace5, 1\rbrace, \lbrace7, 1\rbrace, \lbrace17, -1\rbrace, \lbrace19, -1\rbrace, \lbrace23, -1\rbrace\rbrace$, where $\lbrace prime, exponent\rbrace$, we get this sequence.
I like the fact that the sequence is finite and am wondering what it is doing in my fraction.
Is this coincidence? Or, what?
 A: The product is clearly divergent (the Möbius function takes the values $-1,\,0,\,1$ and we will multiply or divide by $n$ for an infinite number of $n$) but we may observe that :
$$\tag{1}\log\left(\prod _{n=1}^{\infty } n^{\mu (n)}\right)=\sum_{n=1}^\infty \mu (n)\log(n)$$ 
and it is well known that (at least for $\Re(s)>1$) we have : 
$$\tag{2}\sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\frac 1{\zeta(s)}$$
This is true too for $s=1$ but according to G. Hardy the proof of this (given by Landau I think) is equivalent to the PNT.
From this we may deduce, using $n^{-s}=e^{-s\log(n)}$ and supposing that $(2)$ may be extended for $s\in [0,1)$, that :
$$\sum_{n=1}^\infty \mu (n)\log(n)=-\lim_{s\to 0^+} \frac d{ds}\sum_{n=1}^\infty \mu(n)\,e^{-s\log(n)}=\frac {\zeta'(0)}{\zeta(0)^2}=\frac{-\log(2\pi)/2}{(-1/2)^2}=-2\,\log(2\pi)$$
so that, if an answer has to be provided, I would suggest :
$$\prod _{n=1}^{\infty } n^{\mu (n)}=\frac 1{4\pi^2}$$
The method used here is named zeta function regularization and was found pretty efficient to get actual values in physics (especially in QFT).
(Hmmm I notice now, following your link, that the same answer was provided in MO ; sorry for the repetition...)
