TITLE: $\prod_{n=1}^{\infty} n^{\mu(n)}=\frac{1}{4 \pi ^2}$
QUESTION [36 upvotes]: When I tested this in Mathematica, I had expected it to say it did not converge.  However, I got this:  
$$\prod_{n=1}^\infty n^{\mu(n)}=\frac{1}{4 \pi ^2}$$
Note: this is the reciprocal of (3) zeta-regularized product over all primes.  
This indicates there is a slight skew toward the square-free numbers that have an odd number of factors.
I haven't been able to find this in the literature, but my guess is that it is already known.
Can someone point me in the right direction?
Edit: We have been told it is a bug. Mathematica V.10 does not produce a symbolic answer. 
Edit 2 We can make this work by creating a function $f(k)$ that returns the $k$-th square-free number (in the constructive? order).  We can find a few ideas from A019565.  
$$\prod_{n=1}^{\infty}f(n)^{\mu(f(n))}$$
$$\sum_{n=1}^\infty \mu(f(n))$$
At the steps when $n>2$ is a power of $2$, sum$=0$ and prod$=1$.
Also, at those steps, $f(n)$ is the primorial with $p_k$ as the greatest factor, with $k=\frac{\log(n)}{\log(2)}$  
There is a revised question over here. It is a variation to Edit 2.

REPLY [12 votes]: This is a discussion following from David Speyer about Cesaro summability of the series. 
This series is not Cesaro summable in any order. 
In "The General Theory of Dirichlet Series" by Hardy and Riesz. Chapter 4.
Let $\kappa\geq 0$. $0\leq \lambda _0<\lambda_1<\cdots<\lambda_n\rightarrow\infty$, and write
$$
A_{\lambda}^{\kappa}(w) = \sum_{\lambda_n \textrm{<}w}(w-\lambda_n)^{\kappa}a_n.
$$
\bf Summability $(\lambda,\kappa)$. \rm If $A_{\lambda}^{\kappa}(w)\sim Aw^{\kappa}$ as $w\rightarrow\infty$, then we say that $\sum a_n$ is summable $(\lambda,\kappa)$ to sum $A$.
The second consistency theorem(we do not need first consistency theorem here) states that 
If $\sum a_n$ is summable $(l,\kappa)$ where $l_n=e^{\lambda_n}$, then it is summable $(\lambda,\kappa)$ to the same sum. In particular, Cesaro summability of order $\kappa$ implies $(\log n, \kappa)$ summability to the same sum. 
We consider the formula
$$
\sum_{n=1}^{\infty}a_ne^{-\lambda_ns}=\frac{1}{\Gamma(\kappa+1)}\int_0^{\infty}s^{\kappa+1}e^{-s\tau}A_{\lambda}^{\kappa}(\tau)d\tau.$$
This is originally valid if $\sigma> max(0,\sigma_c)$ where $\sigma_c$ is the abscissa of convergence of the left series. On the other hand, the right allows an analytic continuation of the function represented by the series on the left up to $\sigma>0$. 
What we use here is $a_n = \mu (n)\log n$ and  $\lambda_n = \log n$ . 
Hence, assuming Cesaro summability of any order $\kappa$, will allow 
$$-\frac{\zeta'(s)}{\zeta^2(s)}$$
to have an analytic continuation to $\sigma>0$ which contradicts the fact that $\zeta(s)$ has zeros on the critical line.