When I tested this in *Mathematica*, I had expected it to say it did not converge. However, I got this:

$$\prod_{n=1}^{\infty} n{}^{\mu(n)}=\frac{1}{4 \pi ^2}$$

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.**

**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 $k=\frac{\text{log}(n)}{\text{log}(2)}$

**Edit 3**

Instead of changing the sequence (as in Edit 2), we can try this: With $\epsilon = 1. \times 10^{-7}\text{ or smaller}$
$$
\sum _{n=1}^{\infty } \frac{\mu(n)+\epsilon }{n^s}/\prod _{n=1}^{\infty } n^{\mu(n)\epsilon }
\equiv \frac{1}{\zeta(s)}\pm\epsilon
$$

The $\sum / \prod$ work in tandem to remove the wheat from the chaff.

When $s=0$, we get $-2\pm\epsilon$, though *Mathematica* complains about it. *fixed below*

**Edit 3a**

Removed the generating function because $(2 \pi )^{2 \epsilon }$ could be any positive number. I have no more questions.