MoebiusMu product

Question

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?

@draks...I took it to WorkingPrecision->500 and then to 5000 and got the same rationalization. — Fred Daniel Kline, Mar 12 '14 at 15:13
@draks...I get the same rationalization from WorkingPrecision->14 and up. — Fred Daniel Kline, Mar 12 '14 at 15:20
The accepted answer in the MO question you linked to begins with "So of course it does not converge." Are you saying it does, or am I misunderstanding something? — Daniel R, Mar 12 '14 at 15:48
@DanielR, It doesn't converge in the usual sense. The sum is a cancellation of the +1s and the -1s. The product is the cancellation of the whole numbers (exponent +1) and fractional numbers (exponent -1). — Fred Daniel Kline, Mar 12 '14 at 16:11
$\prod_{i=1}^{26} n^{\mu(n)} = \frac{5040}{7429}$ but 26 is still pretty far from infinity — Myself, Mar 12 '14 at 16:13
$\prod_{i=1}^{100} n^{\mu(n)} = \frac{84040181877991379040000}{3749562977351496827}\cong 22413$. — Myself, Mar 12 '14 at 16:16
@Fred. You're not talking to yourself but to myself now, right ? :-) Poor joke, I know. However, could you elaborate a bit on how you calculated the numerical convergence? Keen to reproduce it. Note that $5040 = 7!$. — Agno, Mar 12 '14 at 19:37
@Agno, NProduct[n^(MoebiusMu[n]), {n, 1, Infinity}, WorkingPrecision -> 500] Rationalize[%] — Fred Daniel Kline, Mar 12 '14 at 20:29

Raymond Manzoni · Accepted Answer · 2014-03-13T07:20:37.763

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...)

On other thread, David Speyer states that the use of $\zeta(0)$ is apparently the approach also taken by Mathematica. However, I think the $\Re(s)>1$ should be $\Re(s)\neq1$, but I don't have the chops to explain my intuition. — Fred Daniel Kline, Mar 12 '14 at 23:32
@FredKline: I share David Speyer's opinion. The temptation (for me too) is to use analytic extension and accept it everywhere (giving infinity at the zeros...) :-). There was an earlier question where the value of $(2)$ at $s=0$ was used (not the derivative). — Raymond Manzoni, Mar 12 '14 at 23:36
Concerning the case $s=1$ some links are given here. The value at $s=0$ was implicitely used here and is clearly linked to the Mertens function. Fine continuation! — Raymond Manzoni, Mar 12 '14 at 23:47

MoebiusMu product

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