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When I run this,

Product[n^MoebiusMu[n],{n,1,Infinity}]  

I get $\frac{1}{4 \pi^{2}}$

Over on Math Overflow they are saying it shouldn't happen. So, how do I determine if it really converges?

Edit If we could re-order the square-free numbers into sequence by their greatest factors, this product would converge absolutely to $1$. This sum:

Sum[MoebiusMu[n],{n,i,Inifinty}]  

would converge absolutely to $0$ with the same re-ordering.

Edit 2

ListPlot[Table[n^MoebiusMu[n], {n, 1, 1000}]]  

MoebiusMu Plot

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I don't know why it converges. But, running Tally[MoebiusMu /@ Range[1, 10^7]] gives {{1, 3040164}, {-1, 3039127}, {0, 3920709}} indicates that the even factors have a slight edge, and would likely not converge unless some interesting cancellation was occurring. I don't know enough about MoebiusMu to know if the interesting cancellation is correct, or not. –  rcollyer Mar 12 '13 at 12:55
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BTW, I love the hamster comment on your user profile. –  rcollyer Mar 12 '13 at 12:56
    
I see. For any finite range, the even factors include nothing near the top of the range, while the odd factors do. That would cause it to converge. How did you generate your ranges for the terms in the numerator and the denominator, empirically? –  rcollyer Mar 12 '13 at 13:15
2  
David's comment on mathoverflow is absolutely correct, there's no way this product can converge. –  Mark McClure Mar 12 '13 at 13:21
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The property holds for arbitrary products, whether the terms are negative or not, they can as well be complex numbers. The reduction to sums via logs is a red herring. If $\prod_na_n$ converges to $a$, then the sequence $A_N=\prod_{n<N}a_n$ has limit $a$, and so does its shift, hence $\lim_na_n=\lim_n(A_{n+1}/A_n)=a/a=1$. –  Emil Jeřábek Mar 12 '13 at 14:33
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1 Answer 1

up vote 6 down vote accepted

I get this result as well in both V8 and V9.

Product[n^MoebiusMu[n], {n, 1, Infinity}]

(* Out: 1/(4*Pi^2) *)

It's a simple fact, though, that an infinite product can converge to a non-zero value only if the general term tends to 1. As MoebiusMu takes each of the values $\pm 1$ (as well as zero) infinitly often, this product simply can't converge.

We can also relate an infinite product to an infinte sum by applying the logarithm - for this particular product:

$$\log\left(\prod_{n=1}^{\infty} n^{\mu(n)}\right) = \sum_{n=1}^{\infty} \mu(n)\log(n).$$

So, maybe we should examine the following sum sum.

Sum[MoebiusMu[n] Log[n], {n, 1, Infinity}]

(* Out: -2*Log[2*Pi] *)

Well, that didn't help! Perhaps these bugs are related? At least the convergence of infinite sums is a bit more elementary; clearly, $\mu(n)\log(n)$ does not converge to zero. Perhaps, we could try a numerical test:

NSum[MoebiusMu[n] Log[n], {n, 1, Infinity}]

(* Out: -0.387985 *)

Damn. Well, let's at least examine some partial sums.

ListLinePlot[Accumulate[Table[MoebiusMu[n] N@Log[n], {n, 1, 1000}]]]

enter image description here

Well, that certainly doesn't look convergent - because, of course, it isn't.

Maybe @DanielLichtblau could shed some light on this. Most likelly I think, he'll say it's just a bug.

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I just added a plot to the OP. Even count square-frees are on the slope and the odd count square-frees tend to zero along the bottom. –  Fred Kline Mar 12 '13 at 14:42
    
My guess is somehow it gives s regularized result. Or else it is just a bug, I don't know. I'm pretty sure the product result just comes from exponentiating the sum result. –  Daniel Lichtblau Mar 12 '13 at 15:17
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I'm told it's a bug and will be treated as such. –  Daniel Lichtblau Mar 12 '13 at 15:36
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