What is the confidence limit on this convergence?

13

1

Bug introduced in 7.0 and fixed in 10.0.0


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

Fred Kline

Posted 2013-03-12T12:30:11.450

Reputation: 2 226

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 – 2013-03-12T12:55:39.217

1BTW, I love the hamster comment on your user profile. – rcollyer – 2013-03-12T12:56:11.603

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 – 2013-03-12T13:15:12.680

2David's comment on mathoverflow is absolutely correct, there's no way this product can converge. – Mark McClure – 2013-03-12T13:21:08.630

@MarkMcClure normally, I'd agree with you, but have you looked at the finite products? (and, no I have no experience with infinite products) Run Module[{mu = MoebiusMu /@ Range[1, 1000]}, Tally@Flatten[ FactorInteger[#][[All, 1]] & /@ Pick[Range[Length@mu], mu, #]] & /@ {1, -1} ] to get a list of factors in the numerator and denominator with multiplicities. – rcollyer – 2013-03-12T13:26:10.790

After inspecting the intermediate products (which oscillate wildly), I'm inclined to agree with Mark. I think by factoring the numerator and denominator, you lose the forest for the trees. For instance, while the numerator only consists of primes $<=n/3$, they have multiplicities greater than 1, while the denominator only consists of factors with multiplicity 1. I think that would ultimately allow the numerator to over power the denominator. As to why Mathematica gives a finite limit, I cannot say. – rcollyer – 2013-03-12T13:37:13.270

@rcollyer: It follows from the definition that if $\prod_na_n$ converges, then $\lim_{n\to\infty}a_n=1$. No amount of data on finite products can change this trivial fact. – Emil Jeřábek – 2013-03-12T13:37:34.360

@EmilJ you mean converge to a non-zero value, of course. I think I understand what the confusion is (see my prior comment) as to why Mathematica thinks differently, I don't know. – rcollyer – 2013-03-12T13:40:09.790

There is no exception with zero. Please read at least the lead section of http://en.wikipedia.org/wiki/Infinite_product to see what is the definition of convergence for products.

– Emil Jeřábek – 2013-03-12T13:44:57.353

@rcollyer I was on iPhone earlier and couldn't respond in detail. As has been pointed out, we definitely need the terms in the product to converge to 1 in order to get convergence. Alternatively, the sum of the logarithms of the product must converge, and an intuitive analysis is typically simpler in the context. I'll post an answer detailing that shortly. – Mark McClure – 2013-03-12T13:56:47.023

@EmilJ one thing stands out from that page: the theorems apply for a finite number of $a_i < 0$, which isn't the case here. This does not preclude them from applying, just care must taken considering that $\sum \log a_i$ is an alternating series. – rcollyer – 2013-03-12T14:01:39.953

@rcollyer, my n/3 and n/2 boundaries are not consistant. So, I'm deleting those comments. – Fred Kline – 2013-03-12T14:08:32.437

It cannot converge. the product converges iff the sum of logs converges. The nth term in the sum of logs is MoebiusMu[n]*Log[n]. These do not converge in size to zero, ence the sum does not converge. – Daniel Lichtblau – 2013-03-12T14:16:04.857

Okay, what i wrote is correct, but the easier way to say it has alre3ady been stated: for convergence, the factors MUST converge to 1, and they do not. – Daniel Lichtblau – 2013-03-12T14:17:32.013

Since it results in an alternating log series, the logs must monotonically converge to 0, which is the equivalent statement of the factors going to 1. It is interesting to note, that this result is stronger than that initially presented by the wikipedia page on infinite products as it allows for an infinite number of $a_i < 1$.

– rcollyer – 2013-03-12T14:26:20.020

1The 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 – 2013-03-12T14:33:32.417

@EmilJ I wouldn't go so far as calling the reduction to sums a red herring; incomplete, but certainly stronger than what is presented in wikipedia. I can see where it would be difficult to prove it is a general result using the series, though. – rcollyer – 2013-03-12T14:46:48.067

The statement on Wikipedia is formulated in a confusing way, the log criterion also applies to arbitrary products. Fix any branch of logarithm which makes $\log(1)=0$. If infinitely many terms of the product fall on the branch cut, the product diverges, otherwise it converges iff the corresponding $\sum_n\log a_n$ converges. – Emil Jeřábek – 2013-03-12T15:08:33.977

I left an answer over on MO explaining what Mathematica may be thinking. – David E Speyer – 2013-03-12T15:15:39.100

Answers

11

Edit: Returns unevaluated in V10, which is an improvement. Ideally, though, the divergence would be identified.


Orginal response

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.

Mark McClure

Posted 2013-03-12T12:30:11.450

Reputation: 31 084

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 – 2013-03-12T14:42:19.803

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 – 2013-03-12T15:17:22.323

8I'm told it's a bug and will be treated as such. – Daniel Lichtblau – 2013-03-12T15:36:45.880