$Π ζ (x)$ Convergent

By Brad Rodgers on Wednesday, February 06, 2002 - 10:40 pm:

Is

$Π_{x = 2}^{\infty} ζ (x)$

Convergent? If so, does it have a closed form value?

I've worked on this a little, and the product seems to at least be convergent. I can't find a proof though. Any thoughts?

Brad

By David Loeffler on Thursday, February 07, 2002 - 09:25 am:

How's this for an idea:

(1 - 1 / 2^{s - 1}) ζ (s) = 1 - 1 / 2^{s} + 1 / 3^{s} - 1 / 4^{s}

(Dirichlet's

β

function).

This is an alternating series so its sum is $< 1$ .

Hence $(1 - 1 / 2^{s - 1}) ζ (s) < 1$ and it isn't hard to prove that $ζ (s) < 1 + 1 / 2^{s - 2}$ .

Hence the terms of the product are bounded above by an obviously convergent product.

As for what this converges to, that is a more interesting question. The limit of this appears in a surprising way in abstract algebra: for any $n$ there exist abelian groups of order $n$ . Let $r (n)$ be the total number of distinct isomorphism classes of abelian groups of order $n$ ; then the average order of $r (n)$ is this product (in the sense that it is equal to the limit of $1 / n \sum_{i = 1}^{n} r (i)$ ). So on average there are about this many abelian groups of order $n$ for any $n$ .

David

By Michael Doré on Thursday, February 07, 2002 - 02:03 pm:

Alternatively, to prove convergence, note that for $x \geq 1$ we have $\ln (x) \leq x - 1$ . Therefore:

$\ln (Π_{r = 2}^{r = N} ζ (r)) \leq \sum_{r = 2}^{N} (ζ (r) - 1) = \sum_{r = 2}^{N} \sum_{s = 2}^{\infty} 1 / s^{r} = \sum_{s = 2}^{\infty} \sum_{r = 2}^{N} 1 / s^{r} \leq \sum_{s = 2}^{\infty} 1 / (s^{2} - s) = 1$

So the sequence $x_{n} = ζ (2) \dots ζ (n)$ is bounded above by $e$ , and it is strictly increasing so it converges (to something less than or equal to $e$ ).

By David Loeffler on Thursday, February 07, 2002 - 02:50 pm:

In fact that's a pretty good upper bound (the limit is about 2.2948565916733137941835158313443) - my method only gives that it converges to something less than e² .

David