Õ
z(x)

Convergent

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

Is

¥
Õ
x=2
z(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)z(s)=1-1/2^s+1/3^s-1/4^s (Dirichlet's b function).

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

Hence (1-1/2^s-1)z(s) < 1 and it isn't hard to prove that z(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 n
å
i=1
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 ³ 1 we have ln(x) £ x-1. Therefore:

ln( r=N
Õ
r=2
z(r)) £ N
å
r=2
(z(r)-1)= N
å
r=2
¥
å
s=2
1/s^r= ¥
å
s=2
N
å
r=2
1/s^r £ ¥
å
s=2
1/(s²-s)=1

So the sequence x_n=z(2)¼z(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