Prod[1-(n+1)^3, n=1,...,Inf]

[David W. Cantrell]

----- Original Message ----- 
From: "Eric W. Weisstein" <eww at wolfram.com>
To: "Dean Hickerson" <dean at math.ucdavis.edu>
Cc: <seqfan at ext.jussieu.fr>
Sent: Tuesday, April 18, 2006 15:00
Subject: Re: Prod[1-(n+1)^3, n=1,...,Inf]


> On Mon, 17 Apr 2006, Dean Hickerson wrote:
>
>> Zak Seidov wrote:
>>
>>> Prod[1-(n+1)^3,n=1,...,Inf]=
>>> Cosh[(Sqrt[3]*Pi)/2]/(3*Pi)=
>>> 0.809396597366290109578680478726.
>>
>> The exponent should be -3, not 3.  (And it could be simplified to
>> prod[1-n^-3, n=2,...,inf].)
>>
>>> Is this cons known?
>
> The product is very similar to 
> http://mathworld.wolfram.com/InfiniteProduct.html eqn (23) and might 
> very well have a similar formula for arbitrary power.  Any takers?

Sure. But before I get to that, I see that (23) is not attributed to 
anyone. However, I assume that it was sent to you by Paul Abbott based 
on my post 
<http://groups.google.com/group/sci.math/msg/de92321b4a19e8e8> on 2006 
Mar. 29 in the sci.math thread "infinite product". Unless that general 
result was stated previously elsewhere (which it certainly may have 
been), shouldn't it be attributed to me?

Now to your current question, Eric. (I had already worked out the 
answer before I saw your post here, BTW.)

Product[1 - 1/n^p, {n, 2, Infinity}]

simplifies, if p is odd, to

1/(p * Product[Gamma[- (-1)^(j*(1 + 1/p))], {j, 1, p - 1}])

and, if p is even, to the elementary


Product[Sin[Pi*(-1)^((2*j)/p)]/(Pi*I), {j, 1, p/2 - 1}] / p.

I will also be posting this general result to the previously mentioned 
sci.math thread soon.

Regards,

David W. Cantrell