Prod[1-(n+1)^3, n=1,...,Inf]
David W. Cantrell
DWCantrell at sigmaxi.org
Tue Apr 18 17:11:58 CEST 2006
----- 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
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