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

Joerg Arndt arndt at jjj.de
Tue Apr 18 16:26:16 CEST 2006

```* Eric W. Weisstein <eww at wolfram.com> [Apr 18. 2006 16:08]:
> 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?
>
> Cheers,
> -E
>

product(1-1/n^2,n=2..infinity) == 1/2

Many relations similar to your (63) can be found
via the basic trigonometric formulas (hand typed
from old scribblings, so beware correctness):

sin(z)/z
= P{n=1..infty}{cos(z/2^n)}
= P{n=1..infty}{1/(1+tan^2(z/2^n))}
= P{n=1..infty}{(1-tan^2(z/2^n))^(2^n-1)}
= 2*P{n=1..infty}{2^n/(tan^(2^n)(z/2^n))}

tan(z)/z
= P{n=1..infty}{1/(1-tan^2(z/2^n))}
= 1/(1-Sum{n=1..infinity}{z/2^n * tan(z/2^n)})

Convergence?

```