Infinite product Π (n^3-1)/(n^3+1)

[marcelB612]

On Wolfram MathWorld the following infinite product is given as example 10:

http://mathworld.wolfram.com/InfiniteProduct.html

\prod\limits_{n = 2}^\infty {\frac{{n^3 - 1}}{{n^3 + 1}}} = \frac{2}{3}

It is given along with examples 9, 11, 12, and 13 as being in the same class, and yet its solution is remarkably simpler than any of the others.

I'm very curious as to why this is, and how one would go about analytically proving the identity on Wolfram for example 10. There must be some simple trick that I'm missing that would give a simple formula for the k^th partial product, but I'm just not seeing how to get that k^th partial product formula out of it.

Any help?

Some algebraic manipulations that might help the thinking process:

\frac{n^3-1}{n^3+1} = 1 - \frac{2}{n^3+1} = \frac{n^3}{n^3+1} - \frac{1}{n^3+1}p.s. A little bit of googling came up with this homework problem, which is pretty much the same as my question from an '05 Complex Analysis course: http://math.georgiasouthern.edu/~asills/teach/spr05/infprod.pdf , 1.c is the problem in question.

 

[marcelB612]

Well, I solved it, sorry for the premature post.

SPOILER ALERT:

It's fairly simple, actually, just factor out the (n-1) in the numerator and the (n+1) in the denominator, then do the partial products for (n-1)/(n+1) and (n^2+n+1)/(n^2-n+1) separately.

Note that (n+1) = ((n+2)-1) and that n^2+n+1 = (n+1)^2-(n+1)+1, so both terms cancel out.

(n-1)/(n+1) gives you a 2, (n^2+n+1)/(n^2-n+1) gives you a 1/3. So your product ends up 2/3.