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Last August I posted this on mathoverflow: http://mathoverflow.net/questions/71856/a-serendipitous-riemann-identity. I show the (slightly revised) equation below:

$$\zeta (3)=\frac{2\pi^4}{315} \prod _{n=1}^{\infty } \left(\frac{1}{(p_n){}^2-p_n}+1\right)$$

Since the constant, $\frac{2\pi^4}{315}$ contains $\pi$, which is known to be transcendental, wouldn't this prove that $\zeta(3)$ is transcendental?

I have calculated the product through the first million primes and Mathematica's Element[product,Rationals] returns True. Also, I built a continued fraction of 18,500,045 elements.

The product converges to http://oeis.org/A082695

A paper that uses the product: http://jtnb.cedram.org/cedram-bin/article/JTNB_2004__16_1_107_0.pdf

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1 Answer 1

up vote 19 down vote accepted

No, because an infinite product of rationals is not necessarily rational.

For instance, $$\prod_{n=1}^\infty \left(1-\frac{1}{4n^2}\right)=\frac{2}{\pi}$$

is not rational.

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Wikipedia states the a number like $4\pi$ is transcendental. My product is made up of numbers like$\left\{\frac{3}{2}*\frac{7}{6}*\frac{21}{20}\right\}$. Would this do it? –  Fred Kline Apr 20 '12 at 8:19
11  
No, it wouldn't. It doesn't matter where your numbers come from - an infinite product of rational numbers need not be rational. (It might be, but in general it won't. If it is then you need to prove it.) –  Bruno Joyal Apr 20 '12 at 8:24
11  
What Bruno is telling you is that $(\pi)((3/4)(15/16)(35/36)\dots)=2$, from which you would like to conclude that 2 is transcendental. –  Gerry Myerson Apr 20 '12 at 12:53
1  
For the whole product or the partial products? Try it with my product and tell me what it gives you. In any case that's not an acceptable "proof" –  Bruno Joyal Apr 21 '12 at 0:00
2  
@Bruno, you are right. At convergence, your product is not Rational and not Algebraic. At intermediate steps, it is both. –  Fred Kline Apr 21 '12 at 20:47

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