I have generalized the product from this thread:

Let $s=2n+1$ for $n\ge1$, $$\zeta (s)=\frac{\zeta (2 s)}{\zeta (2)} \prod _{n=1}^{\infty } \frac{\sum _{i=0}^{s-1}(-p_n){}^i}{(p_{n}-1)p_{n}^{s-2}}$$

This is a $\zeta(s)$ identity for each odd $s$.

Does it have a name? I can only find $\zeta(3)$ in the literature.


1 Answer 1


Using the Euler product,

$$\zeta(s)=\prod_p \big(1-p^{-s}\big)^{-1},$$

this identity amounts to rearranging polynomials. Indeed, we have

$$\frac{\zeta(s)\zeta(2)}{\zeta(2s)}=\prod_p \frac{(1-p^{-2s})}{(1-p^{-s})(1-p^{-2})}=\prod_p\frac{1+p^{-s}}{1-p^{-2}}$$


$$\prod_p \frac{1-p\cdots\pm p^{s-1}}{p^{s-1}-p^{s-2}}=\prod_p \frac{\frac{1-(-p)^s}{1-(-p)}}{p^s(1/p)(1-1/p)}=\prod_p \frac{1+p^{-s}}{1-p^{-2}}$$

when $s$ is an odd integer. The actual form of the identity doesn't look familiar to me, and I don't see any obvious use for it. It looks like it might have been the result of rearranging the terms for the very sake of rearranging them and getting something new, not any particular external goal, and at least to me there's nothing especially aesthetically appealing about it so I'm going to go out on a limb and say no, this identity doesn't have a name.

This sort of rearrangement is standard fare when deducing relations between $L$-functions with simple Euler products. For example, one can decompose $L$-functions associated to quadratic fields using quadratic reciprocity. That's relatively deeper in number theory, though; simpler examples of applying this method can be seen in this Wikipedia section.

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    $\begingroup$ For $s=7$,$$\frac{(p_n){}^6-(p_n){}^5+(p_n){}^4-(p_n){}^3+(p_n){}^2-p_n+1}{(p_n){}^6-(p_n){}^5}$$ I feel this has a certain beauty. $\endgroup$ Commented Apr 22, 2012 at 7:38
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    $\begingroup$ For $s=19$ $$(p_n)^{18}-(p_n)^{17}+(p_n)^{16}-(p_n)^{15}+(p_n)^{14}-(p_n)^{13}+(p_n)^{12}-(p_n)^{11}\\ +(p_n)^{10}-(p_n)^9+(p_n)^8-(p_n)^7+(p_n)^6-(p_n)^5+(p_n)^4-(p_n)^3+(p_n)^2-p_n+1$$ divided by $$(p_n){}^{18}-(p_n){}^{17}$$ $\endgroup$ Commented Apr 22, 2012 at 7:41
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    $\begingroup$ my motivation is to determine the structure so I can determine if the product is rational. The structure might lend itself to prove one---you prove them all. $\endgroup$ Commented Apr 22, 2012 at 7:46
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    $\begingroup$ I'm looking for odd zeta values to be transcendental because I multiply my rational product by the $\frac{\zeta(2s)}{\zeta(2)}$ ratio. I'm still in the exploratory stage. $\endgroup$ Commented Apr 22, 2012 at 8:01
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    $\begingroup$ Careful what you mean by "rational product." As discussed in the previous question, an infinite product of rational numbers isn't necessarily rational. I don't want to be mean, butting into your personal exploration and all, I just don't want to see you churn out a lot of effort predicated on a misunderstanding (which may or may not be present here, I can't say for certain what you're doing). $\endgroup$
    – anon
    Commented Apr 22, 2012 at 8:04

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