Does this $\zeta(s)$ identity have a name? 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.
 A: Using the Euler product,
$$\zeta(s)=\prod_p \big(1-p^{-s}\big)^{-1},$$
this identity amounts to rearranging polynomials. Indeed, we have
$$\frac{\zeta(s)\zeta(2)}{\zeta(2s)}=\prod_p \frac{(1-p^{-2s})}{(1-p^{-s})(1-p^{-2})}=\prod_p\frac{1+p^{-s}}{1-p^{-2}}$$
and
$$\prod_p \frac{1-p\cdots\pm p^{s-1}}{p^{s-1}-p^{s-2}}=\prod_p \frac{\frac{1-(-p)^s}{1-(-p)}}{p^s(1/p)(1-1/p)}=\prod_p \frac{1+p^{-s}}{1-p^{-2}}$$
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.
