I am trying to figure out if the infinite product $$\omega=\frac{5\sqrt{3}}{12}\prod\limits\_{\substack{p\equiv 1\pmod3 \\
p\ge 13}}\left(\frac{p-2}{p-1}\right)\prod\limits\_{\substack{p\equiv 2\pmod3 \\
p\ge 13}}\left(\frac{p}{p-1}\right)$$ is asymptotically equal to the infinite product $$c=\frac{5775}{2592\pi}\prod\limits\_{\substack{p\equiv 1\pmod3 \\
p\ge 13}}\left(\frac{p(p-2)}{(p-1)^2}\right)\prod\limits\_{\substack{p\equiv 2\pmod3 \\
p\ge 13}}\left(\frac{p^2}{p^2-1}\right)$$.


So, I reformulated the products as $$\omega(x)=\frac{5\sqrt{3}}{12}\prod\limits\_{\substack{p\equiv 1\pmod3 \\
13\leq p\leq x}}\left(\frac{p-2}{p-1}\right)\prod\limits\_{\substack{p\equiv 2\pmod3 \\
13\leq p\leq x}}\left(\frac{p}{p-1}\right)$$ and $$c(x)=\frac{5775}{2592\pi}\prod\limits\_{\substack{p\equiv 1\pmod3 \\
13\leq p\leq x}}\left(\frac{p(p-2)}{(p-1)^2}\right)\prod\limits\_{\substack{p\equiv 2\pmod3 \\
13\leq p\leq x}}\left(\frac{p^2}{p^2-1}\right)$$.
Then, taking the quotient we get
$$\frac{\omega(x)}{c(x)}=\frac{216\sqrt{3}\pi}{1155}\prod\limits\_{\substack{p\equiv 1\pmod3 \\
13\leq p\leq x}}\left(1-\frac{1}{p}\right)\prod\limits\_{\substack{p\equiv 2\pmod3 \\
13\leq p\leq x}}\left(1+\frac{1}{p}\right)$$ $$=\frac{216\sqrt{3}\pi}{1155}\prod\limits\_{\substack{p\equiv 2\pmod3 \\
13\leq p\leq x}}\left(1-\frac{1}{p^2}\right)\prod\limits\_{\substack{p\equiv 1\pmod3 \\
13\leq p\leq x}}\left(1-\frac{1}{p}\right)\prod\limits\_{\substack{p\equiv 2\pmod3 \\13\leq p\leq x}}\left(\frac{1}{1-\frac{1}{p}}\right).$$


As $x\to\infty$, from [A. Languasco's paper](https://core.ac.uk/reader/81188410), we observe that the last two products in the quotient above are asymptotically equal. So, the limit of the quotient depends on the product $$\frac{216\sqrt{3}\pi}{1155}\prod\limits\_{\substack{p\equiv 2\pmod3 \\
13\leq p\leq x}}\left(1-\frac{1}{p^2}\right)$$, but I don't know if this converges to $1$ or not as $x\to\infty$. I can see that the last product (except the constant) has something to do with $\frac{1}{\zeta(2)}$ but I don't know if it's actually smaller or greater than that.


I am essentially trying to check if eventually the products would be same or not?


I would really appreciate it if somebody could provide some hints or ideas on how to progress from here.



Asymptotic equivalence of two infinite products of prime numbers in residue classes
I don't know why you are restricting the products to $p \geq 13$ or where the factor $5\sqrt{3}/12$ is coming from. I am going to ignore that and discuss the following product over all primes $p$:
$$
C = \prod\_{p}\left(1- \frac{\chi(p)}{p-1}\right)
$$
where the terms in the product are in order of increasing $p$ and $\chi$ is the nontrivial Dirichlet character mod $3$, so $\chi(p) = 1$ if $p \equiv 1 \bmod 3$, $\chi(p) = -1$ if $p \equiv 2 \bmod 3$, and $\chi(3) = 0$.


When $p \equiv 1 \bmod 3$ we have
$$
1- \frac{\chi(p)}{p-1} = 1 - \frac{1}{p-1} = \frac{p-2}{p-1}
$$
and when $p \equiv 2 \bmod 3$ we have
$$
1- \frac{\chi(p)}{p-1} = 1 + \frac{1}{p-1} = \frac{p}{p-1}
$$
When $p = 3$, $\chi(p) = 0$, so
$$
1- \frac{\chi(p)}{p-1} = 1.
$$


Therefore the factors in $C$ are the same as what you wrote, but you broke up the product into separate products over $p \equiv 1 \bmod 3$ and $p \equiv 2 \bmod 3$ (I ignore the condition $p \equiv 13$). That is a subtle issue because *your products don't converge*. I am multiplying the terms at all primes in $C$ together in increasing order, not using separate products depending on $p \bmod 3$.


As an analogue, consider the alternating harmonic series $S = \sum\_{n \geq 1} (-1)^{n-1}/n$. This converges (and equals $\log 2$), but you can't write
$S$ with two separate sums over odd and even $n$:
$$
S \not= \sum\_{{\rm odd } \ n} \frac{1}{n} - \sum\_{{\rm even } \ n} \frac{1}{n}
$$
because those separate sums individually do not converge.


Returning to $C$, we want to insert factors into each term to improve the convergence (to make it absolutely convergent). To do that, write
$$
1 - \frac{\chi(p)}{p-1} = 1-\frac{\chi(p)/p}{1-1/p} = 
1 - \frac{\chi(p)}{p} + \frac{\chi(p)}{p^2} + O\left(\frac{1}{p^3}\right)
$$
by expanding $1/(1-1/p)$ into a geometric series in powers of $1/p$. Since $1-\chi(p)/p \sim 1$ as $p \to \infty$, dividing by $1-\chi(p)/p$ tells us
$$
\frac{1-\chi(p)/(p-1)}{1-\chi(p)/p} = 1 + O\left(\frac{1}{p^2}\right),
$$
so the product
$$
\prod\_{p} \left(\frac{1-\chi(p)/(p-1)}{1-\chi(p)/p}\right)
$$
is absolutely convergent.


Now we can rewrite $C$:
$$
C = \prod\_p \left(1 - \frac{\chi(p)}{p-1}\right) = \prod\_p \left(1-\frac{\chi(p)}{p}\right)\left(\frac{1-\chi(p)/(p-1)}{1-\chi(p)/p}\right),
$$
where the product is in order of increasing $p$. Can we split apart this product?


Yes! When ${\rm Re}(s) > 1$, the $L$-function of $\chi$ has the absolutely convergent Euler product representation
$$
L(s,\chi) = \prod\_{p} \frac{1}{1-\chi(p)/p^s}
$$
and it can be shown, with some nontrivial work, that this product remains valid on the line ${\rm Re}(s) = 1$ *provided* the product is taken in order of increasing $p$ (on that line the product is no longer absolutely convergent). Taking $s = 1$ and reciprocating,
$$
\prod\_p \left(1 - \frac{\chi(p)}{p}\right) = \frac{1}{L(1,\chi)}.
$$
Feeding that into the formula for $C$ above,
$$
C = \frac{1}{L(1,\chi)}\prod\_p \frac{1-\chi(p)/(p-1)}{1-\chi(p)/p}.
$$


It can be shown that $L(1,\chi) = \pi/(3\sqrt{3})$, so
$$
C = \frac{3\sqrt{3}}{\pi}\prod\_p \frac{1-\chi(p)/(p-1)}{1-\chi(p)/p},
$$
where the $p$-th factor in this product is $1+O(1/p^2)$.


What is the $p$-th factor here? When $\chi(p) = 1$,
$$
\frac{1-\chi(p)/(p-1)}{1-\chi(p)/p} = \frac{1-1/(p-1)}{1-1/p} = \frac{p(p-2)}{(p-1)^2}
$$
and when $\chi(p) = -1$,
$$
\frac{1-\chi(p)/(p-1)}{1-\chi(p)/p} = \frac{1+1/(p-1)}{1+1/p} = \frac{p^2}{p^2-1}.
$$
And when $\chi(p) = 0$ (namely when $p = 3$,
$$
\frac{1-\chi(p)/(p-1)}{1-\chi(p)/p} = 1.
$$


Because these terms are $1 + O(1/p^2)$, the order of multiplication now does not matter and we can split apart this product depending on $p \bmod 3$:
$$
C = \frac{3\sqrt{3}}{\pi}\prod\_{p\equiv 1 \bmod 3}\frac{p(p-2)}{(p-1)^2} \prod\_{p \equiv 2 \bmod 3} \frac{p^2}{p^2-1}.
$$


On the MO page [here](https://mathoverflow.net/questions/31150/calculating-the-infinite-product-from-the-hardy-littlewood-conjecture-f?rq=1) I discuss such a product with $\chi$ replaced by a more general Legendre symbol.


