Is there a limit/convergence in that sequence of ratios of primeproducts?

Question

In another question I discuss the function on primes $$ r(p) = { p \over 2^{E_2} 3^{E_3} 5^{E_5} ... q^{E_q} } \tag 1 $$ where $p \in \mathbb P$ and $q$ is the next smaller prime before $p$.
Here $$E_q = { q \over (q-1)^2 } \tag 2 $$ Looking at $p \lt 1e8$ it seems, this could converge to some value above $0.5221...$ but it might as well increase unboundedly with increasing $p$.

Q: Does a limit $ \lim_{p \to \infty} r(p)$ exist? And in case it exists, what value does it assume?

---

(Graphic from linked question): Another graphic with x-scale doubly logarithmic: A graphic showing even more values (up to $p \lt 10^8$) but the upper and lower hullcurves giving some suggestion for convergence:

Answer 1

Yes, your suspicion is right; the limit exists and it converges to $0.5221$.

Let $p_n$ be the $n$-th prime. We have $E_{p_n} = \frac{p_n}{(p_n-1)^2} = \frac{1}{p_n} + \frac{2}{p_n^2} + \frac{3}{p_n^3} + \cdots = \frac{1}{p_n} + f_{p_n}$

where $f_{p_n} = \frac{2}{p_n^2} + \frac{3}{p_n^3} + \cdots$. We have

$$ 2^{E_2} 3^{E_3} ... q^{E_{p_n}} = (2^{\frac{1}{2}} 3^{\frac{1}{3}} ... p_n^{\frac{1}{p_{n}}}) . (2^{f_2} 3^{f_3} ... p_n^{f_{p_n}}) \tag 1 $$

Let us look at the two parts $2^{\frac{1}{2}} 3^{\frac{1}{3}} ... p_n^{\frac{1}{p_n}}$ and $2^{f_2} 3^{f_3} ... p_n^{f_{p_n}}$ separately.

For the first part, Mertens first theorem states that $ \sum_{ p \le p_n } \frac{\log p}{p} = \log p_n + B_3 + o(1)$ where $B_3 \approx 1.332582$ is the Merten's constant. Hence

$$ 2^{1/2} 3^{1/3} ... p_n^{1/p_n} \approx p_n e^{-B_3} \tag 2 $$

For the second part, we have

$$ 2^{f_2} 3^{f_3} 5^{f_5}... p_n^{f_{p_n}} = \prod_{i = 1}^n p_i^{\sum_{k = 2}^{\infty} \frac{k}{p_i^k}} = \prod_{i = 1}^n p_i^{\frac{2p_i - 1}{p_i(p_i-1)^2}} \tag 3 $$

For the closed form of the sum on the above exponent, refer to this link. Hence,

$$ \lim_{p \to \infty}r(p) = \lim_{n \to \infty}{ p_{n+1} \over 2^{1/2} 3^{1/3} ... p_n^{1/p_n}} \cdot { 1 \over 2^{f_2} 3^{f_3} ... p_n^{f_{p_n}}} = \lim_{n \to \infty} \frac{e^{B_3} p_{n+1}}{p_n} \cdot \prod_{i = 1}^n p_i^{-\frac{2p_i - 1}{p_i(p_i-1)^2}} \tag 4 $$

Since $\frac{p_{n+1}}{p_n} \to 1$ as $n \to \infty$, $(4)$ reduces to

$$ \lim_{p \to \infty}r(p) = e^{B_3} \prod_{p} p^{-\frac{2p - 1}{p(p-1)^2}} \approx 0.522174 $$