Let $E(p)= \prod_{p}\frac{p+1}{p}$ for prime $p$.
Then $E(2)=\frac{2}{1}$, $E(3)=\frac{2}{1}*\frac{3}{2}$, $E(5)=\frac{2}{1}*\frac{3}{2}*\frac{5}{4}$, $E(7)=\frac{2}{1}*\frac{3}{2}*\frac{5}{4}*\frac{7}{6}$, $E(11)=\frac{2}{1}*\frac{3}{2}*\frac{5}{4}*\frac{7}{6}*\frac{11}{10}$, and so on.
It is known that $E(p)$ diverges. For any prime $p$, is there a method to approximate the value of $E(p)$?
For instance, $p=3$, then $E(p) = \frac{3}{1}$, and for $p=13$, then $E(p) = \frac{1001}{192} \approx 5.214$. What is an approximation of $E(p)$ when $p=2^{31}-1$?
Thanks for help in advance.
Edit: Didn't realized I needed the reciprocal $E(p)$ when I could change $\frac{p-1}{p} to \frac{p+1}{p}$.
I assume you mean to define something like $E(x) = \prod_{p\le x} \frac{p+1}p$. In that case, we have \begin{align*} E(x) &= \bigg( \prod_{p\le x} \frac{p^2-1}{p^2} \bigg) \bigg( \prod_{p\le x} \frac p{p-1} \bigg) \\ &= \bigg( \frac1{\zeta(2)} + O\bigg( \frac1x \bigg) \bigg) \big( e^\gamma \log x + O(1) \big) = \frac{6e^\gamma}{\pi^2} \log x + O(1), \end{align*} where $\gamma$ is Euler's constant. (The evaluation of the second product is one of Mertens's theorems.) In particular, when $p=2^{31}-1$, the main term in this asymptotic formula is about $23.2659$.