In the excellent answer at https://math.stackexchange.com/a/22435/86846, the identity $\log\left(1-\frac{2}{p}+\frac{1}{p^2}\right) +\log\left(1- \frac{1}{(p-1)^2} \right)=\log\left(1- \frac{2}{p} \right)$ was put forward as a starting point. I thought I would try a similar thing for estimating $\prod_{3\lt p\le x}\left(1-\frac 3p\right)$ and thus use some identity which might be along the lines of

$$\log\left(1-\frac 3p\right)=\log\left(1-\frac 1p\right)+\log\left(1-\frac 2p\right)+\log\left(1-\frac 2{p-1}-\frac 2{p-2}\right)$$

but as I work with these quantities I realize that no identity will completely remove the factor $p-3$ from the results entirely. In my search results I keep finding the same "Euler Product" and "Mertens' Theorem" pages on Wikipedia and Mathworld, but none of those pages has a $p-3$ factor listed that I have noticed, nor any suggestion as to how to approach the broader question.

I am not especially worried about the accuracy of the error term, as long as the error is not asymptotically as big as the rest of the result.

Are there known results that my primitive searches haven't found yet? Are there any reference materials I could look at that might help me work out some of these bounds myself, or perhaps some simple observation that has a relatively large error term but simplifies the approach in general?


Is your goal to estimate $\prod_{p \le x} (1-\frac{n}{p})$ as $x \to \infty$ ? It follows from the PNT error term for the Mertens 2nd theorem that $$-\log \prod_{p \le x} (1-\frac{n}{p}) = n \log \log x + n M +o(\frac{n}{\log^r x})+ A_n$$ Where $M $ is the Mertens constant and $A_n = \sum_p (-\log (1-\frac{n}{p})-\frac{n}{p})$

(I used that the series for $A_n$ converges in $\mathcal{O}(n/x)$ and $\sum_{p \le x} \frac{1}{p} = \log \log x + M + o(1/\log^r x)$ which is the PNT in one of its well-known forms)

| cite | improve this answer | |
  • $\begingroup$ Yes, this is my goal. Is there a reference I could look at with this info? $\endgroup$ – abiessu Dec 6 '17 at 19:40
  • $\begingroup$ I see that the product appears to be taken over $p\le x$; is this actually $n\lt p\le x$? $\endgroup$ – abiessu Dec 6 '17 at 19:44
  • $\begingroup$ for $p \le N$ you define $\log(1-n/p)$ as you want, but it must be the same on each side, for $p > N$ you need to take the principal branch $-\log(1-z) \sim z$ $\endgroup$ – reuns Dec 6 '17 at 19:47
  • $\begingroup$ If the product is defined here for $n\lt p\le x$ then that satisfies what I'm looking for. Thank you! $\endgroup$ – abiessu Dec 6 '17 at 19:50
  • $\begingroup$ I appreciate your work here, I found that my question is a duplicate of this one math.stackexchange.com/q/741633/86846 $\endgroup$ – abiessu Dec 7 '17 at 3:45

Not the answer you're looking for? Browse other questions tagged or ask your own question.