In Dusart papers he proves that $\prod \limits_{p \leq x} \frac{p_i}{p_i-1} \leq e^\gamma \ln (x) \left(1+\frac{0.2}{\ln ^2 (x)} \right)$ for large numbers.

What I am asking is could we make the bound a little bit sharper by making it like this $$ \prod \limits_{p \leq x} \frac{p_i}{p_i-1} \leq e^\gamma \ln (x) \left(1+\frac{0.06}{\ln ^3 (x)} \right)$$ for sufficiently large numbers?


1 Answer 1


Yes, although writing down explicitly what "sufficiently large" means might be a challenge.

The prime number theorem implies that there exists a constant $c>0$ such that $$ \prod_{p\le x} \frac p{p-1} = e^\gamma \ln (x) \big( 1 + O\big( \exp(-c\sqrt{\ln x}) \big) \big). $$ From this it is easy to deduce, for every $\varepsilon>0$ and every $A>0$, that $$ \prod_{p\le x} \frac p{p-1} \le e^\gamma \ln (x) \bigg( 1 + O\bigg( \frac{\varepsilon}{\ln^A(x)} \bigg) \bigg) $$ holds when $x$ is sufficiently large in terms of $\varepsilon$ and $A$. (Also, the right-hand side with $+$ changed to $-$ is a lower bound for the prime product for sufficiently large $x$.)

  • $\begingroup$ What is the "truth" under GRH? $\endgroup$
    – Igor Rivin
    Oct 5, 2016 at 9:26
  • 2
    $\begingroup$ Igor Rivin: Under RH, one gets $e^{\gamma} \ln(x)(1 + O(\epsilon(x)))$, with $\epsilon(x) = x^{-\frac{1}{2}} \ln(x)$ (and the implicit constant can be made explicit). $\endgroup$
    – js21
    Oct 5, 2016 at 10:05

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