4
$\begingroup$

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?

$\endgroup$
11
$\begingroup$

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 $x$.)

$\endgroup$
  • $\begingroup$ What is the "truth" under GRH? $\endgroup$ – Igor Rivin Oct 5 '16 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 '16 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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