2
$\begingroup$

Let $n$ be an initially arbitrarily large variable, but always decreasing (and more specifically non-increasing) to exactly $1$ when $p$ is the largest prime in the product. Then, denoting with $\gamma$ the Euler-Mascheroni constant, do we have $$\frac{1}{e^\gamma\log x} \prod_{\substack{p < x \\ \text{p prime}}} \frac{p}{p-1}<1+ \prod_{\substack{p<x \\ p \ \text{prime}}}\frac{1}{p^{n+1}-1} \ \ ?$$

As $x \to \infty$, the LHS tends to $1$ by Mertens' third theorem, and Robin showed that $$1-e^\gamma \log x \prod_{\substack{p\le x \\ p \ \text{prime}}} \frac{p-1}{p}$$ changes sign infinitely often, so that it applies to the LHS too.

$\endgroup$
  • $\begingroup$ You're right my proof don't guarantees that $c>0$. I've deleted it. I try to find another way. $\endgroup$ – Marco Cantarini Nov 26 '14 at 12:53
  • $\begingroup$ @The_Cam You might as well want to try with this related question, which I derived from that theorem of the paper you provided. $\endgroup$ – Vincenzo Oliva Nov 26 '14 at 13:15
1
$\begingroup$

It is not a rigorous proof, but I think it can gives an idea. It known that $$\underset{p\leq x}{\prod}\frac{p}{p-1}=e^{\gamma}\log\left(x\right)+2e^{-c\sqrt{\log\left(x\right)}}$$ where $c>0$ is the constant of the Zero Free Region of the Riemann zeta function. So $$\frac{1}{e^{\gamma}\log\left(x\right)}\underset{p\leq x}{\prod}\frac{p}{p-1}=1+\frac{2}{e^{\gamma+c\sqrt{\log\left(x\right)}}\log\left(x\right)}.$$So your inequality is true if $$\frac{2}{e^{\gamma+c\sqrt{\log\left(x\right)}}\log\left(x\right)}<\underset{p\leq x}{\prod}\frac{1}{p^{n+1}-1}.$$ Take logs. You obtain $$\log(2)-\gamma-c\sqrt{\log\left(x\right)}-\log\left(\log\left(x\right)\right)<-\underset{p\leq x}{\sum}\log\left(p^{n+1}-1\right)$$ hence your inequality is true if $$\underset{p\leq x}{\sum}\log\left(p^{n+1}-1\right)<\gamma+c\sqrt{\log\left(x\right)}+\log\left(\log\left(x\right)\right)-\log(2).\,\,\,(1)$$ Now analize sum. Note that is about $\underset{p\leq x}{\sum}\log\left(p^{n+1}-1\right)\approx\underset{p\leq x}{\sum}\log\left(p^{n+1}\right)$, but, for PNT that sum is $$\underset{p\leq x}{\sum}\log\left(p^{n+1}\right)=\left(n+1\right)\left(c_{1}x+c_{2}\frac{x}{\log\left(x\right)}+o\left(\frac{x}{\log\left(x\right)}\right)\right)$$with $c_{1},c_{2}>0$. So you have that the sum grow up essentialy like $x$ and in the right of $(1)$ you have a function that grow up like $\sqrt{\log\left(x\right)}$, which is a contradiction if $x$ is large enough. It is not rigorous, but I think it's point in the right way. I hope can be useful.

$\endgroup$
  • $\begingroup$ Thanks for your answer. The LHS of your second equality tends to $1$ as $x\to \infty$, whereas the RHS tends to $\frac{1}{2}$. How can the equality be true? $\endgroup$ – Vincenzo Oliva Nov 27 '14 at 18:35
  • $\begingroup$ I've found the identity here math.stackexchange.com/questions/22411/… point two of the proof. There is an reference too. $\endgroup$ – Marco Cantarini Nov 27 '14 at 19:16
  • $\begingroup$ Now I see. He wrote $\prod_{2<p\leq n} \left( 1-\frac{1}{p}\right)^{-1}=\frac{1}{2}e^\gamma \log n + e^{-c\sqrt{\log n}}$. For your first and second equalities to be true you need to multiply the RHS by $2$. $\endgroup$ – Vincenzo Oliva Nov 27 '14 at 19:45
  • $\begingroup$ How is the first equation known? Could you give a reference? And could you help with this question on $\prod_{2 < p \le n} \left( 1 - \frac{1}{p-1}\right)$? $\endgroup$ – ShreevatsaR May 27 '18 at 21:05

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.