# Is $\frac{1}{e^\gamma\log x} \prod\limits_{p < x,p\,\text{prime}} \frac{p}{p-1}<1+ \prod\limits_{p<x,p\,\text{prime}}\frac{1}{p^{n+1}-1}?$

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.

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

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.
• 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? – Vincenzo Oliva Nov 27 '14 at 18:35
• 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 $2$. – Vincenzo Oliva Nov 27 '14 at 19:45
• How is the first equation known? Could you give a reference? And could you help with this question on $\prod_{2 < p \le n} \left( 1 - \frac{1}{p-1}\right)$? – ShreevatsaR May 27 '18 at 21:05