7
$\begingroup$

Let $N_n=\prod_{k=1}^np_k$ be the primorial of order $n$,$\gamma$ be the Euler-Mascheroni constant and $\varphi$ denote the Euler phi function. Nicolas showed that if the Riemann Hypothesis is true, then $$\forall \ n\in\mathbb{Z^+}, \ \frac{N_n}{\varphi(N_n)}=\prod_{k=1}^n\frac{p_k}{p_k-1}>e^\gamma \log\log N_n.\tag{NI}$$He also proved that the falsity of RH would imply the existence of infinitely many $n$ violating NI, as well as infinitely many $n$ satisfying it. Now, the combination of the following statement, equivalent to PNT:$$\lim_{n\to\infty}\frac{p_n}{\log N_n}=1$$ and Mertens' third theorem $$\lim_{n\to\infty}\frac{1}{\log p_n}\prod_{k=1}^n\frac{p_k}{p_k-1}=e^\gamma$$ yields $$\lim_{n\to\infty}\frac{1}{\log\log N_n}\prod_{k=1}^n\frac{p_k}{p_k-1}=e^\gamma,$$ thus NI would follow from the decreasingness, even only for large enough $n$, of the sequence $$u_n=\frac{1}{\log\log N_n}\prod_{k=1}^n\frac{p_k}{p_k-1}.$$ Therefore, since $\displaystyle \log \log 2 $ is negative, let us consider for $n>1$ $$\frac{u_{n+1}}{u_n}=\frac{\log\log N_n}{\log\log N_{n+1}}\frac{p_{n+1}}{p_{n+1}-1}<1 $$ $$\frac{p_{n+1}}{p_{n+1}-1}<\frac{\log\log N_{n+1}}{\log \log N_n}=\frac{\log\log N_{n+1}}{\log(\log N_{n+1}-\log p_{n+1})}.\tag{$\star$}$$ We have $\log N_{n+1}<p_{n+1}$, wherefrom, for any real $r>\log p_{n+1}+1,$ $$\frac{1}{\log(r-\log\log N_{n+1})}<\frac{1}{\log(r-\log p_{n+1})},$$ so as long as $\log N_{n+1}>\log p_{n+1}+1$ for $n>1$, $(\star)$ is weaker than $$\frac{p_{n+1}}{p_{n+1}-1}<\frac{\log\log N_{n+1}}{\log(\log N_{n+1}-\log \log N_{n+1})}.\tag{1}$$

But $\displaystyle f(x)=\frac{\log x}{\log(x-\log x)}$ is decreasing on $(1,+\infty)$: $$\require\cancel f'(x)= \frac{\log(x-\log x)/x - \log x \cdot D(\log(x-\log x))}{\cancel{\log^2(x-\log x)}}<0 \\ \frac{\log(x-\log x)}{x}<\frac{(1-1/x)\log x}{x-\log x}=\frac{(x-1)\log x}{{x}(x-\log x)} \\ (x-\log x)\log(x-\log x)<(x-1)\log x \\ (x-\log x)^{x-\log x}<x^{x-1},$$ which follows from $$(x-\log x)^x < x^{x-1} \\ \left(1-\frac{\log x}{x}\right)^x<x^{-1} \\ \left(1-\frac{\log x}{x}\right)^{x/\log x}<x^{-1/\log x} =e^{-1},$$ which, setting $\displaystyle t=\frac{x}{\log x}$, reduces to the familiar $$\left(1-\frac{1}{t}\right)^t<e^{-1}.\tag{2}$$ With $x$ in place of $t$ we would say $(2)$ holds for $x>1$, but given that the range of $\displaystyle \frac{x}{\log x}$ does not include $[0,e)$, we really need $t>0$, or equivalently, again, $x>1$.

Hence $(1)$ is in turn implied by $$\frac{p_{n+1}}{p_{n+1}-1}<\frac{\log p_{n+1}}{\log(p_{n+1}-\log p_{n+1})} \\ p_{n+1}\log(p_{n+1}-\log p_{n+1})<(p_{n+1}-1)\log p_{n+1} \\ (p_{n+1}-\log p_{n+1})^{p_{n+1}}<p_{n+1}^{p_{n+1}-1},$$ and this we have already showed to ensue from $(2)$. Thence, $(\star)$ is true and so is NI.


However, not only such a result would prove RH, but in this paper it is shown that such a result would confute Cramér's conjecture. On account of this I expect to have made some stupid mistake, I just cannot see it. Thank you for any observations.

$\endgroup$
4
  • $\begingroup$ @MarcoCantarini I think you misunderstood the process of getting to $(1)$ because of the expression in fact, which means in reality, not the Italian infatti. $\endgroup$ Commented Apr 8, 2015 at 10:53
  • $\begingroup$ No, I only read too fast :) My professor used to say "the devil is in the details". Try to rewrite every single passage, even the most obvious. Later I'll try to work with more attention on your proof. $\endgroup$ Commented Apr 8, 2015 at 11:15
  • $\begingroup$ @Marco Oh I see, will do, he's so right! :D Ok, thanks! $\endgroup$ Commented Apr 8, 2015 at 11:38
  • $\begingroup$ I'm not sure if it's critical, but I don't think $\log N_{n+1} < p_{n+1}$ is always true. $\endgroup$
    – Zander
    Commented Apr 8, 2015 at 23:31

2 Answers 2

2
$\begingroup$

The problem is the claim $\log\left(N_{n+1}\right)=\theta\left(p_{n+1}\right)<p_{n+1} $ where $\theta\left(x\right)$ is the Chebyshev theta function. We knows the bounds $$n\left(\log\left(n\right)+\log\left(\log\left(n\right)\right)-1+\frac{\log\left(\log\left(n\right)\right)-2.050735}{\log\left(n\right)}\right)\leq\theta\left(p_{n}\right)\leq n\left(\log\left(n\right)+\log\left(\log\left(n\right)\right)-1+\frac{\log\left(\log\left(n\right)\right)-2}{\log\left(n\right)}\right) $$ and $$n\left(\log\left(n\right)+\log\left(\log\left(n\right)\right)-1+\frac{\log\left(\log\left(n\right)\right)-2.1}{\log\left(n\right)}\right)\leq p_{n}\leq n\left(\log\left(n\right)+\log\left(\log\left(n\right)\right)-1+\frac{\log\left(\log\left(n\right)\right)-2}{\log\left(n\right)}\right) $$ for every sufficiently large $n $ and these limitations don't guarantee your inequality.

$\endgroup$
0
$\begingroup$

Indeed, $\log N_{n+1}=\theta(p_{n+1})<p_{n+1}$ doesn't always hold. In particular, here is shown that $\theta(x)-x$ changes sign infinitely often. Since for $p_n<x<p_{n+1}$, $\theta(x)-x=\theta(p_n)-x<\theta(p_n)-p_n$, my inequality fails infinitely many times.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .