Deje $N_n=\prod_{k=1}^np_k$ ser el primorial de orden $n$,$\gamma$ ser el de Euler-Mascheroni constante y $\varphi$ el valor del phi de Euler de la función. Nicolas mostró que si la Hipótesis de Riemann es cierta, entonces $$\forall \ n\in\mathbb{Z^+}, \ \frac{N_n}{\varphi(N_n)}=\prod_{k=1}^n\frac{p_k}{p_k-1}>e^\gamma \log\log N_n.\tag{NI}$$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 $n>1$ $$\frac{u_{n+1}}{u_n}=\frac{\log\log N_n}{\log\log N_{n+1}}\frac{p_{n+1}}{p_{n+1}-1}<1 $$ $$\frac{p_{n+1}}{p_{n+1}-1}<\frac{\log\log N_{n+1}}{\log \log N_n}=\frac{\log\log N_{n+1}}{\log(\log N_{n+1}-\log p_{n+1})}.\tag{$\estrella de$}$$ Tenemos $\log N_{n+1}<p_{n+1}$, de lo cual, para cualquier 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 $\registro N_{n+1}>\log p_{n+1}+1$ for $n>1$, $(\estrellas)$ is weaker than $$\frac{p_{n+1}}{p_{n+1}-1}<\frac{\log\log N_{n+1}}{\log(\log N_{n+1}-\log \log N_{n+1})}.\tag{1}$$
Pero $\displaystyle f(x)=\frac{\log x}{\log(x-\log x)}$ es decreciente en $(1,+\infty)$: $$\require\cancel f'(x)= \frac{\log(x-\log x)/x - \log x \cdot D(\log(x-\log x))}{\cancel{\log^2(x-\log x)}}<0 \\ \frac{\log(x-\log x)}{x}<\frac{(1-1/x)\log x}{x-\log x}=\frac{(x-1)\log x}{{x}(x-\log x)} \\ (x-\log x)\log(x-\log x)<(x-1)\log x \\ (x-\log x)^{x-\log x}<x^{x-1},$$ 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, $x>1$.
Por lo tanto $(1)$ es a su vez implica por $$\frac{p_{n+1}}{p_{n+1}-1}<\frac{\log p_{n+1}}{\log(p_{n+1}-\log p_{n+1})} \\ p_{n+1}\log(p_{n+1}-\log p_{n+1})<(p_{n+1}-1)\log p_{n+1} \\ (p_{n+1}-\log p_{n+1})^{p_{n+1}}<p_{n+1}^{p_{n+1}-1},$$ and this we have already showed to ensue from $(2)$. Thence, $(\star)$ es verdad y así es NI.
Sin embargo, no sólo un resultado demostraría RH, pero en este artículo se demuestra que este resultado podría confundir a Cramér de la conjetura. A cuenta de esto espero haber hecho algún estúpido error, simplemente no puedo ver. Gracias por las observaciones.
