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Does the inequality

$$1 + \dfrac{\log p_{k+1}}{\log N_k} > (\log N_k)^{\dfrac{1}{p_{k+1} -1}}$$

hold for all integers $k\geq 2$, where $N_k$ denotes the $k$-th primorial number $N_k=p_1p_2\cdots p_k$ and $p_j$ the $j$-th prime number?

What I think:

By the prime number theorem, we know that there exists some constant $$1/2 \leq \theta <1$$ such that $$\log N_k = p_k + O({p_{k}^{\theta}\log p_k})$$

and upon invoking this together with Erdos' result that $p_{k+1}$ tends to $p_k$ for large $k$, one notices that the RHS of the inequality converges faster with respect to the LHS, so that if it holds for some $k=k_{0}$, then it holds for all $k\geq k_{0}$. Taking $k_{0} = 2$, we obtain the desired result ?

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  • $\begingroup$ If you rewrite the inequality then the question amounts to $\log N_{k+1}>(\log N_k)^{\frac{p_{k+1}}{p_{k+1}-1}}$. This seems believable, at least for $k\gg1$. $\endgroup$
    – T. Amdeberhan
    Commented Jan 29, 2017 at 15:14
  • $\begingroup$ "Erdos' result that $p_{k+1}$ tends to $p_k$ for large $k$" – what? $p_{k+1}-p_k$ certainly does not tend to zero, and if what you mean is that $p_{k+1}/p_k$ tends to one, I think that was known before Erdos. $\endgroup$
    – Gerry Myerson
    Commented Jan 29, 2017 at 22:13
  • $\begingroup$ I don't think your last display is known. It would be known under the hypothesis that $\zeta(s)\neq 0$ for $\Re s>\theta$, but this hypothesis is far from being proven. $\endgroup$
    – GH from MO
    Commented Jan 29, 2017 at 22:27
  • $\begingroup$ The answer is probably no. See my response. $\endgroup$
    – GH from MO
    Commented Jan 29, 2017 at 23:58
  • $\begingroup$ @GerryMyerson, yes i mean $p_{k+1}/p_k$ tends to $1$. And as far as i know, this is due to Erdos (research a certain paper outlining the Erdos-Selberg dispute). $\endgroup$
    – Community user.
    Commented Jan 30, 2017 at 10:39

2 Answers 2

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The answer is probably no. Your inequality implies $$ \frac{\log p_{k+1}}{\log N_k} > \log\left(1+\frac{\log p_{k+1}}{\log N_k}\right)>\frac{\log\log N_k}{p_{k+1}-1}>\frac{\log\log N_k}{p_{k+1}}.$$ In particular, $$ p_{k+1}\log p_{k+1}>\theta(p_k)\log\theta(p_k),$$ whence $$ p_{k+1}>\theta(p_k). $$ By Littlewood's theorem, the right hand side exceeds $p_k+p_k^{1/2}$ infinitely often, while it is widely believed that the left hand side is smaller than $p_k+p_k^{1/3}$ for every large $k$. So, if we believe the last upper bound, there are infinitely many counterexamples. By quoting Littlewood's theorem a bit more precisely, we get a contradiction already by Legendre's conjecture that there is a prime number between any two large consecutive squares.

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  • $\begingroup$ Note that your answer would imply the reverse inequality $$1 + \dfrac{\log p_{k+1}}{\log N_k} < (\log N_k)^{\dfrac{1}{p_{k+1} - 1}}$$ for some $k$, which is known to be a contradiction to Cramer's conjecture, see arXiv:1012.3613[math.NT] $\endgroup$
    – Community user.
    Commented Jan 30, 2017 at 12:10
  • $\begingroup$ @Communityuser: I think Propositions 3-4 in arXiv:1012.3613 say much the same as me, namely your inequality fails for infinitely many $k$'s if the Cramer conjecture is true. $\endgroup$
    – GH from MO
    Commented Jan 30, 2017 at 15:12
  • $\begingroup$ would you explain your recent comment ? Because it seems you're saying that the Cramer conjecture implies the falsehood of either inequality ? $\endgroup$
    – Community user.
    Commented Jan 30, 2017 at 15:14
  • $\begingroup$ @Communityuser: By Proposition 3, your inequality implies (a slight variant of) $u_k>u_{k+1}$. On the other hand, Proposition 4 says that $u_k>u_{k+1}$ fails for infinitely many $k$'s if Cramer's conjecture holds. (In fact one only needs Legendre's conjecture, which is weaker than Cramer's, as my response explained.) $\endgroup$
    – GH from MO
    Commented Jan 30, 2017 at 15:18
  • $\begingroup$ in either case, could we just judge the above argument without basing on unproven conjectures ? $\endgroup$
    – Community user.
    Commented Jan 30, 2017 at 15:22
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Not an answer yet.

As in my comment, we may rewrite the inequality as $\log N_{k+1}>\left(\log N_k\right)^{\frac{p_{k+1}}{p_{k+1}-1}}$. Introduce the function $f_a(x)=\log(ax)-\left(\log a\right)^{\frac{x}{x-1}}$ so that the problem amounts to $f_{N_k}(p_{k+1})>0$. Notice $$\frac{d}{dx}f_a(x)=\frac1x+\frac{(\log a)\,a^{\frac{x}{x-1}}}{(x-1)^2}>0$$ (of course $x, a>1$) shows $f_a(x)$ is an increasing function of $x$. That means, once $x$ is large enough we conquer positivity as the OP noted.

I claim a stronger inequality $f_{N_k}(p_k)>0$ which improves $f_{N_k}(p_{k+1})>0$.

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