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In page 197, Equivalents of the Riemann Hypothesis Vol 1, the following statement caught my eye

There is an editorial comment in [102] that includes an observation by the GCHQ Problem Solving Group. They contest that one could replace the number 2 in inequality (7.94) by any constant $K > 1$ at the cost of having to check more cases by hand when $K$ is close to one. An example is cited: when $K = 1.2$ one needs to verify inequality (7.94) for $1 \leq n \leq 10^6$.

referring the inequality 7.94

$\sigma(n) \leq H_n +2\exp(H_n) \log(H_n)$, $n \geq 1$,

and the reference 102 being

[102] J. C. Lagarias and W. Janous, A generous bound for divisor sums: problem 10949, Amer. Math. Monthly 111 (2004), 264–265.

According to An Elementary Problem Equivalent to the Riemann Hypothesis by J. C. Lagarias, $K = 1$ is equivalent to RH. I understand the claim is not that RH can be proved by checking more cases by hand, but I want to understand what are the methods by which $K$ can be improved to any $1 + \epsilon$ by checking more cases? What are the barriers that prevent it from proving that $K \leq 1$ by checking for more cases?

Edit: Edited my question with more details after looking at the answer by @Charles

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1. By an inequality due to Robin (see (2.2) in Lagarias's paper), $$\sigma(n)\leq e^\gamma n\log\log n+\frac{n}{\log\log n},\qquad n\geq 3.$$ By Lemma 3.1 in Lagarias's paper, we also know that $$\exp(H_n)\log(H_n)\geq e^\gamma n\log\log n,\qquad n\geq 3.$$ Combining these two estimates, we infer that $$\sigma(n)\leq\left(1+\frac{1}{(\log\log n)^2}\right)\exp(H_n)\log(H_n),\qquad n\geq 3.$$ The fraction on the right hand side tends to zero (effectively), confirming the observation by the GCHQ Problem Solving Group. Explicitly, given any $\varepsilon>0$, we have $$\sigma(n)\leq(1+\varepsilon)\exp(H_n)\log(H_n),\qquad n\geq\exp\exp(\varepsilon^{-1/2}).$$

2. The barrier to prove Lagarias's inequality is the same as the barrier to prove the Riemann Hypothesis, since these two statements are equivalent. It is a difficult problem, and perhaps it is even undecidable from the ZFC axioms (of course most of us believe it is decidable).

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    $\begingroup$ +1. Prashanth, this is the answer to accept. $\endgroup$
    – Charles
    Commented Aug 12, 2021 at 18:56
  • $\begingroup$ Thanks for the explanation @GHfromMO . It's very clear for me now. Also, shouldn't there be $\exp(\gamma)$ in the denominator of the fraction on the RHS ? Your next statement that it tends to zero is still valid as $\exp(\gamma)$ is a constant, but I'm just making sure my understanding is correct as I don't really have a math background. $\endgroup$
    – npcr
    Commented Aug 12, 2021 at 19:20
  • $\begingroup$ Is there a way to check if the attempt of proof of some statement (not necessarily RH) is decidable from ZFC axioms? $\endgroup$
    – Sylvain JULIEN
    Commented Aug 12, 2021 at 19:44
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    $\begingroup$ @PrashanthNCR: The third display can be strengthened by including a factor of $e^\gamma$ in the denominator. Note that $e^\gamma$ exceeds $1$. In fact this factor could be enlarged slightly, because in Robin's bound there is a constant $0.6482$ present in front of the fraction. I omitted these constants for elegance of presentation. $\endgroup$
    – GH from MO
    Commented Aug 12, 2021 at 19:53
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    $\begingroup$ @SylvainJULIEN: If ZFC is consistent, then there is no algorithm that could tell about every statement in ZFC whether it is decidable or not from the ZFC axioms. $\endgroup$
    – GH from MO
    Commented Aug 12, 2021 at 19:57
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The case $K=1$ is equivalent to the Riemann Hypothesis. [1] The GCHQ Problem Solving Group is not claiming that RH can be proved by checking a finite number of cases of this problem. I don't know what the barriers are that prevent it, though; I don't even know the method they used to extend $K$.

[1] Jeffrey C. Lagarias, An Elementary Problem Equivalent to the Riemann Hypothesis, American Mathematical Monthly 109 (2002), pp. 534–543.

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  • $\begingroup$ Among those $n$ that I could reach, the largest value of $\frac{\sigma(n)-H_n}{\exp(H_n)\log(H_n)}$ is $\approx0.987238$, attained at $n=12$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 12, 2021 at 14:03
  • $\begingroup$ Thanks @Charles , I understood that they are not claiming RH can be proved by checking a finite number of cases, I wanted to understand how can we reduce K to any 1+ϵ by checking more cases and why can't we go beyond $1$ ? Perhaps, I should edit my question. $\endgroup$
    – npcr
    Commented Aug 12, 2021 at 17:11
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    $\begingroup$ Well for $K<1$ it’s simply false: combining Thm. 2.2 and L. 3.2 in Lagarias’s paper, it follows immediately that $\limsup_{n\to\infty}\frac{\sigma(n)-H_n}{\exp(H_n)\log(H_n)}=\limsup_{n\to\infty}\frac{\sigma(n)}{\exp(H_n)\log(H_n)}\ge1$. $\endgroup$
    – Emil Jeřábek
    Commented Aug 12, 2021 at 18:07
  • $\begingroup$ @EmilJeřábek Is any $n$ known with $\frac{\sigma(n)-H_n}{\exp(H_n)\log(H_n)}>\frac{\sigma(12)-H_{12}}{\exp(H_{12})\log(H_{12})}$? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 12, 2021 at 19:03
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    $\begingroup$ @მამუკა ჯიბლაძე sure, take the 100th number from this sequence: oeis.org/A004490 and you'll get $\approx 0.9929$ $\endgroup$
    – Alexander Kalmynin
    Commented Aug 13, 2021 at 7:15

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