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I hope 2015 doesn't start with gross nonsense question for me.

Robins equivalence for RH is

$$ \frac{\sigma(n)}{n \log\log n} < e^{\gamma}\qquad(1)$$

for $n \ge 5041$. We have $ \limsup \frac{\sigma(n)}{n \log\log n} = e^{\gamma}$.

Lagarias equivalence is

$$ \sigma(n) < H_n + \exp(H_n) \log(H_n) \qquad(2)$$

for $n \ge 2$ and $H_n$ the harmonic numbers.

(1) and (2) are closely related.

According to "Sharp Bounds for the Harmonic Numbers"

$$ \frac{1}{2n +\frac{1}{1-\gamma} - 2} \le H_n - \log n - \gamma$$

and a simpler bound for $n$ large enough is $\log{n} + \gamma \le H_n $

Assume $\log\log{n} > 1$. Substituting the simpler bound in (2) and dividing by $n \log{(\log{n}+\gamma)}$:

$$\frac{\sigma(n)}{n\log{(\log{n}+\gamma})} < e^{\gamma}+\frac{\log{n}+\gamma}{n\log{(\log{n}+\gamma)}} \qquad (3)$$

Since we substitued the lower bound for $H_n$, (3) implies RH.

The LHS of (3) is strictly smaller than the LHS of (1) since $ \frac{\sigma(n)}{n \log\log n} > \frac{\sigma(n)}{n\log{(\log{n}+\gamma})}$ and the RHS of (3) is larger than RHS of (1) since $\frac{\log{n}+\gamma}{n\log{(\log{n}+\gamma)}}$ is positive.

Because of this (3) might be easier to prove and $(1) \implies (3)$.

It can't happen (3) to be true and (1) false, since in this case RH will be both true and false, so $(1) \iff (3)$.

Is it true that (3) is equivalent to RH?

In case of positive answer:

Define $G(n)=\frac{\sigma(n)}{n \log\log n}$.

(1) and (3) are logically equivalent though technically different. Does this imply $G(n)$ can't take non-trivial values/ranges because of contradiction?

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    $\begingroup$ Your question is answered by the results of Robin that are quoted in Lagarias's paper. (You can find a version of Lagarias's paper on the arXiv here.) $\endgroup$ Jan 1, 2015 at 14:40
  • $\begingroup$ @JeremyRouse Thank you. Including the second question about forbidden values of G(n)? $\endgroup$
    – joro
    Jan 1, 2015 at 15:41
  • $\begingroup$ I confess that I don't understand your second question. Are you looking for results about bounds on $G(n)$ that are unconditional (and rely, for example, on the fact that (1) and (3) are equivalent)? (Or are you looking for conditional results.) $\endgroup$ Jan 1, 2015 at 16:22
  • $\begingroup$ @JeremyRouse My second question might be not clear enough currently. Basically log(log(n)) and log(log(n)+gamma) are very close. Assume $\sigma(n)$ can take arbitrary values (this is false). The technical differences between (1) and (3) might give unconditional forbidden values of G(n), since both are logically equivalent. $\endgroup$
    – joro
    Jan 1, 2015 at 16:32
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    $\begingroup$ I've sent you an e-mail about this. I think that discussing this further in the comments is not the best place to do so. $\endgroup$ Jan 1, 2015 at 16:53

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