I've formulated two conjectures that seems to imply a strong result when are combined with well known equivalences of the Riemann hypothesis, and I would like to know how get a disproof of such statements.

Let $S(n)=\sum_{k=1}^n\text{n mod k}$ the sum of remainder function, then it is know that for each $n>1$ $$\sigma(n)+S(n)=S(n-1)+2n-1,$$ where $\sigma(n)$ is the sum of divisors function, and let $H_n=1+1/2+\ldots+1/n$ the nth harmonic number.

Conjecture 1. The following asymptotic equivalence holds $$\frac{S(n)}{e^{H_n}}\sim\frac{n}{10},$$ as $n\to\infty$.

On assumption of this Conjecture 1 one has that using Robin's criterion for the Riemann's hypothesis (or also Lagarias equivalence, the computations are the same), $\forall n>5040$ $$\sigma(n)<e^\gamma n\log\log n$$ can be evaluated as $$\frac{S(n-1)}{e^{\frac{1}{n}}e^{H_{n-1}}}+\frac{2n}{ne^\gamma e^{O(\frac{1}{n})}}<\frac{1}{e^{H_{n}}}+\frac{S(n)}{e^{H_{n}}}+\frac{e^\gamma n\log\log n}{ne^\gamma e^{O(\frac{1}{n})}},$$ since $H_n=\log n+\gamma+O(\frac{1}{n})$, where $\gamma$ is Euler's constant, thus $LHS\sim\frac{n-1}{10}+\frac{2}{e^\gamma}$ and by comparison with $RHS\sim\frac{n}{10}+\log\log n$ one conclude this equivalence with Riemann hypothesis holds asymptotically.

On the other hand, one has for $M(n)=\sum_{k=1}^n\mu(k)$ is the Mertens function, where $\mu(k)$ is Mobius function, then

Conjecture 2. The following holds $$(M(n))^2=o\left(e^{H_n}\right)$$ as $n\to\infty$.

Thus by comparison with the equivalence of Riemann hypothesis stated as $M(x)=O(x^{\frac{1}{2}+\epsilon}),\forall\epsilon>0$, on assumption of this Conjecture 2 one get a contradiction since there is a constant $C>0$ such that $\lim_{n\to\infty}\frac{(M(n))^2}{e^{H_n}}$ is computed as $$\lim_{n\to\infty}\frac{C\cdot n^{2(\frac{1}{2}+\epsilon)}}{ne^\gamma e^{O(\frac{1}{n})}}=\lim_{n\to\infty}\frac{Cn^{2\epsilon}}{e^\gamma}$$ that is infnite for any $\epsilon>0$, a contradiction with the assumption that $(M(n))^2=o\left(e^{H_n}\right)$ that means $\lim_{n\to\infty}\frac{(M(n))^2}{e^{H_n}}=0$.

I did graphs of such ratios, but I'm assuming that perhaps there are mistakes in my computations or facts that I don't understand, since I understand that my reasoning are very soft for such great unsolved problem. I would like to ask how can we refute these conjectures.

Question. Can you refute each of these conjecture? Then I can learn how you give a mathematical reasoning for my question. Thanks in advance.


Conjecture 2 is false. Since $H_n = \log n + \gamma + O(1/n)$, we have that $e^{H_n} = e^{\gamma} n + O(1)$, and so your conjecture $M(n)^2 = o(e^{H_n})$ is equivalent to $M(x) = o(\sqrt{x})$, which is well known to be false.

  • $\begingroup$ Very thanks much @PeterHumphries I choose close the question, I say choose an aswer since I believe that since $S(n)=O(n^2)$ then Conjecture 1 also is false.. Very thanks much, $\endgroup$ – user243301 Mar 5 '16 at 7:42
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    $\begingroup$ @user243301 : if $M(x) = o(x^a)$ then because $1/\zeta(s) = s \int_1^\infty M(x) x^{-s-1} dx$ is meromorphic (hence its poles are at least of order $1$ ) if $1/\zeta(s)$ had a pole at $s=\rho$ of order $1$ then $M(x) = \frac{x^{\rho}}{\zeta'(\rho) } + \ldots$ which implies $Re(\rho) < a$, but we know there are poles such that $Re(\rho) = 1/2$ hence $M(x) \ne o(x^{1/2})$ $\endgroup$ – reuns Mar 8 '16 at 12:10
  • $\begingroup$ Very thanks much for the explanation, Best wishes this week @user1952009 $\endgroup$ – user243301 Mar 8 '16 at 16:50

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