# How is this elementary problem equivalent to the Riemann Hypothesis?

The problem I'm referring to is: Lagarias' Elementary Version of the Riemann Hypothesis, which states:

For a positive integer $n$, let $\sigma(n)$ be the sum of all of its positive divisors. Let $H_n$ denote the $n$-th Harmonic number. ($\sum_{k=1}^{n}\frac{1}{k}$).

Is the inequality true for all $n$ greater than or equal to $1$?

$$\sigma(n)\leq H_n+\ln(H_n)e^{H_n}$$

I want to gain some insight as to how the problems are equivalent.

• See this, you need to see how the Riemann hypothesis is related with the prime number theorem, and then to study the Robin/Nicolas/Lagarias papers. – reuns Dec 23 '16 at 7:18
• That's not an equality, that's an inequality. – Jose Arnaldo Bebita-Dris Dec 23 '16 at 7:29
• "seemingly simple problem": haha. – Yves Daoust Dec 23 '16 at 8:53
• The interesting question is why. With this I want to do feedback from a different viewpoint trying a comparison with Prime Number Theorem. For each perfect number $m$ one can define a sequence $a_n(m)=\sum_{1<d\mid m}\frac{1}{d^n-1}$ depending on $m$, that satisfies $\forall k\geq 1$, $a_k>0$, and $\sum_{n=1}^\infty \mu(n)a_n=1$ (using Lambert series). The second condition is equivalent to $\sum_{n=1}^\infty\frac{\mu(n)}{n}(na_n+\log n)=0$. But I don't know how interpret it, thus don't ask me about it. With respect your question I would like encourage to you to read Lagarias paper. Thanks – user243301 Dec 23 '16 at 15:42