Let $n={\prod }_{i=1}^r{p}_i^{a_i}$ be the prime factorization of n, where $p_1<\dots <p_r$ are distinct prime numbers and $a_1,\dots ,a_r$ are positive integers. Then [7, Lemma 1 (2) pp. 2]:

For $n>1$ [8, (2.7) pp. 362]:

We say that $\mathsf{Robin}\left(n\right)$ holds provided that

For $x\ge 7232121212$ [9, Lemma 2.7 (4) pp. 19]:

For $x\ge 2278382$ [9, Lemma 2.7 (5) pp. 19]:

Let n be a superabundant number. Denoting its largest prime factor by p, we have [6, Theorem 7 pp. 454]:

The $\mathit{p}-\mathit{adic}$ order of an integer n is the exponent of the highest power of the prime number p that divides n. It is denoted ${\nu }_p\left(n\right)$.

Suppose $p>q^{{\nu }_q\left(n\right)+1}$. We evaluate $I\left(s\right)$ for $s=n$ and $s=m={\displaystyle \frac{n·q^{{\nu }_q\left(n\right)+1}}{p}}$. Given the multiplicative property of $I\left(s\right)$, we focus on distinct factors. Moreover, n is superabundant and $m<n$. By Proposition 2.1, we obtain: Therefore, ${\displaystyle \frac{1}{q^{{\nu }_q\left(n\right)+1}}}<{\displaystyle \frac{1}{p}}$, implying $p<q^{{\nu }_q\left(n\right)+1}$, contradicting our initial assumption. Clearly, we can observe that

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By joining Proposition 2.8 and Lemma 3.2, we obtain the present result. □

This follows directly from Propositions 2.6, 2.9, and 2.11. □

For each prime number $p_k>2$, we define the sequence

This Lemma is straightforward. □

Consider the first k consecutive primes $p_1,p_2,\dots ,p_k$ such that $p_1<p_2<\dots <p_k$ and $p_k>7232121212$. It follows that

Proposition 2.4 implies that Therefore, it is possible to prove that Thus, we can demonstrate that Consequently, we can establish that According to Proposition 2.5, we can show that when $p_k>7232121212$. □

For every possible superabundant number $n>5040$ violating $\mathsf{Robin}\left(n\right)$, we have

Suppose that $\mathsf{Robin}\left(n\right)$ does not hold. Let $n={\prod }_{i=1}^k{p}_i^{a_i}$ be an Hardy-Ramanujan integer, where the primes $p_1<\dots <p_k$ are the first k consecutive primes, $p_k>e^{31.018189471}$ and $a_1\ge a_2\ge \dots \ge a_k\ge 1$: This follows directly from Propositions 2.7 and 2.10. Given our assumption, we can conclude that Additionally, we can infer that for all $p_k>e^{31.018189471}$, as per Lemma 3.7 with $log\left({\left(N_k\right)}^{Y_k}\right)=Y_k·\theta \left(p_k\right)$. By combining Proposition 2.2 with the inequalities (1) and (2), we obtain the following result: This leads to ${\left(N_k\right)}^{Y_k}>n$ by transitivity. □

We use a proof by contradiction, assuming that the Riemann hypothesis is not true. Given Lemma 3.4, there are infinitely many superabundant numbers n that do not satisfy $\mathsf{Robin}\left(n\right)$. Let $n_{k_i}$ be the infinite sequence of superabundant numbers where $\mathsf{Robin}\left(n_{k_i}\right)$ is false and $p_{k_i}$ is the largest prime factor of $n_{k_i}$. Proposition 2.10 additionally implies that $p_{k_i}>e^{31.018189471}$ for all i. Equivalently, Lemma 3.6 implies that $Y_{k_i}<1.00021$. By Lemma 3.3, we conclude that there exists a finite, strictly increasing subsequence $n_{k_j^{\prime }}$ of m superabundant numbers, taken from the general sequence $n_{k_i}$, such that where $N_{k_j^{\prime }}={\prod }_{s=1}^{k_j^{\prime }}{p}_s$ is the primorial number of order $k_j^{\prime }$. Lemma 3.3 guarantees the existence of this subsequence of superabundant numbers $n_{k_j^{\prime }}$ that can be made arbitrarily long, because for every fixed prime p, ${\nu }_p\left(n_{k_j^{\prime }}\right)$ approaches infinity as $n_{k_j^{\prime }}$ grows larger. To illustrate, consider a finite, strictly increasing subsequence $n_{k_j^{\prime }}$ of m superabundant numbers, satisfying the condition that for $1<j\le m$ and where $n_{k_1^{\prime }}$ and $n_{k_m^{\prime }}$ are the minimum and maximum elements of the subsequence, respectively. Utilizing inequalities (3) and (4), we deduce that: This yields the trivial inequality $1.00022>1.00021$. While this is just one example, many such cases may exist. Therefore, by Lemmas 3.6 and 3.8, it suffices to show that the inequality leads to a contradiction, because $1.00021$ is always greater than $Y_{k_1^{\prime }}$ when $n_{k_1^{\prime }}$ is the smallest element in the subsequence. Definitely, that’s the same as saying and it is clear that due to the fact that $Y_{k_j^{\prime }}$ decreases as the subsequence $n_{k_j^{\prime }}$ grows, and ${\left(N_{k_j^{\prime }}\right)}^{Y_{k_j^{\prime }}}>n_{k_j^{\prime }}$ according to Lemmas 3.6 and 3.8. Hence, our initial assumption has been contradicted. This proof by contradiction, combined with Proposition 2.6, establishes the truth of the Riemann hypothesis. □