[Corrigendum] Note on the Riemann Hypothesis: International Conference on Recent Developments in Mathematics (ICRDM 2022)

Robin's criterion states that the Riemann hypothesis is true if and only if the inequality σ ( n ) < e γ ⋅ n ⋅ loglog n holds for all natural numbers n > 5040, where σ ( n ) is the sum-of-divisors function of n and γ ≈ 0.57721 is the Euler-Mascheroni constant. We require the properties of superabundant numbers, that is to say left to right maxima of n ↦ σ ( n ) n . In this note, using Robin's inequality on superabundant numbers, we prove that the Riemann hypothesis is true.

The Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1 2 .It is considered by many to be the most important unsolved problem in pure mathematics.It was proposed by Bernhard Riemann (1859).The Riemann hypothesis belongs to the Hilbert's eighth problem on David Hilbert's list of twenty-three unsolved problems.This is one of the Clay Mathematics Institute's Millennium Prize Problems.As usual σ(n) is the sum-of-divisors function of n  [3] , Theorem In 1997, Ramanujan's old notes were published where he defined the generalized highly composite numbers, which include the superabundant and colossally abundant numbers [4] .Superabundant numbers were also defined by Leonidas Alaoglu and Paul Erdős (1944).Let q 1 = 2, q 2 = 3, …, q k denote the first k consecutive primes, then an integer of the form [5] , pp. 367].A natural number n is called superabundant precisely when, for all natural numbers m < n We know the following property for the superabundant numbers: If n is superabundant, then n is a Hardy-Ramanujan integer [ [6] , Theorem 1 pp.450].
A number n is said to be colossally abundant if, for some ϵ > 0, There is a close relation between the superabundant and colossally abundant numbers. [6], pp. 455].

Proposition 3. Every colossally abundant number is superabundant [
Several analogues of the Riemann hypothesis have already been proved.Many authors expect (or at least hope) that it is true.However, there are some implications in case of the Riemann hypothesis might be false.
Putting all together yields the proof of the Riemann hypothesis.The following is a key Lemma.

Lemma 1.
If the Riemann hypothesis is false, then there are infinitely many superabundant numbers n such that Robin(n) fails.
Proof.This is a direct consequence of Propositions 1, 3 and 4. ◻ For every prime number q k > 2, we define the sequence: ) .
As the prime number q k increases, the sequence Y k is strictly decreasing [ [2] , Lemma 6.1 pp.750].We use the following Propositions: Proposition 5. [ [2] , Theorem 6.6 pp.752].Let ∏ k i= 1 q a i i be the representation of a superabundant number n > 5040 as the product of the first k consecutive primes q 1 < … < q k with the natural numbers a 1 ≥ a 2 ≥ … ≥ a k ≥ 1 as exponents.
Suppose that Robin(n) fails.Then, where N k = ∏ k i= 1 q i is the primorial number of order k and α n = ∏ k i= 1 . Proposition 6. [ [7] , Lemma 3.3 pp.8].Let x ≥ 11.For y > x, we have This is the main insight.
i be the representation of a superabundant number n > 5040 as the product of the first k consecutive primes q 1 < … < q k with the natural numbers a 1 ≥ a 2 ≥ … ≥ a k ≥ 1 as exponents.Suppose that Robin(n) fails.Then, where N k = ∏ k i= 1 q i is the primorial number of order k and α n = ∏ k i= 1 Proof.When n > 5040 is a superabundant number and Robin(n) fails, then we have n by Proposition 6.As result, we obtain that and thus, the proof is done.◻ This is the main theorem.
Theorem 1.The Riemann hypothesis is true.
Proof.We know there are infinitely many superabundant numbers [ [6] , Theorem 9 pp.454].In number theory, the p-adic order of an integer n is the exponent of the highest power of the prime number p that divides n.It is denoted ν p (n).
Equivalently, ν p (n) is the exponent to which p appears in the prime factorization of n.For every prime q, ν q (n) goes to infinity as long as n goes to infinity when n is superabundant [ [7] , Theorem 4.4 pp.12], [ [6] , Theorem 7 pp.454].Let n k > 5040 be a large enough superabundant number such that q k is the largest prime factor of n k .Suppose that Robin(n k ) fails and thus, we have that necessarily Y k < 1.03352795481 [ [2] , Theorem 6.7 pp.753].In the same way, let n k ′ be another superabundant number such that n k ′ ≫ n k , ν q k (n k ′ ) ≫ 3 and Robin(n k ′ ) fails too: The symbol ≫ means "much greater than".By Lemma 2, we have Hence, Consequently, However, we know that Moreover, we can see that since the following inequalities In this way, we obtain the contradiction 1 < 1 under the assumption that Robin(n k ) fails.To sum up, the study of this arbitrary large enough superabundant number n k > 5040 reveals that Robin(n k ) holds on anyway.Accordingly, Robin(n) holds for all large enough superabundant numbers n.This contradicts the fact that there are infinite superabundant numbers n, such that Robin(n) fails when the Riemann hypothesis is false according to Lemma 1.By reductio ad absurdum, we prove that the Riemann hypothesis is true.◻

Conclusions
Practical uses of the Riemann hypothesis include many propositions that are known to be true under the Riemann hypothesis and some that can be shown to be equivalent to the Riemann hypothesis.Indeed, the Riemann hypothesis is -BY 4.0 • Article, May 31, 2023 Qeios ID: 6H8DX7.2• https://doi.org/10.32388/6H8DX7.2 4/6 -BY 4.0 • Article, May 31, 2023 Qeios ID: 6H8DX7.2• https://doi.org/10.32388/6H8DX7.2 5/6 ≈ 0.57721 is the Euler-Mascheroni constant and log is the natural logarithm.The Ramanujan's Theorem stated that if the Riemann hypothesis is true, then the previous inequality holds for large enough n.