Arguments in Favor of the Riemann Hypothesis

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 12 . The Riemann hypothesis belongs to the David Hilbert’s list of 23 unsolved problems. Besides, it is one of the Clay Mathematics Institute’s Millennium Prize Problems. This problem has remained unsolved for many years. The Robin criterion states that the Riemann hypothesis is true if and only if the inequality σ ( n ) < e γ × n × log log n holds for all natural numbers n > 5040, where σ ( x ) is the sum-of-divisors function and γ ≈ 0 . 57721 is the Euler-Mascheroni constant. The Nicolas criterion states that the Riemann hypothesis is true if and only if the inequality (cid:81) q ≤ q n qq − 1 > e γ × log θ ( q n ) is satisfied for all primes q n > 2, where θ ( x ) is the Chebyshev function. Using both inequalities, we show some arguments in favor of the Riemann hypothesis is true.


Introduction
In mathematics, the Chebyshev function θ(x) is given by where q ≤ x means all the prime numbers q that are less than or equal to x. Let N n = 2 × 3 × 5 × 7 × 11 × · · · × q n denotes a primorial number of order n such that q n is the n th prime number. Thus, θ(q n ) = log N n . We define a sequence based on this function: Definition 1.1. For every prime number q n , we define the sequence of real numbers: X n = q≤q n q+1 q log θ(q n ) .
The constant γ ≈ 0.57721 is the Euler-Mascheroni constant and log is the natural logarithm. The importance of this inequality is: Nicolas(q n ) holds for all prime numbers q n > 2 if and only if the Riemann hypothesis is true [2].
As usual σ(n) is the sum-of-divisors function of n [1]: where d | n means the integer d divides n and d ∤ n signifies that the integer d does not divide n.
Define f (n) to be σ(n) n . Say Robins(n) holds provided f (n) < e γ × log log n.
The importance of this inequality is: Theorem 1.4. Robins(n) holds for all natural numbers n > 5040 if and only if the Riemann hypothesis is true [3]. If the Riemann hypothesis is false, then there are infinitely many natural numbers n > 5040 such that Robins(n) does not hold [3].
It is known that Robins(n) holds for many classes of numbers n. We recall that an integer n is said to be square free if for every prime divisor q of n we have q 2 ∤ n [1]. Theorem 1.5. Robins(n) holds for all natural numbers n > 5040 that are square free [1].
Let q 1 = 2, q 2 = 3, . . . , q m be the first m consecutive primes, then an integer of the form m i=1 q a i i with a 1 ≥ a 2 ≥ · · · ≥ a m ≥ 0 is called an Hardy-Ramanujan integer [1]. Based on the theorem 1.4, we know this result: Theorem 1.6. If the Riemann hypothesis is false, then there exist infinitely many natural numbers n > 5040 which are an Hardy-Ramanujan integer and Robins(n) does not hold [1].
We define H = γ − B such that B ≈ 0.2614972128 is the Meissel-Mertens constant [4]. For all real numbers x ≥ 2, the function u(x) is defined as follows For all real numbers x > 1, we define: Definition 1.7. We define another function: Putting all together we provide some arguments in favor of the Riemann hypothesis is true.

Known Results
We know from the constant H, the following formula: We know this property for the Chebyshev function: Mertens second theorem states that: We know these properties for the function f (n): Theorem 2.4. [6]. Let m i=1 q a i i be the representation of n as a product of primes q 1 < · · · < q m with natural numbers as exponents a 1 , . . . , a m . Then, Theorem 2.5. [1]. For all natural numbers n > 1: We know these results for the Riemann zeta function: Theorem 2.7. [7]. For a ≥ 1: .

Ancillary lemmas
The following is a key lemma. It gives an upper bound on f (n) that holds for all natural numbers n. The bound is too weak to prove Robins(n) directly, but is critical because it holds for all natural numbers n. Further the bound only uses the primes that divide n and not how many times they divide n.
Lemma 3.1. Let n > 1 and let all its prime divisors be q 1 < · · · < q m . Then, Proof. Putting together the theorems 2.5 and 2.6 yields the proof: The following is another key lemma.
There exists a natural number N such that X n < e γ ×6 π 2 + ε for all natural numbers n > N and for a positive real number ε < 6 π 2 . Only a finite number of elements of the sequence are greater than e γ ×6 π 2 + ε (this could be an empty set). Proof. The limit superior of a sequence of real numbers y n is the smallest real number b such that, for any positive real number ε, there exists a natural number N such that y n < b + ε for all natural numbers n > N. Only a finite number of elements of the sequence are greater than b + ε (this could be an empty set). Therefore, this is a consequence of the theorem 1.2. This is also a helpful lemma.
Proof. If we add H to then we obtain that according to the theorems 2.1 and 2.6. Therefore, the proof is done.

A Simple Case
We can easily prove that Robins(n) is true for certain kind of numbers: Lemma 4.1. Robins(n) holds for all natural numbers n > 5040 when q ≤ 5, where q is the largest prime divisor of n.
Proof. Let n > 5040 and let all its prime divisors be q 1 < · · · < q m ≤ 5, then we need to prove For all natural numbers n > 5040, we note that e γ × log log(5040) < e γ × log log n and therefore, the proof is complete when q 1 < · · · < q m ≤ 5.

The Function ϖ(x)
Lemma 5.1. The inequality ϖ(p) > u(p) is satisfied for a prime number p ≥ 3 if and only if Nicolas(p) holds.
Proof. We start from the inequality: which is equivalent to We add the following formula to the both sides of the inequality, and due to the theorem 2.1, we obtain that We distribute it and remove B from the both sides: If we apply the exponentiation to the both sides of the inequality, then we have that which means that Nicolas(p) holds. The same happens in the reverse implication. Proof. This is a direct consequence of theorems 1.3 and 5.1.
Proof. We know that lim x→∞ ϖ(x) = 0 for the limits lim x→∞ δ(x) = 0 and lim x→∞ θ(x) x = 1. In this way, this is a consequence from the theorems 2.2 and 2.3. 6 6. Inequalities on Hardy-Ramanujan integers Lemma 6.1. Let m i=1 q a i i be the representation of an Hardy-Ramanujan integer n > 5040 as a product of the first m primes q 1 < · · · < q m with natural numbers as exponents a 1 ≥ a 2 ≥ · · · ≥ a m ≥ 0. If Robins(n) does not hold, then Nicolas(q m ) holds indeed.
Proof. When Robins(n) does not hold, then Let's assume that Nicolas(q m ) does not hold as well. Consequently, According to the theorem 2.5, However, this implies that N m > n which is a contradiction since n > 5040 is an Hardy-Ramanujan integer.

When the Nicolas inequality may fail
Lemma 7.1. If some prime number q n > 2 complies with X n ≤ e γ × 6 π 2 then Nicolas(q n ) does not hold.
Proof. If we have the inequality X n ≤ e γ × 6 π 2 then this is equivalent to If we multiply the both sides by π 2 6 , so We use that theorem 2.6 to show that Besides, Consequently, we obtain that and therefore, Nicolas(q n ) does not hold.

Main Insight
The next lemma is a main insight.
Lemma 8.1. Let π 2 6 × log log n ′ ≤ log log n for some natural number n > 5040 such that n ′ is the square free kernel of the natural number n. Then Robins(n) holds.
Proof. Let n ′ be the square free kernel of the natural number n, that is the product of the distinct primes q 1 , . . . , q m . By assumption we have that × log log n ′ ≤ log log n.
We claim that Since otherwise we would have a contradiction. This shows that This is a contradiction since f (n ′ ) is equal to (q 1 + 1) × · · · × (q m + 1) q 1 × · · · × q m according to the formula f (x) for the square free numbers [1].

Pros for the Riemann Hypothesis
Theorem 9.1. The Riemann hypothesis is possibly true.
Proof. Let m i=1 q a i i be the representation of a sufficiently large Hardy-Ramanujan integer n > 5040 as a product of the first m primes q 1 < · · · < q m with natural numbers as exponents a 1 ≥ a 2 ≥ · · · ≥ a m ≥ 0. We claim that for every sufficiently large Hardy-Ramanujan integer n > 5040, then Robins(n) could always hold. Suppose that Robins(n) does not hold and so, the Riemann hypothesis would be false. Hence, f (n) ≥ e γ × log log n.
We use that theorem 2.4, If we divide the both sides by log log N m , then we obtain where N m is the primorial number of order m. We know that X m ≤ e γ ×6 π 2 is false according to the lemmas 6.1 and 7.1. From the lemma 3.2, we know that there exists a natural number N such that X m < e γ ×6 π 2 + ε for all natural numbers m > N and for a positive real number ε < 6 π 2 . Moreover, only a finite number of elements of the sequence are greater than e γ ×6 π 2 + ε (this could be an empty set). Under our assumption, there exist infinitely many Hardy-Ramanujan integers n > 5040 such that Robins(n) does not hold and X m < e γ ×6 π 2 + ε. In addition, q m cannot have an upper bound under our assumption. In general, if q m would have 9 an upper bound, then our assumption fails as a consequence of the lemma 8.1. In this way, we obtain that log log n log log N m which is the same as for a sufficiently small positive value of c = ε × π 2 6 . That would be equivalent to Since n is an Hardy-Ramanujan integer, then because of the theorem 2.7, where a 1 is the highest exponent such that 2 a 1 | n. Therefore, for a sufficiently small positive value of 0 < c < 1. However, this could be false for a sufficiently small positive value of ε < 6 π 2 that we could choose, where c = ε × π 2 6 would be a very small positive value as well. In addition, we know that log log n log log N m > 1 due to the theorem 1.5. Furthermore, from the paper [6], we know that Robins(n) holds for all natural numbers n > 5040 when In conclusion, for every sufficiently large Hardy-Ramanujan integer n > 5040, then Robins(n) could always hold. By contraposition, the Riemann hypothesis is possibly true, because of the theorems 1.4 and 1.6.
Theorem 9.2. The Riemann hypothesis is possibly true.
Proof. We claim that for every sufficiently large Hardy-Ramanujan integer n > 5040, then Robins(n) could always hold. Let m i=1 q a i i be the representation of a sufficiently large Hardy-Ramanujan integer n > 5040 as a product of the first m primes q 1 < · · · < q m with natural numbers as exponents a 1 ≥ a 2 ≥ · · · ≥ a m ≥ 0. Suppose that Robins(n) does not hold and so, the Riemann hypothesis would be false. Hence, f (n) ≥ e γ × log log n.
We use that theorem 2.4, This is equivalent to where N m is the primorial number of order m. If we apply the logarithm to the both sides of the inequality, then because of log N m = θ(q m ). Let's multiply by −1 the both sides of the inequality, after adding q≤q m 1 q to the both sides of the inequality. This the same as log( log log n log log N m ) − log In general, if q m would have an upper bound, then our assumption fails as a consequence of the lemma 8.1: our assumption is that there would be infinitely many natural numbers n > 5040 which are an Hardy-Ramanujan integer and counterexample of the Robin inequality. We know that = log( π 2 6 ) using the theorem 2.6. It is enough to distribute and remove the value of log( π 2 6 ) from the both sides to show that log( log log n log log N m ) + log However, this could be false for a sufficiently small positive value of ε, since we know that ε tends to 0 as n grows. In addition, we know that log log n log log N m > 1 due to the theorem 1.5. In conclusion, for every sufficiently large Hardy-Ramanujan integer n > 5040, then Robins(n) could always hold. By contraposition, the Riemann hypothesis is possibly true, because of the theorems 1.4 and 1.6.