The Nicolas criterion for the Riemann Hypothesis

For every prime number p n , we deﬁne the sequence X n = (cid:81) q ≤ p n qq − 1 − e γ × log θ ( p n ), where θ ( x ) is the Chebyshev function and γ ≈ 0 . 57721 is the Euler-Mascheroni constant. The Nicolas theorem states that the Riemann hypothesis is true if and only if the X n > 0 holds for all prime p n > 2. For every prime number p k , X k > 0 is called the Nicolas inequality. We show if the sequence X n is strictly decreasing for n big enough, then the Riemann hypothesis must be true. For every prime number p n > 2, we deﬁne the sequence Y n = e 1 2 × log( pn ) (1 − 1 log( pn ) ) and show that Y n is strictly decreasing for p n > 2. For all prime p n > 286, we demonstrate that the inequality X n < e γ × log Y n is always satisﬁed. We prove that lim n →∞ X n = lim n →∞ (log Y n ) = 0.


Introduction
In mathematics, 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 [1]. In mathematics, the Chebyshev function θ(x) is given by θ(x) = p≤x log p with the sum extending over all prime numbers p that are less than or equal to x [2]. For every prime p n , we define the sequence The constant γ ≈ 0.57721 is the Euler-Mascheroni constant and log is the natural logarithm. The importance of this property is: [4]. X n > 0 holds for all prime p n > 2 if and only if the Riemann hypothesis is true. Moreover, the Riemann hypothesis is false if and only if there are infinitely many prime numbers q i for which X i ≤ 0 and infinitely many prime numbers r j for which X j > 0.
We also use the Mertens' theorem which states: We prove if the sequence X n is strictly decreasing for n big enough, then the Riemann hypothesis must be true. For every prime number p n > 2, we define the sequence Y n = e 1 2×log(pn ) (1− 1 log(pn ) ) and show that Y n is strictly decreasing for p n > 2. Finally, for all prime p n > 286, we demonstrate that the inequality X n < e γ × log Y n is always satisfied.

Results
Proof. We know by the theorem 1.5: and we have by the theorem 1.2: Putting all this together yields the proof: Theorem 2.2. If X n is strictly decreasing for n big enough, then the Riemann hypothesis must be true.
Proof. Suppose that p n > 2 is the smallest prime number such that the Nicolas inequality is false under the assumption that X i is strictly decreasing (that is X i > X i+1 ). In this way, we have X n ≤ 0 and thus X n+1 < X n ≤ 0.
This implies lim sup n→∞ X n < 0 which is a contradiction with the theorem 2.1. By contraposition, the Nicolas inequality would be satisfied for all prime p n big enough. Consequently, there would be not infinitely many prime numbers for which the Nicolas inequality is unsatisfied. In this way, using the theorem 1.1, we can conclude that the Riemann hypothesis must be true when X n is strictly decreasing for n big enough.
For every prime number p n > 2, we define sequence Y n = e 1 2×log(pn ) (1− 1 log(pn ) ) . Theorem 2.3. For every prime number p n > 2, the sequence Y n is strictly decreasing.
Proof. For every real value x ≥ 3, we state the function where the derivative of f (x) is Consequently, the function f (x) is monotonically decreasing for every real value x ≥ 3 and therefore, the sequence Y n is monotonically decreasing as well. Indeed, a function f (x) of a real variable x is monotonically decreasing in some interval if the derivative of f (x) is lesser than zero and the function f (x) is continuous over that interval [7]. Certainly, the function f (x) is lesser than zero for all values x ≥ 3 where f (x) is continuous. In addition, Y n is essentially a strictly decreasing sequence, since there is not any natural number n > 1 such that Y n = Y n+1 .
We will prove another important result: Theorem 2.4. Let q 1 , q 2 , . . . , q m denote the first m consecutive primes such that q 1 < q 2 < · · · < q m and q m > 286. Then Proof. From the theorem 1.3, we know that ) × q m .
In this way, we can show that ) .
We know that .
Consequently, we obtain that ).
Due to the theorem 1.4, we prove that ) < e γ × log (Y m × θ(q m )) when q m > 286.
We finally obtain the main result: Theorem 2.5. For all prime p n > 286, we show that the inequality X n < e γ × log Y n is always satisfied.
Proof. According to the theorem 2.4, we have that for all prime p n > 286: which is equivalent to q≤p n q i q i − 1 − e γ × log θ(p n ) < e γ × log Y n and thus, X n < e γ × log Y n .