The probability of Riemann's hypothesis being true is equal to 1
Abstract
Let P be the set of all prime numbers, q1,q2, ·s ,qm ∈ P, Pk be the k-th (k = 1,2, ·s m) element of P in ascending order of size, α 1,α 2, ·s ,α m be positive integers, and β 1,β 2, ·s ,β m is a permutation of α 1,α 2, ·s ,α m with β 1 β 2 ·s β m, The following results are given in this paper: (i) The following inequality is true: eγ Πk = 1m qkα k - Πk = 1m qk - 1 qkα kqk - 1 eγ Πk = 1m pkβ k - Πk = 1m pk - 1 pkβ kpk - 1. (ii) If n = Πk = 1m pkβ k= ( Πk = 1m pk )1 + m(n), m ∞ m(n) > 0 or m ∞ m(n) = + ∞, then m ∞ (eγ n n - σ (n)) > 0 . Where \ β k\ is a sequence, β k ∈ N, β 1 β 2 ·s β m, σ (n) = Σ. d |n d, and γ is the Euler constant. (iii) The probability of Riemann's hypothesis being true is equal to 1. In addition, two results are given when m ∞ m(n) = 0.
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