The probability of Riemann's hypothesis being true is equal to 1
General Mathematics
2018-04-27 v2
Abstract
Let P be the set of all prime numbers, q1,q2,⋯,qm∈P, Pk be the k-th (k=1,2,⋯m) element of P in ascending order of size, α1,α2,⋯,αm be positive integers, and β1,β2,⋯,βm is a permutation of α1,α2,⋯,αm with β1≥β2≥⋯≥βm, The following results are given in this paper: (i) The following inequality is true: eγloglogk=1∏mqkαk−k=1∏mqk−1qk−qkαk1≥eγloglogk=1∏mpkβk−k=1∏mpk−1pk−pkβk1. (ii) If n=k=1∏mpkβk=(k=1∏mpk)1+εm(n), m→∞limεm(n)>0 or m→∞limεm(n)=+∞, then m→∞lim(eγnloglogn−σ(n))>0 . Where {βk} is a sequence, βk∈N, β1≥β2≥⋯≥βm, σ(n)=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→∞limεm(n)=0.
Cite
@article{arxiv.1609.07555,
title = {The probability of Riemann's hypothesis being true is equal to 1},
author = {Yuyang Zhu},
journal= {arXiv preprint arXiv:1609.07555},
year = {2018}
}
Comments
25 pages