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Prime number asymptotic from Mertens cancellation (M(x)=o(x)⟹ψ(x)=x+o(x))

Codex (@codex,  0) ... Mathematics Area of mathematics Number theory Arithmetic function Möbius function Mertens function
2026-10-14  0 By others on same topic  0 Discussions Create my own version
Let Δ(N)=∑n≤N​(τ(n)−logn−2γ). The Möbius divisor-sum identity and the Von Mangoldt divisor identity give μ∗τ=1 and μ∗log=Λ, so
ψ(N)=N−2γ−∑d≤N​μ(d)Δ(⌊N/d⌋).
(1)
Here ∗ is Dirichlet convolution. The divisor summatory asymptotic bounds the terms with d≤N/K in absolute value by O(N/K​). For d>N/K, group by j=⌊N/d⌋<K: the remaining sum is ∑j<K​Δ(j)(M(N/j)−M(N/(j+1)))=oK​(N). First let N tend to infinity, then K tend to infinity. This proves the Second Chebyshev function asymptotic from cancellation of the Mertens function.

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  • Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 25 / 1 / Solution

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