Divergence of the Prime Reciprocal Series #
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This file proves Theorem 81 from the 100 Theorems List. The theorem states that the sum of the reciprocals of all prime numbers diverges. The formalization follows Erdős's proof by upper and lower estimates.
Proof outline #
- Assume that the sum of the reciprocals of the primes converges.
- Then there exists a
k : ℕsuch that, for anyx : ℕ, the sum of the reciprocals of the primes betweenkandx + 1is less than 1/2 (sum_lt_half_of_not_tendsto). - For any
x : ℕ, we can partitionrange xinto two subsets (range_sdiff_eq_bUnion):M x k, the subset of thoseefor whiche + 1is a product of powers of primes smaller than or equal tok;U x k, the subset of thoseefor which there is a primep > kthat dividese + 1.
- Then
|U x k|is bounded by the sum over the primesp > kof the number of multiples ofpin(k, x], which is at mostx / p. It follows that|U x k|is at mostxtimes the sum of the reciprocals of the primes betweenkandx + 1, which is less than 1/2 as noted in (2), so|U x k| < x / 2(card_le_mul_sum). - By factoring
e + 1 = (m + 1)² * (r + 1),r + 1squarefree andm + 1 ≤ √x, and noting that squarefree numbers correspond to subsets of[1, k], we find that|M x k| ≤ 2 ^ k * √x(card_le_two_pow_mul_sqrt). - Finally, setting
x := (2 ^ (k + 1))²(√x = 2 ^ (k + 1)), we find that|M x k| ≤ 2 ^ k * 2 ^ (k + 1) = x / 2. Combined with the strict bound for|U k x|from (4),x = |M x k| + |U x k| < x / 2 + x / 2 = x.
References #
https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes
The primes in (k, x].
Equations
- theorems_100.P x k = {p ∈ finset.range (x + 1) | k < p ∧ nat.prime p}
The union over those primes p ∈ (k, x] of the sets of e < x for which e + 1 is a multiple
of p, i.e., those e < x for which there is a prime p ∈ (k, x] that divides e + 1.
Equations
- theorems_100.U x k = (theorems_100.P x k).bUnion (λ (p : ℕ), {e ∈ finset.range x | p ∣ e + 1})
If the sum of the reciprocals of the primes converges, there exists a k : ℕ such that the sum of
the reciprocals of the primes greater than k is less than 1/2.
More precisely, for any x : ℕ, the sum of the reciprocals of the primes between k and x + 1
is less than 1/2.
Removing from {0, ..., x - 1} those elements e for which e + 1 is a product of powers of primes
smaller than or equal to k leaves those e for which there is a prime p > k that divides
e + 1, or the union over those primes p > k of the sets of es for which e + 1 is a multiple
of p.
The number of e < x for which e + 1 has a prime factor p > k is bounded by x times the sum
of reciprocals of primes in (k, x].
The number of e < x for which e + 1 is a squarefree product of primes smaller than or equal to
k is bounded by 2 ^ k, the number of subsets of [1, k].
The number of e < x for which e + 1 is a product of powers of primes smaller than or equal to
k is bounded by 2 ^ k * nat.sqrt x.