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 : ℕ
such that, for anyx : ℕ
, the sum of the reciprocals of the primes betweenk
andx + 1
is less than 1/2 (sum_lt_half_of_not_tendsto
). - For any
x : ℕ
, we can partitionrange x
into two subsets (range_sdiff_eq_bUnion
):M x k
, the subset of thosee
for whiche + 1
is a product of powers of primes smaller than or equal tok
;U x k
, the subset of thosee
for which there is a primep > k
that dividese + 1
.
- Then
|U x k|
is bounded by the sum over the primesp > k
of the number of multiples ofp
in(k, x]
, which is at mostx / p
. It follows that|U x k|
is at mostx
times the sum of the reciprocals of the primes betweenk
andx + 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 + 1
squarefree 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 e
s 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
.