On the sum of the reciprocals of the differences between consecutive ...

ON THE SUM OF THE RECIPROCALS OF THE DIFFERENCES BETWEEN CONSECUTIVE PRIMES

arXiv:1803.03377v2 [math.NT] 12 Mar 2018

NIANHONG ZHOU Abstract. Let pn denote the n-th prime number, and let dn = pn+1 − pn . Under the Hardy–Littlewood prime-pair conjecture, we prove ( X log log log X X logα dn α = −1, log X ∼ (log log X)1+α X dn α > −1, log X 1+α n≤X

and establish asymptotic properties for some series of dn without the Hardy–Littlewood prime-pair conjecture.

1. Introduction Let pn denote the n-th prime number, and let dn = pn+1 − pn . In [1], Erd¨os and Nathanson show that for c > 2, ∞ X 1 < +∞. (1.1) d n(log log n)c n=3 n

The authors give a heuristic argument explaining why the series must diverge for c = 2. We will prove the above (1.1) by some conclusions of the sieve method. Let H = {0, h1 , . . . , hk−1 } be a set of k(k ≥ 2) distinct integers satisfying 0 < h1 < h2 < · · · < hk−1 and not covering all residue classes to any prime modulus. Also, denote π(x; H) = #{n ∈ N : n + hk−1 ≤ x, n, n + h1 , . . . , n + hk−1 are all primes}. The Hardy–Littlewood prime k-tuple conjecture is that, for X → +∞, X π(X; H) = S(H) k (1 + o(1)) , log X where the singular series  −k Y vH (p) 1 S(H) = 1− 1− , p p p

with p running through all the primes and

vH (p) = #{m (mod p) : m(m + h1 ) . . . (m + hk−1 ) ≡ 0 mod p}. We will also need the following well-known sieve bound, for X sufficiently large, X (1.2) π(X; {0, h, d}) ≤ 23 × 3!S({0, h, d}) 3 (1 + o(1)) , log X when S({0, h, d}) 6= 0. (See Iwaniec and Kowalski’s excellent monograph [2].) To prove our main theorem, we will need the following Hardy–Littlewood prime-pair conjecture. 2010 Mathematics Subject Classification. Primary: 11N05, Secondary: 11N36, 11A41. Key words and phrases. Differences between consecutive primes; Hardy–Littlewood prime-pair conjecture; Applications of sieve methods. 1

Conjecture 1.1. Let X be sufficiently large and d ≪ log X be a natural number. Then X (1.3) π(X, {0, d}) = S({0, d}) 2 (1 + o(1)) , log X where  Q    Q p−1 1 2 1 − (p−1) if d is even, 2 p−2 p>2 p|d,p>2 S({0, d}) =  0 if d is odd. with the product extending over all primes p > 2.

Our main result can be summarized as follows under the above conjecture. Theorem 1.1. Assume that the Hardy–Littlewood prime-pair conjecture holds for all sufficiently large X. Then we have ( X (log log X)1+α X logα dn α > −1, X 1+α ∼ log X log log log X dn α = −1. log X n≤X

Letting α = 0 in above theorem and using Abel’s summation formula, one can obtain the following corollary. Corollary 1.2. Let X be sufficiently large. Then   γc + o(1)  X 1 = log log log X (1 + o(1)) c  d n(log log n) n  (log log X)2−c 3≤n≤X (1 + o(1)) 2−c

c > 2, c = 2, c < 2,

where γc is a constant.

Without the Hardy–Littlewood prime-pair conjecture, using the same idea one can obtain the following result. Theorem 1.3. Let X be sufficiently large. Then ( X (log log X)1+α X logα dn X 1+α ≪ log X log log log X d n log X n≤X

α > −1, α = −1.

Similar to Corollary 1.2, one can obtain the following corollary.

Corollary 1.4. Let X be sufficiently large. Then  γc + o(1)   X 1 X) = O (log   log log c  d n(log log n) 2−c n (log log X)  3≤n≤X O 2−c

c > 2, c = 2, c < 2,

where γc is a constant.

2. Basic Lemma To prove Theorem 1.1, we need the following lemmas. Lemma 2.1. (See [3, Proposition 1]). Let X be sufficiently large. Then X 2 log X S({0, d}) − X + ≪ log 3 X. 2 d≤X As a special case of [4, Lemma 2], we have 2

Lemma 2.2. Let d be an even integer. Then d−1 X

S({0, h, d}) = S({0, d})d(1 + od (1)).

h=1

The following lemma is important in this paper. Lemma 2.3. Let f (x) ∈ C 1 [2, +∞) be strictly monotonically decreasing to 0, and divergence. Also, let X sufficiently large, y = o(log X) and y ≫ log log X. (a) Using Conjecture 1.1, we have Z y X X f (dn ) ∼ f (t) dt. log X 2

R∞ 2

f (t) dt

dn ≤y pn+1 ≤X

(b). Without using Conjecture 1.1, we have Z y X X f (dn ) ≪ f (t) dt. log X 2 d ≤y n

pn+1 ≤X

Proof. The proof of parts (a) and (b) are essentially the same. Therefore, we prove part (a) only. Firstly, we have X X X 1 f (d) f (dn ) = dn ≤y pn+1 ≤X

dn =d pn+1 ≤X

d≤y

=

X

f (d)π(X; {0, d}) +

X d≤y

d≤y





  X 1 − π(X; {0, d}) f (d)  .  dn =d pn+1 ≤X

By the inclusion-exclusion principle, it is easy to see that π(X; {0, d}) −

d−1 X

π(X; {0, h, d}) ≤

X

1 ≤ π(X; {0, d}).

dn =d pn+1 ≤X

h=1

Hence (2.1)

X

dn ≤y pn+1 ≤X

f (dn ) =

X

f (d)π(X; {0, d}) + O

d≤y

X d≤y

f (d)

d−1 X

!

π(X; {0, h, d}) .

h=1

Combining (1.2) with Lemma 2.2, we see that the error term in (2.1) is d−1 X X X X X S({0, h, d}) ≪ ≪ f (d) f (d)dS({0, d}). log3 X d≤y log3 X d≤y h=1

Using Abel’s summation formula, noting that f (x) ∈ C 1 [2, +∞) is strictly monotonically decreasing to 0 and y ≫ log log X, we have ! Z y Z y X X f (d)dS({0, d}) = f (x)x d S({0, d}) ≪ f (x)x dx. d≤y

2

d≤x

3

2

Together with (1.1), we have   Z y X X X X f (dn ) = (2.2) (1 + o(1)) f (d)S({0, d}) + O f (x)x dx . log2 X log3 X 2 d ≤y d≤y n

pn+1 ≤X

Combining Lemma 2.1 and using Abel’s summation formula again, we obtain ! Z y X X f (d)S({0, d}) = f (x) d S({0, d}) 2

d≤y

(2.3) Since

=

Z

d≤x

y

f (x) dx + O(1) + O(f (y) log y) + O 2

Z

y

2

 f (x) dx . x

y

Z y f (x) dx = f (y) log y − f ′ (x) log x dx + O(1) x 2 2 Ry and − 2 f ′ (x) log x dx > 0 by the assumption on f , hence by (2.2) and (2.3) we have Ry  Z y  Z y X f (x)x dx X f (x) 2 f (dn ) = (1 + o(1)) f (x) dx + O . dx + x log X log2 X 2 2 Z

dn ≤y pn+1 ≤X

By using L’Hospital’s rule, we get Ry Ry 2 f (x)x dx f (x)x−1 dx 2R ≤ 1. = 0 and lim R y lim y y→+∞ y y→+∞ f (x) dx f (x) dx 2 2

Hence

X

dn ≤y pn+1 ≤X

X f (dn ) = log2 X

  1 + o(1) + O

y log X

 Z

y

f (x) dx.

2

On noting that y = o(log X), we obtain the proof of part (a).

 R∞

Lemma 2.4. Let f (x) ∈ C 1 [2, +∞) be strictly monotonically decreasing to 0, and 2 f (t) dt 1 divergence. Also, let X sufficiently large and log 2 X ≤ x < y ≤ log2 X. Then we have Z y  X X f (t) dt + f (x) log log X . f (dn ) ≪ log2 X x x