New Estimates for the nth Prime Number

Christian Axler Department of Mathematics Heinrich-Heine-University 40225 Düsseldorf Germany

                    Christian.Axler@hhu.de

Abstract

In this paper we establish new upper and lower bounds for the nth prime number p_{n}, which improve several existing bounds of similar shape. As the main tool, we use some explicit estimates recently obtained for the prime counting function. A further main tool is the use of estimates concerning the reciprocal of log p_{n}. As an application, we derive new estimates for ϑ(p_{n}), where ϑ(x) is Chebyshev’s ϑ-function.

1 Introduction

Let p_{n} denote the nth prime number and let π(x) be the number of primes not exceeding x. In 1896, Hadamard [10] and de la Vallée-Poussin [19] independently proved the asymptotic formula π(x) ∼ x/ log x as x → ∞, which is known as the prime number theorem. (Here log x is the natural logarithm of x.) As a consequence of the prime number theorem, one gets the asymptotic expression

p_{n} ∼ n log n (n → ∞).  (1.1)

Here p_{n} is the nth prime. Cipolla [5] found a more precise result. He showed that for every positive integer m there exist unique monic polynomials T_{1}, …, T_{m} with rational coefficients and deg(T_{k}) = k with

p_{n} = n (log n + log log n − 1 + ∑_{k=1}^{m} \frac{(−1)^{k+1}T_{k}(log log n)}{k log^{k} n}) + O (\frac{n(log log n)^{m+1}}{log^{m+1} n} ) . (1.2)

The polynomials T_{k} can be computed explicitly. In particular, T_{1}(x) = x − 2 and T_{2}(x) = x^{2} − 6x + 11 (see Cipolla [5] or Salvy [18] for further details). Since the computation of the nth prime number is difficult for large n, we are interested in explicit estimates for p_{n}. The asymptotic formula (1.2) yields

p_{n} > n log n,  (1.3)
p_{n} < n(log n + log log n),  (1.4)
p_{n} > n(log n + log log n − 1)  (1.5)

for all sufficiently large values of n. The first result concerning a lower bound for the nth prime number is due to Rosser [15, Theorem 1]. He showed that the inequality (1.3) holds for every positive integer n. In the literature, this result is often called Rosser’s theorem. Moreover, he proved [15, Theorem 2] that

p_{n} < n(log n + 2 log log n)  (1.6)

for every n ≥ 4. The next results concerning the upper and lower bounds that correspond to the first three terms of (1.2) are due to Rosser and Schoenfeld [16, Theorem 3]. They refined Rosser’s theorem and the inequality (1.6) by showing that

p_{n} > n(log n + log log n − 1.5)

for every n ≥ 2 and that the inequality

p_{n} < n(log n + log log n − 0.5)  (1.7)

holds for every n ≥ 20. The inequality (1.7) implies that (1.4) is fulfilled for every n ≥ 6. Based on their estimates for the Chebyshev functions ψ(x) and ϑ(x), Rosser and Schoenfeld [17] announced to have new estimates for the nth prime number p_{n} but they have never published the details. In the direction of (1.5), Robin [14, Lemme 3, Théorème 8] showed that

p_{n} ≥ n(log n + log log n − 1.0072629)  (1.8)

for every n ≥ 2, and that the inequality (1.5) holds for every integer n such that 2 ≤ n ≤ π(10^{11}). Massias and Robin [11, Théorème A] gave a series of improvements of (1.7) and (1.8). For instance, they have found that p_{n} ≥ n(log n+log log n−1.002872) for every n ≥ 2. Dusart [6, p. 54] showed that the inequality

p_{n} ≤ n (log n + log log n − 1 + \frac{log log n − 1.8}{log n})  (1.9)

holds for every n ≥ 27 076. Further, he [7, Theorem 3] made a breakthrough concerning the inequality (1.5) by showing that this inequality holds for every n ≥ 2. The current best estimates for the nth prime, which correspond to the first terms in (1.2), are also given by Dusart [8, Propositions 5.15 and 5.16]. He used explicit estimates for Chebyshev’s ϑ-function to show that the inequality

p_{n} ≤ n (log n + log log n − 1 + \frac{log log n − 2}{log n}) ,  (1.10)

which corresponds to the first four terms of (1.2), holds for every n ≥ 688 383 and that

p_{n} ≥ n (log n + log log n − 1 + \frac{log log n − 2.1}{log n})  (1.11)

for every n ≥ 3. The goal of this paper is to improve the inequalities (1.10) and (1.11) with regard to Cipolla’s asymptotic expansion (1.2). For this purpose, we use estimates for the quantity 1/ log p_{n} and some estimates [3] for the prime counting function π(x) to obtain the following refinement of (1.10).

Theorem 1. For every integer n ≥ 46 254 381, we have

p_{n} < n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2} − 6 log log n + 10.667}{2 log^{2} n}) . (1.12)

Under the assumption that the Riemann hypothesis is true, Dusart [9, Theorem 3.4] found that

p_{n} < n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2} − 6 log log n}{2 log^{2} n}) .  (1.13)

for every integer n ≥ 3468. Using Theorem 1 and a computer for smaller values of n, we get

Corollary 2. The inequality (1.13) holds unconditionally for every n ≥ 3468.

In the other direction, we find the following result which yields a lower bound for the nth prime number in a bounded range.

Theorem 3. For every integer n satisfying 2 ≤ n ≤ π(10^{19}) = 234 057 667 276 344 607, we have

p_{n} > n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2} − 6 log log n + 11.25}{2 log^{2} n}) .

Finally, we use Theorem 3 to give the following improvement of (1.11).

Theorem 4. For every integer n ≥ 2, we have

p_{n} > n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2} − 6 log log n + 11.321}{2 log^{2} n}) . (1.14)

We get the following corollary which was already known under the assumption that the Riemann hypothesis is true (see Dusart [9, Theorem 3.4]).

Corollary 5. For every n ≥ 2, we have

p_{n} > n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2}}{2 log^{2} n} ) .

In Section 6 we apply the Theorems 1 and 4 to find some refined estimates for ϑ(p_{n}), where ϑ(x) = ∑_{p≤x} log p is Chebyshev’s ϑ-function.

Notation 6. Throughout this paper, let n denote a positive integer. For better readability, in the majority of the proofs we use the notation

w = log log n, y = log n, z = log p_{n}.

2 Effective estimates for the reciprocal of log p_{n}

Let m be a positive integer. Using Panaitopol’s asymptotic formula for the prime counting function π(x) — see [12] — we see that

p_{n} = n (log p_{n} − 1 − \frac{1}{log p_{n}} − \frac{3}{log^{2} p_{n}} − ⋯ − \frac{k_{m}}{log^{m} p_{n}} ) + O (\frac{n}{log^{m+1} n}) ,  (2.1)

where the positive integers k_{1}, …, k_{m} are given by the recurrence formula

k_{m} + 1!k_{m−1} + 2!k_{m−2} + ⋯ + (m − 1)!k_{1} = m · m!.

So, in order to prove Theorems 1 and 4, we first use some results of [3] concerning effective estimates for π(x) which imply estimates for the nth prime number p_{n} in the direction of (2.1). Then we apply the estimates for the quantity 1/ log p_{n} obtained in this section. Cipolla [5, p. 139] showed that

\frac{1}{log p_{n}} = \frac{1}{log n} − \frac{log log n}{log^{2} n} + o (\frac{1}{log^{2} n}) .

Concerning this asymptotic formula, we give the following inequality involving 1/ log p_{n}, where the polynomials P_{1}, …, P_{4} ∈ ℤ[x] are given by

P_{1}(x) = 3x^{2} − 6x + 5,
P_{2}(x) = 5x^{3} − 24x^{2} + 39x − 14,
P_{3}(x) = 7x^{4} − 48x^{3} + 120x^{2} − 124x + 51,
P_{4}(x) = 9x^{5} − 80x^{4} + 280x^{3} − 480x^{2} + 405x − 124.

Proposition 7. For every integer n ≥ 688 383, we have

\frac{1}{log p_{n}} ≥ \frac{1}{log n} − \frac{log log n}{log^{2} n} + \frac{(log log n)^{2} − log log n + 1}{log^{2} n log p_{n}} + \frac{1}{log p_{n}} ∑_{k=1}^{4} \frac{(−1)^{k+1}P_{k}(log log n)}{k(k + 1) log^{k+2} n}.

Proof. We just give a sketch of the proof. For details, see [2, Proposition 2.2]. We write w = log log n, y = log n, and z = log p_{n}. By (1.10), the inequality log(1+x) ≤ ∑_{k=1}^{7}(−1)^{k+1}x^{k}/k, which holds for every x > −1, and the fact that (w − 1)/y + (w − 2)/y^{2} > −1, we see that

−y^{2} + (y − w)z ≤ −w^{2} + (y − w) ∑_{k=1}^{7} \frac{(−1)^{k+1}}{k} (\frac{w − 1}{y} + \frac{w − 2}{y^{2}})^{k} .

Finally, we extend the right-hand side of the last inequality to complete the proof.

Corollary 8. For every integer n ≥ 456 914, we have

\frac{1}{log p_{n}} ≥ \frac{1}{log n} − \frac{log log n}{log^{2} n} + \frac{(log log n)^{2} − log log n + 1}{log^{2} n log p_{n}} + \frac{P_{1}(log log n)}{2 log^{3} n log p_{n}} − \frac{P_{2}(log log n)}{6 log^{4} n log p_{n}} .

Proof. See [2, Korollar 2.6].

Corollary 9. For every integer n ≥ 71, we have

\frac{1}{log p_{n}} ≥ \frac{1}{log n} − \frac{log log n}{log^{2} n} + \frac{(log log n)^{2} − log log n + 1}{log^{2} n log p_{n}}.

Proof. Since the inequality

\frac{P_{1}(log log n)}{2 log n} − \frac{P_{2}(log log n)}{6 log^{2} n} ≥ 0  (2.2)

holds for every n ≥ 3, Corollary 8 implies the validity of the required inequality for every n ≥ 456 914. We finish by checking the remaining cases with a computer.

Using a similar method as in the proof of Proposition 7, we find the following inequality involving the reciprocal of log p_{n}. Here, we have

P_{5}(x) = 3x^{2} − 6x + 5.2,
P_{6}(x) = x^{3} − 6x^{2} + 11.4x − 4.2,
P_{7}(x) = 2x^{3} − 7.2x^{2} + 8.4x − 4.41,
P_{8}(x) = x^{3} − 4.2x^{2} + 4.41x.

Proposition 10. For every integer n ≥ 2, we have

\frac{1}{log p_{n}} ≤ \frac{1}{log n} − \frac{log log n}{log^{2} n} + \frac{(log log n)^{2} − log log n + 1}{log^{2} n log p_{n}} + \frac{P_{5}(log log n)}{2 log^{3} n log p_{n}} −∑_{k=4}^{6} \frac{P_{k+2}(log log n)}{2 log^{k} n log p_{n}}.

Proof. First, we consider the case where n ≥ 33. We write again w = log log n, y = log n, and z = log p_{n}. Notice that log(1 + t) ≥ t − t^{2}/2 for every t ≥ 0. If we combine the last fact with (1.11) and (w − 1)/y + (w − 2.1)/y^{2} ≥ 0, we obtain the inequality

−y^{2} + (y − w)z ≥ −w^{2} + (y − w) ∑_{k=1}^{2} \frac{(−1)^{k+1}}{k} (\frac{w − 1}{y} + \frac{w − 2.1}{y^{2}})^{k}

which implies the required inequality. A computer check completes the proof.

Proposition 10 implies the following both corollaries.

Corollary 11. For every integer n ≥ 2, we have

\frac{1}{log p_{n}} ≤ \frac{1}{log n} − \frac{log log n}{log^{2} n} + \frac{(log log n)^{2} − log log n + 1}{log^{2} n log p_{n}} + \frac{P_{5}(log log n)}{2 log^{3} n log p_{n}} −∑_{k=4}^{5} \frac{P_{k+2}(log log n)}{2 log^{k} n log p_{n}}.

Proof. See [2, Korollar 2.20].

Corollary 12. For every integer n ≥ 2, we have

\frac{1}{log p_{n}} ≤ \frac{1}{log n} − \frac{log log n}{log^{2} n} + \frac{(log log n)^{2} − log log n + 1}{log^{2} n log p_{n}} + \frac{P_{5}(log log n)}{2 log^{3} n log p_{n}} − \frac{P_{6}(log log n)}{2 log^{4} n log p_{n}} .

Proof. See [2, Korollar 2.21].

3 Proof of Theorem 1

First, we introduce the following notation. Let the polynomials P_{1}, …, P_{4} ∈ ℤ[x] are given as in the beginning of Section 2. Let A_{0} be a real number with 0.75 ≤ A_{0} < 1 and let F_{0} : ℕ → ℝ be defined by

F_{0}(n) = log n − A_{0} log p_{n}.

From (1.1), it follows that F_{0}(n) is nonnegative for all sufficiently large values of n. Let N_{0} be a positive integer so that F_{0}(n) ≥ 0 for every n ≥ N_{0}. Furthermore, let A_{1} be a real number with 0 < A_{1} ≤ 458.7275, and for w = log log n let F_{1} : ℕ_{≥2} → ℝ be given by

F_{1}(n) = \frac{A_{1}}{log^{5} p_{n}} + \frac{(w^{2} − 3.85w + 14.15)(w^{2} − w + 1)}{log^{4} n log p_{n}} + \frac{2.85P_{1}(w)}{2 log^{3} n log^{2} p_{n}} + \frac{2.85P_{1}(w)}{2 log^{4} n log p_{n}}

+ (\frac{13.15(w^{2} − w + 1)}{log^{2} n log^{2} p_{n}} − \frac{70.7w}{log^{2} n log^{2} p_{n}} ) (\frac{1}{log n} + \frac{1}{log p_{n}} ) − \frac{P_{2}(w)}{6 log^{4} n log p_{n}} .

Then F_{1}(n) is nonnegative for all sufficiently large values of n, and we can define N_{1} to be a positive integer so that F_{1}(n) ≥ 0 for every n ≥ N_{1}. Further we set A_{2} = (458.7275 − A_{1})A_{0}^{5} and A_{3} = 3428.7225A_{0}^{6}. To prove Theorem 1, we first use a recently obtained estimate [3] for the prime counting function π(x) and some results from the previous section to construct a positive integer n_{0} and an arithmetic function b_{0} : ℕ_{≥2} → ℝ, both depending on some parameters, with b_{0}(n) → 10.7 as n → ∞ so that

p_{n} < n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2} − 6 log log n + b_{0}(n)}{2 log^{2} n})

for every n ≥ n_{0}. In order to do this, let a_{0} : ℕ_{≥2} → ℝ be an arithmetic function satisfying

a_{0}(n) ≥ −(log log n)^{2} + 6 log log n,  (3.1)

and let N_{2} be a positive integer depending on the arithmetic function a_{0} so that the inequalities

− 1 < \frac{log log n − 1}{log n} + \frac{log log n − 2}{log^{2} n} − \frac{(log log n)^{2} − 6 log log n + a_{0}(n)}{2 log^{3} n} ≤ 1,  (3.2)

\frac{log log n − 2}{log^{2} n} − \frac{(log log n)^{2} − 6 log log n + a_{0}(n)}{2 log^{3} n} ≥ 0, and  (3.3)

p_{n} < n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2} − 6 log log n + a_{0}(n)}{2 log^{2} n}) (3.4)

hold simultaneously for every n ≥ N_{2}. Now we set

G_{0}(x) = \frac{2x^{3} − 21x^{2} + 82.2x − 98.9}{6e^{3x}} − \frac{x^{4} − 14x^{3} + 53.4x^{2} − 100.6x + 17}{4e^{4x}}

+ \frac{2x^{5} − 10x^{4} + 35x^{3} − 110x^{2} + 150x − 42}{10e^{5x}} − \frac{3x^{4} − 44x^{3} + 156x^{2} − 96x + 64}{24e^{6x}},

and for w = log log n we define

b_{0}(n) = 10.7 + \frac{2A_{2}}{log^{3} n} + \frac{2A_{3}}{log^{4} n} + \frac{a_{0}(n)}{log n} (1 − \frac{w − 1}{log n} − \frac{w − 2}{log^{2} n} + \frac{2w^{2} − 12w + a_{0}(n)}{4 log^{3} n})

− 2 G_{0}(w) log^{2} n + \frac{A_{0}((5.7A_{0} + 8.7)w^{2} − (32A_{0} + 38)w + 147.1A_{0} + 10.7)}{log^{2} n}

+ \frac{2 · 70.7A_{0}^{3}(w^{2} − w + 1)}{log^{4} n} + \frac{2 · 70.7A_{0}^{4}(w^{2} − w + 1)}{log^{4} n}.  (3.5)

Then we obtain the following

Proposition 13. For every integer n ≥ max{N_{0}, N_{1}, N_{2}, 841 424 976}, we have

p_{n} < n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2} − 6 log log n + b_{0}(n)}{2 log^{2} n}) .

In order to prove this proposition, we need the following lemma. Its proof is left to the reader.

Lemma 14. For every x ≥ 2.103, we have

0 ≤ \frac{(x^{2} − 3.85x + 14.15)P_{1}(x)}{2} − \frac{2.85P_{2}(x)}{3} + \frac{P_{3}(x)}{12}
− \frac{(x^{2} − 3.85x + 14.15)P_{2}(x)}{6e^{x}} − \frac{P_{4}(x)}{20e^{x}}.  (3.6)

Now we give a proof of Proposition 13.

Proof of Proposition 13. Let n ≥ max{N_{0}, N_{1}, N_{2}, 841 424 976}. Using [3, Theorem 3] with x = p_{n}, we see that

p_{n} < n (log p_{n} − 1 − \frac{1}{log p_{n}} − \frac{2.85}{log^{2} p_{n}} − \frac{13.15}{log^{3} p_{n}} − \frac{70.7}{log^{4} p_{n}} − \frac{458.7275}{log^{5} p_{n}} − \frac{3428.7225}{log^{6} p_{n}}) .

(3.7)

For convenience, we write w = log log n, y = log n, and z = log p_{n}. By Corollary 8, we have

\frac{1}{z^{2}} ≥ \frac{1}{yz} − \frac{w}{y^{2}z} + \frac{w^{2} − w + 1}{y^{2}z^{2}} + \frac{P_{1}(w)}{2y^{3}z^{2}} − \frac{P_{2}(w)}{6y^{4}z^{2}}.  (3.8)

Again using Corollary 8, we get

\frac{1}{yz} ≥ Φ_{1}(n) = \frac{1}{y^{2}} − \frac{w}{y^{3}} + \frac{w^{2} − w + 1}{y^{3}z} + \frac{P_{1}(w)}{2y^{4}z} − \frac{P_{2}(w)}{6y^{5}z}.  (3.9)

Applying (3.9) to (3.8), we see that

\frac{1}{z^{2}} ≥ Φ_{2}(n),  (3.10)

where

Φ_{2}(n) = \frac{1}{y^{2}} − \frac{w}{y^{3}} − \frac{w}{y^{2}z} + \frac{w^{2} − w + 1}{y^{3}z} + \frac{w^{2} − w + 1}{y^{2}z^{2}} + (\frac{P_{1}(w)}{2y^{3}z} − \frac{P_{2}(w)}{6y^{4}z}) (\frac{1}{y} + \frac{1}{z}) .

Now (2.2) implies that

\frac{1}{z^{2}} ≥ Φ_{3}(n) = \frac{1}{y^{2}} − \frac{w}{y^{3}} − \frac{w}{y^{2}z} + \frac{w^{2} − w + 1}{y^{3}z} + \frac{w^{2} − w + 1}{y^{2}z^{2}}.  (3.11)

We assumed n ≥ N_{0}. Hence F_{0}(n) ≥ 0, which is equivalent to

\frac{A_{0}}{y} ≤ \frac{1}{z}.  (3.12)

From (3.12) and the fact that 2.85x^{2} − 16x + 73.55 ≥ 0 for every x ≥ 0, it follows

\frac{2.85w^{2} − 16w + 73.55}{z^{2}} ≥ \frac{A_{0}(5.7w^{2} − 32w + 147.1)}{2yz}.  (3.13)

Let f(x) = (5.7A_{0} + 8.7)x^{2} − (32A_{0} + 38)x + 147.1A_{0} + 10.7. Since 0.75 ≤ A_{0} < 1, we get f(x) ≥ 12.975x^{2} − 70x + 121.025 ≥ 0 for every x ≥ 0. Using (3.12) and (3.13), we get

\frac{2.85w^{2} − 16w + 73.55}{z^{2}} + \frac{8.7w^{2} − 38w + 10.7}{2yz} ≥ \frac{A_{0}f(w)}{2y^{2}}.  (3.14)

We recall that A_{2} = (458.7275 − A_{1})A_{0}^{5} and A_{3} = 3428.7225A_{0}^{6}. Hence (3.12) implies that

\frac{A_{2}}{y^{5}} + \frac{A_{3}}{y^{6}} + \frac{70.7A_{0}^{3}}{y^{6}} + \frac{70.7A_{0}^{4}}{y^{6}} ≤ \frac{458.7275 − A_{1}}{z^{5}} + \frac{3428.7225}{z^{6}} + \frac{70.7}{y^{3}z^{3}} + \frac{70.7}{y^{2}z^{4}} .  (3.15)

Now we apply (3.14) and (3.15) to (3.5) and see that

\frac{10.7 − b_{0}(n)}{2y^{2}} + \frac{2.85(w^{2} − w + 1)}{y^{2}z^{2}} − \frac{13.15w}{y^{2}z^{2}} + \frac{70.7}{y^{2}z^{2}} + \frac{8.7w^{2} − 38w + 10.7}{2y^{3}z}

+ \frac{458.7275 − A_{1}}{z^{5}} + \frac{3428.7225}{z^{6}} + \frac{70.7(w^{2} − w + 1)}{y^{2}z^{3}} (\frac{1}{y} + \frac{1}{z})

≥ G_{0}(w) − \frac{a_{0}(n)}{2y^{3}} (1 − \frac{w − 1}{y} − \frac{w − 2}{y^{2}} + \frac{2w^{2} − 12w + a_{0}(n)}{4y^{3}}) . (3.16)

The inequality (2.2) tells us that

\frac{13.15}{z} (\frac{P_{1}(w)}{2y^{3}z} − \frac{P_{2}(w)}{6y^{4}z}) (\frac{1}{y} + \frac{1}{z}) ≥ 0.  (3.17)

Adding the left-hand side of (3.17) and the right-hand side of (3.6) with x = w to the left-hand side of (3.16), we get

\frac{5.35}{y^{2}} − \frac{b_{0}(n)}{2y^{2}} + \frac{2.85(w^{2} − w + 1)}{y^{2}z^{2}} − \frac{13.15w}{y^{2}z^{2}} + \frac{70.7}{y^{2}z^{2}} + \frac{8.7w^{2} − 38w + 10.7}{2y^{3}z} + \frac{458.7275 − A_{1}}{z^{5}}

+ \frac{3428.7225}{z^{6}} + \frac{70.7(w^{2} − w + 1)}{y^{2}z^{3}} (\frac{1}{y} + \frac{1}{z}) + \frac{13.15}{z} (\frac{P_{1}(w)}{2y^{3}z} − \frac{P_{2}(w)}{6y^{4}z}) (\frac{1}{y} + \frac{1}{z})

− \frac{2.85P_{2}(w)}{6y^{5}z} − \frac{2.85P_{2}(w)}{6y^{4}z^{2}} + \frac{(w^{2} − 3.85w + 14.15)P_{1}(w)}{2y^{5}z} + \frac{P_{3}(w)}{12y^{5}z} − \frac{P_{4}(w)}{20y^{6}z}

− \frac{(w^{2} − 3.85w + 14.15)P_{2}(w)}{6y^{6}z}

≥ G_{0}(w) − \frac{a_{0}(n)}{2y^{3}} (1 − \frac{w − 1}{y} − \frac{w − 2}{y^{2}} + \frac{2w^{2} − 12w + a_{0}(n)}{4y^{3}}) .

Since n ≥ N_{1}, we have F_{1}(n) ≥ 0. Now we add F_{1}(n) to the left-hand side of the last inequality, use the identity 8.7w^{2} −38w +10.7 = P_{1}(w)+2·2.85(w^{2} −w +1)−2·13.15w, and collect all terms containing the number 70.7 and the term w^{2} − 3.85w + 14.15, respectively, to get

\frac{5.35}{y^{2}} − \frac{b_{0}(n)}{2y^{2}} + \frac{2.85(w^{2} − w + 1)}{y^{2}z^{2}} − \frac{13.15w}{y^{2}z^{2}} + \frac{70.7}{z^{2}} · Φ_{3}(n) + \frac{458.7275}{z^{5}} + \frac{3428.7225}{z^{6}}

+ \frac{2.85(w^{2} − w + 1)}{y^{3}z} − \frac{13.15w}{y^{3}z} + (2.85 + \frac{13.15}{z}) (\frac{P_{1}(w)}{2y^{3}z} − \frac{P_{2}(w)}{6y^{4}z}) (\frac{1}{y} + \frac{1}{z})

+ \frac{w^{2} − 3.85w + 14.15}{y} · Φ_{1}(n) + \frac{P_{1}(w)}{2y^{3}z} − \frac{P_{2}(w)}{6y^{4}z} + \frac{P_{3}(w)}{12y^{5}z} − \frac{P_{4}(w)}{20y^{6}z}

+ \frac{13.15(w^{2} − w + 1)}{y^{2}z^{2}} (\frac{1}{y} + \frac{1}{z}) − \frac{2.85w}{y^{3}}

≥ G̃_{0}(w) − \frac{a_{0}(n)}{2y^{3}} (1 − \frac{w − 1}{y} − \frac{w − 2}{y^{2}} + \frac{2w^{2} − 12w + a_{0}(n)}{4y^{3}}) ,

where Φ_{1}(n) and Φ_{3}(n) are given as in (3.9) and (3.11), respectively, and

G̃_{0}(x) = G_{0}(x) + \frac{x^{2} − 3.85x + 14.15}{e^{3x}} − \frac{x^{3} − 3.85x^{2} + 14.15x}{e^{4x}} − \frac{2.85x}{e^{3x}}.

Now we use (3.9) and (3.11) and collect all terms containing the numbers 2.85 and 13.15 to see that

\frac{2.5}{y^{2}} − \frac{b_{0}(n)}{2y^{2}} + (2.85 + \frac{13.15}{z}) Φ_{2}(n) + \frac{70.7}{z^{4}} + \frac{458.7275}{z^{5}} + \frac{3428.7225}{z^{6}} + \frac{w^{2} − w + 1}{y^{2}z}

+ \frac{P_{1}(w)}{2y^{3}z} − \frac{P_{2}(w)}{6y^{4}z} + \frac{P_{3}(w)}{12y^{5}z} − \frac{P_{4}(w)}{20y^{6}z}

≥ G̃_{0}(w) − \frac{a_{0}(n)}{2y^{3}} (1 − \frac{w − 1}{y} − \frac{w − 2}{y^{2}} + \frac{2w^{2} − 12w + a_{0}(n)}{4y^{3}}) .

Applying (3.10) and Proposition 7, we get

\frac{2.5}{y^{2}} − \frac{b_{0}(n)}{2y^{2}} + \frac{2.85}{z^{2}} + \frac{13.15}{z^{3}} + \frac{70.7}{z^{4}} + \frac{458.7275}{z^{5}} + \frac{3428.7225}{z^{6}} − \frac{1}{y} + \frac{w}{y^{2}} + \frac{1}{z}

≥ G̃_{0}(w) − \frac{a_{0}(n)}{2y^{3}} (1 − \frac{w − 1}{y} − \frac{w − 2}{y^{2}} + \frac{2w^{2} − 12w + a_{0}(n)}{4y^{3}}) .

A straightforward calculation shows that the last inequality is equivalent to

−\frac{1}{y} − \frac{w^{2} − 4w − (4 − b_{0}(n))}{2y^{2}} + \frac{1}{z} + \frac{2.85}{z^{2}} + \frac{13.15}{z^{3}} + \frac{70.7}{z^{4}} + \frac{458.7275}{z^{5}} + \frac{3428.7225}{z^{6}}

≥ −\frac{w^{2} − 6w + a_{0}(n)}{2y^{3}} − \frac{1}{2} (\frac{w − 1}{y} + \frac{w − 2}{y^{2}} − \frac{w^{2} − 6w + a_{0}(n)}{2y^{3}})^{2}

+ \frac{1}{3} (\frac{w − 1}{y} + \frac{w − 2}{y^{2}})^{3} − \frac{1}{4} (\frac{w − 1}{y})^{4} + \frac{1}{5} (\frac{w − 1}{y})^{5} .

We add (w − 1)/y + (w − 2)/y^{2} to both sides of this inequality. Since log(1 + x) ≤ ∑_{k=1}^{5}(−1)^{k+1}x/k for every x > −1, g(x) = x^{3}/3 is increasing, and h(x) = −x^{4}/4 + x^{5}/5 is decreasing on the interval [0, 1], we can use (3.1)–(3.3) to get

y + w − 1 + \frac{w − 2}{y} − \frac{w^{2} − 6w + b_{0}(n)}{2y^{2}} + \frac{1}{z} + \frac{2.85}{z^{2}} + \frac{13.15}{z^{3}} + \frac{70.7}{z^{4}} + \frac{458.7275}{z^{5}} + \frac{3428.7225}{z^{6}}

≥ y + w − 1 + log (1 + \frac{w − 1}{y} + \frac{w − 2}{y^{2}} − \frac{w^{2} − 6w + a_{0}(n)}{2y^{3}}) .

Finally, we use (3.4) and (3.7) to arrive at the desired result.

Next we use Proposition 13 and the following both lemmata to prove Theorem 1. In the first lemma we determine a suitable value of N_{0} for A_{0} = 0.87.

Lemma 15. For every integer n ≥ 1 338 564 587, we have

log n ≥ 0.87 log p_{n}.

Proof. We set

f(x) = e^{x} − 0.87 (e^{x} + x + log (1 + \frac{x − 1}{e^{x}} + \frac{x − 2}{e^{2x}})) .

Since f′(x) ≥ 0 for every x ≥ 2.5 and f(3.046) ≥ 0.00137, we see that f(x) ≥ 0 for every x ≥ 3.046. Substituting x = log log n in f(x) and using (1.10), we see that the desired inequality holds for every n ≥ exp(exp(3.046)). We check the remaining cases with a computer.

Now we use Lemma 15 to find a suitable value of N_{1} for A_{1} = 155.32.

Lemma 16. Let A_{1} = 155.32. Then F_{1}(n) ≥ 0 for every n ≥ 100 720 878.

Proof. First, let n ≥ exp(exp(3.05)). We have

F_{1}(n) = \frac{155.32}{z^{5}} + \frac{f(w)}{6y^{4}z} + \frac{34.85w^{2} − 184.8w + 40.55}{2y^{3}z^{2}} + \frac{13.15w^{2} − 83.85w + 13.15}{y^{2}z^{3}}.

where f(x) = 6x^{4} − 34.1x^{3} + 163.65x^{2} − 198.3x + 141.65. Since f(x) ≥ 0 for every x ≥ 3.05, it suffices to show that

\frac{155.32}{z^{5}} + \frac{6w^{4} − 34.1w^{3} + 268.2w^{2} − 752.7w + 263.3}{6y^{3}z^{2}} + \frac{13.15w^{2} − 83.85w + 13.15}{y^{2}z^{3}} ≥ 0.

(3.18)

In order to do this, we set

g(x) = (6x^{4} − 34.1x^{3} + 268.2x^{2} − 752.7x + 263.3)(e^{x} + x)
+ 6e^{x}(13.15x^{2} − 83.85x + 13.15 + 155.32 · 0.87^{2}).

It is easy to see that h_{1}(x) = 6x^{4} − 10.1x^{3} + 244.8x^{2} − 561.6x − 208.229752 ≥ 0 for every x ≥ 2.6 and that h_{2}(x) = 30x^{4} − 136.4x^{3} + 804.6x^{2} − 1505.4x + 263.3 ≥ 0 for every x ≥ 2.2. Hence g′(x) = h_{1}(x)e^{x} + h_{2}(x) ≥ 0 for every x ≥ 2.6. We also have g(3.05) ≥ 0.9. Therefore, g(x) ≥ 0 for every x ≥ 3.05. Since 6x^{4} − 34.1x^{3} + 268.2x^{2} − 752.7x + 263.3 ≥ 0 for every x ≥ 3.05, we can use (1.3) to get g(w)/(6y^{3}z^{3}) ≥ 0. Now we apply Lemma 15 to obtain (3.18). We finish by direct computation.

Finally, we give a proof of Theorem 1.

Proof of Theorem 1. For convenience, we write w = log log n and y = log n. Setting A_{0} = 0.87 and A_{1} = 155.32, we use Lemma 15 and Lemma 16 to get N_{0} = 1 338 564 587 and N_{1} = 100 720 878, respectively. The proof of this theorem goes in two steps.

Step 1. We set a_{0}(n) = −w^{2} + 6w. Then N_{2} = 688 383 is a suitable choice for N_{2}. By (3.5), we get

b_{0}(n) ≥ 10.7 + g(n),  (3.19)

where

g(n) = −\frac{2w^{3} − 18w^{2} + 64.2w − 98.9}{3y} + \frac{w^{4} − 12w^{3} + 63.16w^{2} − 203.17w + 258.29}{2y^{2}}

− \frac{2w^{5} − 10w^{4} + 30w^{3} − 70w^{2} + 90w − 1554.24}{5y^{3}}

− \frac{8w^{3} − 2137.44w^{2} + 2185.45w − 37836.25}{12y^{4}}.

We define

g_{1}(x, t) = 3.54e^{4x} + 20(18x^{2} + 98.9)e^{3x} − 20(2t^{3} + 64.2t)e^{3t}
+ 30(x^{4} + 63.16x^{2} + 258.29)e^{2x} − 30(12t^{3} + 203.17t)e^{2t}
+ 12(10x^{4} + 70x^{2} + 1554.24)e^{x} − 12(2t^{5} + 30t^{3} + 90t)e^{t}
+ 5(2137.44x^{2} + 37836.25) − 5(8t^{3} + 2185.45t).

If t_{0} ≤ x ≤ t_{1}, then g_{1}(x, x) ≥ g_{1}(t_{0}, t_{1}). We check with a computer that g_{1}(i · 10^{−5}, (i + 1) · 10^{−5}) ≥ 0 for every integer i with 0 ≤ i ≤ 699 999. Therefore,

g(n) + 0.059 = \frac{g_{1}(w, w)}{60y^{4}} ≥ 0 (0 ≤ w ≤ 7).  (3.20)

Next we prove that g_{1}(x, x) ≥ 0 for every x ≥ 7. For this purpose, let W_{1}(x) = 3.54e^{x} − 20(2x^{3} − 18x^{2} + 64.2x − 98.9). It is easy to show that W_{1}(x) ≥ 792 for every x ≥ 7. Hence we get

g_{1}(x, x) ≥ (792e^{x} + 30(x^{4} − 12x^{3} + 63.16x^{2} − 203.17x + 258.29))e^{2x}
− 12(2x^{5} − 10x^{4} + 30x^{3} − 70x^{2} + 90x − 1554.24)e^{x}
− 5(8x^{3} − 2137.44x^{2} + 2185.45x − 37836.25).

Since 792e^{t} + 30(t^{4} − 12t^{3} + 63.16t^{2} − 203.17t + 258.29) ≥ 875 011 for every t ≥ 7, we obtain g(n) + 0.059 = g_{1}(w, w)/(60y^{4}) ≥ 0 for w ≥ 7. Combined with (3.20), it gives that g(n) ≥ −0.059 for every n ≥ 3. Applying this to (3.19), we get b_{0}(n) ≥ 10.641 for every n ≥ 3. Hence, by Proposition 13, we get

p_{n} < n (y + w − 1 + \frac{w − 2}{y} − \frac{w^{2} − 6w + 10.641}{2y^{2}})

for every n ≥ 1 338 564 587. For every integer n such that 39 529 802 ≤ n ≤ 1 338 564 586 we check the last inequality with a computer.

Step 2. We set a_{0}(n) = 10.641. Using the first step, we can choose N_{2} = 39 529 802. By (3.5), we have

b_{0}(n) ≥ 10.7 + h(n),  (3.21)

where h(n) is given by

h(n) = −\frac{2w^{3} − 21w^{2} + 82.2w − 130.823}{3y} + \frac{w^{4} − 14w^{3} + 77.16w^{2} − 236.45w + 279.57}{2y^{2}}

− \frac{2w^{5} − 10w^{4} + 35w^{3} − 110w^{2} + 203.205w − 1660.65}{5y^{3}}

+ \frac{3w^{4} − 44w^{3} + 2309.28w^{2} − 2568.52w + 38175.947}{12y^{4}}.

We set

h_{1}(x, t) = 1.98e^{4x} + 20(21x^{2} + 130.823)e^{3x} − 20(2t^{3} + 82.2t)e^{3t}
+ 30(x^{4} + 77.16x^{2} + 279.57)e^{2x} − 30(14t^{3} + 236.45t)e^{2t}
+ 12(10x^{4} + 110x^{2} + 1660.65)e^{x} − 12(2t^{5} + 35t^{3} + 203.205t)e^{t}
+ 5(3x^{4} + 2309.28x^{2} + 38175.947) − 5(44t^{3} + 2568.52t).

Clearly, h_{1}(x, x) ≥ h_{1}(t_{0}, t_{1}) for every x such that t_{0} ≤ x ≤ t_{1}. We use a computer to verify that h_{1}(i · 10^{−6}, (i + 1) · 10^{−6}) ≥ 0 for every integer i with 0 ≤ i ≤ 7 999 999. Therefore,

h(n) + 0.033 = \frac{h_{1}(w, w)}{60y^{4}} ≥ 0 (0 ≤ w ≤ 8).  (3.22)

We next show that h_{1}(x, x) ≥ 0 for every x ≥ 8. Since 1.98e^{t} − 20(2t^{3} − 21t^{2} + 82.2t − 130.823) ≥ 1766 for every t ≥ 8, we have

h_{1}(x, x) ≥ 1766e^{3x} + 30(x^{4} − 14x^{3} + 77.16x^{2} − 236.45x + 279.57)e^{2x}
− 12(2x^{5} − 10x^{4} + 35x^{3} − 110x^{2} + 203.205x − 1660.65)e^{x}
+ 5(3x^{4} − 44x^{3} + 2309.28x^{2} − 2568.52x + 38175.947).

Note that 1766e^{t} + 30(t^{4} − 14t^{3} + 77.16t^{2} − 236.45t + 279.57) ≥ 5 271 998 for every t ≥ 8. Hence h(n) + 0.033 = h_{1}(w, w)/(60y^{4}) ≥ 0 for w ≥ 8. Combined with (3.22) and (3.21), this gives b_{0}(n) ≥ 10.667 for every n ≥ 3. Applying this to Proposition 13, we complete the proof of the required inequality for every n ≥ 1 338 564 587. We verify the remaining cases with a computer.

Denoting the right-hand side of (1.10) by D_{up}(n) and the right-hand side (1.12) by A_{up}(n), we use \underline{A006988} to compare the error term of the approximation from Theorem 1 with Dusart’s approximation from (1.10) for the 10^{n}th prime number:

\begin{matrix}n & p_{n} & ⌈D_{up}(n) − p_{n}⌉ & ⌈A_{up}(n) − p_{n}⌉ \\ 10^{10} & 252 097 800 623 & 20 510 784 & 4 613 984 \\ 10^{11} & 2 760 727 302 517 & 172 884 400 & 38 768 198 \\ 10^{12} & 29 996 224 275 833 & 1 469 932 710 & 311 593 524 \\ 10^{13} & 323 780 508 946 331 & 12 732 767 836 & 2 542 231 421 \\ 10^{14} & 3 475 385 758 524 527 & 112 026 014 682 & 21 049 069 521 \\ 10^{15} & 37 124 508 045 065 437 & 998 861 791 991 & 176 995 293 694 \\ 10^{16} & 394 906 913 903 735 329 & 9 004 342 407 404 & 1 507 803 850 451 \\ 10^{17} & 4 185 296 581 467 695 669 & 81 924 060 077 026 & 12 998 658 322 559 \\ 10^{18} & 44 211 790 234 832 169 331 & 751 154 982 343 786 & 113 204 602 033 556 \\ 10^{19} & 465 675 465 116 607 065 549 & 6 932 757 377 044 654 & 994 838 584 902 026 \\ 10^{20} & 4 892 055 594 575 155 744 537 & 64 346 895 915 006 577 & 8 812 315 669 274 243 \end{matrix}

4 Proof of Theorem 3

In order to do prove Theorem 3, we introduce the logarithmic integral li(x) which is defined for every real x ≥ 0 as

li(x) = ∫_{0}^{x} \frac{dt}{log t} = lim_{ε→0+} \{∫_{0}^{1−ε} \frac{dt}{log t} + ∫_{1+ε}^{x} \frac{dt}{log t}\} .

Proof of Theorem 3. Let x_{0} = 3 273 361 096. First, we verify the required inequality for every integer n with x_{0} ≤ n ≤ π(10^{19}). For x > 1, the logarithmic integral li(x) is increasing with li((1, ∞)) = ℝ. Thus, we can define the inverse function li^{−1} : ℝ → (1, ∞) by

li(li^{−1}(x)) = x.  (4.1)

Further, let

f(x) = x−li (x (log x + log log x − 1 + \frac{log log x − 2}{log x} − \frac{(log log x)^{2} − 6 log log x + 11.25}{2 log^{2} x})) .

We show that f(x) > 0 for every x ≥ x_{0}. We have f(x_{0}) > 0.000001. So it suffices to show that f′(x) ≥ 0 for every x ≥ x_{0}. Setting

g_{1}(a, b) = log (1 + \frac{log a − 1}{a} + \frac{log a − 2}{a^{2}} − \frac{log^{2} b − 6 log b + 11.25}{2b^{3}})

and g(z) = g_{1}(z, z), we see that (z + log z + g(z))f′(e^{z}) = h(z), where

h(z) = g(z) − \frac{log z − 1}{z} + \frac{log^{2} z − 4 log z + 5.25}{2z^{2}} − \frac{log^{2} z − 7 log z + 14.25}{z^{3}}.

Since z + log z + g(z) > 0 for every z ≥ 2.1, it suffices to verify that h(z) ≥ 0 for every z ≥ log x_{0}. We have h(log x_{0}) ≥ 0.000026 and

(−4)z^{7}e^{g(z)}h′(z) = z^{4} + (4 log^{3} z − 46 log^{2} z + 197 log z − 323.5)z^{3}
+ (−6 log^{3} z + 60 log^{2} z − 175.5 log z + 90)z^{2}
+ (−2 log^{4} z + 10 log^{3} z + 19 log^{2} z − 183.5 log z + 234.876)z
+ 6 log^{4} z − 82 log^{3} z + 443 log^{2} z − 1114.5 log z + 1119.375. (4.2)

In order to show that h′(z) > 0 for every z ∈ J = [log x_{0}, 29.8], it suffices to show that the right-hand side of (4.2) is negative. Since z + 4 log^{3} z − 46 log^{2} z + 197 log z − 325.5 < 1.43 for every z ∈ J, we get

(−4)z^{7}e^{g(z)}h′(z) < 1.43z^{3} + (−6 log^{3} z + 60 log^{2} z − 175.5 log z + 90)z^{2}
+ (−2 log^{4} z + 10 log^{3} z + 19 log^{2} z − 183.5 log z + 234.876)z
+ 6 log^{4} z − 82 log^{3} z + 443 log^{2} z − 1114.5 log z + 1119.375.

Notice that 1.43z − 6 log^{3} z + 60 log^{2} z − 175.5 log z + 90 ≤ −0.444 for every z ∈ J. Hence

(−4)z^{7}e^{g(z)}h′(z) < −0.444z^{2} + (−2 log^{4} z + 10 log^{3} z + 19 log^{2} z − 183.5 log z + 234.876)z
+ 6 log^{4} z − 82 log^{3} z + 443 log^{2} z − 1114.5 log z + 1119.375.

We have −0.444z − 2 log^{4} z + 10 log^{3} z + 19 log^{2} z − 183.5 log z + 234.876 ≤ −47.701 for every z ∈ J. Hence (−4)z^{7}e^{g(z)}h′(z) < 0 for every z ∈ J which yields that h′(z) > 0 for every z ∈ J. Combined with h(log x_{0}) > 0, it turns out that h(z) > 0 for every z ∈ [log x_{0}, 29.8]. Similar, we get h′(z) < 0 for every z ≥ 29.88. Together with lim_{z→∞} h(z) = 0, we see that h(z) ≥ 0 for every z ≥ 29.88. It remains to consider the case where z ∈ (29.8, 29.88). If a ≤ z ≤ b, then

h(z) ≥ h_{1}(a, b) = g_{1}(a, b) − \frac{log b − 1}{b} + \frac{log^{2} a − 4 log a + 5.25}{2a^{2}} − \frac{log^{2} b − 7 log b + 14.25}{b^{3}}.

Now we check with a computer that h_{1}(29.8, 29.88) > 0. Hence f(x) > 0 for every x ≥ x_{0}. Since li(x) is increasing for x > 1, we can use (4.1) to get

x (log x + log log x − 1 + \frac{log log x − 2}{log x} − \frac{(log log x)^{2} − 6 log log x + 11.25}{2 log^{2} x}) < li^{−1}(x)

for every x ≥ x_{0}. Applying [13, Lemma 7] to the last inequality, we see that the desired inequality holds for every integer n satisfying 3 273 361 096 ≤ n ≤ π(10^{19}). For every integer n such that 2 ≤ n < 3 273 361 096 we check the desired inequality with a computer.

5 Proof of Theorem 4

Compared with the proof of Theorem 3, the proof of Theorem 4 is rather technical and we need to introduce some notation. First, let

P_{9}(x) = P_{5}(x) + 2 · 3.15(x^{2} − x + 1),
P_{10}(x) = (x^{2} − x + 1)P_{9}(x) + (x^{2} − x + 1)P_{5}(x) − 3.15P_{6}(x) − P_{7}(x) + 12.85P_{5}(x),
P_{11}(x) = 3.15P_{7}(x) + 12.85P_{6}(x),
P_{12}(x) = 2(x^{2} − x + 1)P_{6}(x) − P_{5}(x)P_{9}(x),

where the polynomials P_{5}, P_{6}, P_{7}, and P_{8} were defined as in Section 2. Let B_{1}, …, B_{10} be real positive constants satisfying

B_{6} + B_{7} + B_{8} + B_{9} + B_{10} ≤ 3.15.  (5.1)

Writing w = log log n, y = log n, and z = log p_{n}, we define H_{i} : ℕ_{≥2} → ℝ, where 1 ≤ i ≤ 10, by

• H_{1}(n) = \frac{B_{1}w}{y^{3}z} − \frac{P_{10}(w)}{2y^{5}z} + \frac{P_{11}(w)}{2y^{5}z^{2}} + \frac{P_{12}(w)}{4y^{6}z} + \frac{12.85P_{6}(w)}{2y^{4}z^{3}},

• H_{2}(n) = \frac{B_{2}w}{y^{3}z} + \frac{12.85w}{y^{2}z^{2}} − \frac{71.3}{z^{4}},

• H_{3}(n) = \frac{B_{3}w}{y^{3}z} − \frac{3.15P_{5}(w)}{2y^{3}z^{2}} − \frac{12.85(w^{2} − w + 1)}{y^{3}z^{2}},

• H_{4}(n) = \frac{B_{4}w}{y^{3}z} + \frac{3.15P_{6}(w) − 12.85P_{5}(w)}{2y^{4}z^{2}},

• H_{5}(n) = \frac{B_{5}w}{y^{3}z} + \frac{P_{6}(w) − 3.15P_{5}(w)}{2y^{4}z} − \frac{12.85(w^{2} − w + 1)}{y^{4}z} − \frac{(w^{2} − w + 1)^{2}}{y^{4}z},

• H_{6}(n) = \frac{B_{6}w}{y^{2}z} + \frac{(12.85 − B_{1} − B_{2} − B_{3} − B_{4} − B_{5})w}{y^{3}z} − \frac{3.15(w^{2} − w + 1)}{y^{2}z^{2}},

• H_{7}(n) = \frac{B_{7}w}{y^{2}z} − \frac{12.85P_{5}(w)}{2y^{3}z^{3}},

• H_{8}(n) = \frac{B_{8}w}{y^{2}z} − \frac{12.85(w^{2} − w + 1)}{y^{2}z^{3}},

• H_{9}(n) = \frac{B_{9}w}{y^{2}z} − \frac{463.2275}{z^{5}},

• H_{10}(n) = \frac{B_{10}w}{y^{2}z} − \frac{4585}{z^{6}}.

Then H_{i}(n), 1 ≤ i ≤ 10, is nonnegative for all sufficiently large values of n. Let K_{1} be a positive integer so that H_{i}(n) ≥ 0, 1 ≤ i ≤ 10, for every n ≥ K_{1}. Let a_{1} : ℕ_{≥2} → ℝ be an arithmetic function and let K_{2} be a positive integer, which depends on a_{1}, so that the inequalities

a_{1}(n) > −(log log n)^{2} + 6 log log n,  (5.2)
0 ≤ \frac{log log n − 1}{log n} + \frac{log log n − 2}{log^{2} n} − \frac{(log log n)^{2} − 6 log log n + a_{1}(n)}{2 log^{3} n} ≤ 1, and (5.3)
p_{n} > n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2} − 6 log log n + a_{1}(n)}{2 log^{2} n}) (5.4)

hold simultaneously for every n ≥ K_{2}. Furthermore, we define the function G_{1} : ℝ → ℝ by

G_{1}(x) = \frac{3.15x}{e^{3x}} − \frac{12.85}{e^{3x}} + \frac{12.85x}{e^{4x}} − \frac{x^{2} − x + 1}{e^{3x}} + \frac{(x^{2} − x + 1)x}{e^{4x}} − \frac{P_{9}(x)}{2e^{4x}} + \frac{P_{9}(x)x}{2e^{5x}}

+ \frac{(x − 1)^{2}}{2e^{2x}} − \frac{x^{2} − 6x}{2e^{3x}} − ∑_{k=2}^{4} \frac{(−1)^{k}}{k} (\frac{x − 1}{e^{x}} + \frac{x − 2}{e^{2x}})^{k} + \frac{(x − 2)^{4}}{4e^{8x}} .

In order to prove Theorem 4, we set

b_{1}(n) = 11.3 − 2 G_{1}(log log n) log^{2} n + \frac{a_{1}(n)}{log n}
− \frac{2A_{0}(3.15 − (B_{6} + B_{7} + B_{8} + B_{9} + B_{10})) log log n}{log n}  (5.5)

and prove the following proposition.

Proposition 17. For every integer n ≥ max{N_{0}, K_{1}, K_{2}, 3520}, we have

p_{n} > n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2} − 6 log log n + b_{1}(n)}{2 log^{2} n}) .

The following lemma is helpful for the proof of Proposition 17. The proof is left to the reader.

Lemma 18. Let w = log log n. For every integer n ≥ 6, we have

\frac{12.85P_{6}(w)}{2 log^{6} n log p_{n}} + \frac{3.15P_{7}(w)}{2 log^{6} n log p_{n}} + \frac{P_{8}(w)}{2 log^{6} n log p_{n}} ≥ 0,

and for every integer n ≥ 17, we have

\frac{P_{6}(w)P_{9}(w)}{4 log^{7} n log p_{n}} + \frac{12.85P_{7}(w)}{2 log^{7} n log p_{n}} + \frac{3.15P_{8}(w)}{2 log^{7} n log p_{n}} + \frac{3.15P_{8}(w)}{2 log^{6} n log^{2} p_{n}} ≥ \frac{(w − 2)^{4}}{4 log^{8} n} .

Now we give a proof of Proposition 17.

Proof of Proposition 17. Let n ≥ max{N_{0}, K_{1}, K_{2}, 3520}. By [3, Theorem 2], we have

p_{n} > n (log p_{n} − 1 − \frac{1}{log p_{n}} − \frac{3.15}{log^{2} p_{n}} − \frac{12.85}{log^{3} p_{n}} − \frac{71.3}{log^{4} p_{n}} − \frac{463.2275}{log^{5} p_{n}} − \frac{4585}{log^{6} p_{n}} ) . (5.6)

For convenience, we write w = log log n, y = log n, and z = log p_{n}. From Corollary 12, it follows that

− \frac{1}{z} ≥ Ψ_{1}(n) = −\frac{1}{y} + \frac{w}{y^{2}} − \frac{w^{2} − w + 1}{y^{2}z} − \frac{P_{5}(w)}{2y^{3}z} + \frac{P_{6}(w)}{2y^{4}z}.  (5.7)

Similarly to the proof of (3.10), we use Proposition 10 to get

− \frac{1}{z^{2}} ≥ Ψ_{2}(n),  (5.8)

where

Ψ_{2}(n) = −\frac{1}{y^{2}} + \frac{w}{y^{3}} + \frac{w}{y^{2}z} − (\frac{1}{y} + \frac{1}{z}) (\frac{w^{2} − w + 1}{y^{2}z} + \frac{P_{5}(w)}{2y^{3}z} − \frac{1}{2z} ∑_{k=4}^{6} \frac{P_{k+5}(w)}{y^{k}}) .

Using P_{8}(log log x) ≥ 0 for every x ≥ 3, P_{7}(log log x) ≥ 0 for every x ≥ 3520, and Corollary 11, we get

−\frac{1}{z^{3}} ≥ Ψ_{3}(n) = −\frac{1}{y^{3}} + \frac{w}{y^{4}} + \frac{w}{y^{3}z} + \frac{w}{y^{2}z^{2}} − \frac{w^{2} − w + 1}{y^{4}z} − \frac{w^{2} − w + 1}{y^{3}z^{2}} − \frac{w^{2} − w + 1}{y^{2}z^{3}}
− \frac{P_{5}(w)}{2y^{5}z} − \frac{P_{5}(w)}{2y^{4}z^{2}} − \frac{P_{5}(w)}{2y^{3}z^{3}} + \frac{P_{6}(w)}{2y^{6}z} + \frac{P_{6}(w)}{2y^{5}z^{2}} + \frac{P_{6}(w)}{2y^{4}z^{3}} + \frac{P_{7}(w)}{2y^{7}z}. (5.9)

By (5.1), 3.15−(B_{6} +B_{7} +B_{8} +B_{9} +B_{10}) ≥ 0. Since n ≥ N_{0} is assumed, we have F_{0}(n) ≥ 0. Hence, by (3.12) and (5.5), we see that

\frac{d(n)}{2y^{2}} ≤ G_{1}(w) − \frac{a_{1}(n)}{2y^{3}} + \frac{(3.15 − (B_{6} + B_{7} + B_{8} + B_{9} + B_{10}))w}{y^{2}z},  (5.10)

where d(n) = 11.3 − b_{1}(n). We have n ≥ K_{1}. This means that ∑_{i=1}^{10} H_{i}(n) ≥ 0. So we can add ∑_{i=1}^{10} H_{i}(n) to the right-hand side of (5.10) and use Lemma 18 to get

\frac{d(n)}{2y^{2}} ≤ G_{1}(w) − \frac{a_{1}(n)}{2y^{3}} + 12.85 (Ψ_{3}(n) + \frac{1}{y^{3}} − \frac{w}{y^{4}} + \frac{P_{5}(w)}{2y^{5}z} − \frac{P_{6}(w)}{2y^{5}z^{2}})

+ 3.15 (Ψ_{2}(n) + \frac{1}{y^{2}} − \frac{w}{y^{3}} + \frac{w^{2} − w + 1}{y^{3}z} − \frac{P_{6}(w)}{2y^{5}z} − \frac{P_{7}(w)}{2y^{5}z^{2}})

− \frac{71.3}{z^{4}} − \frac{463.2275}{z^{5}} − \frac{4585}{z^{6}} − \frac{(w^{2} − w + 1)^{2}}{y^{4}z} + \frac{P_{6}(w)}{2y^{4}z} + \frac{P_{6}(w)P_{9}(w)}{4y^{7}z}

+ \frac{P_{8}(w)}{2y^{6}z} − \frac{P_{10}(w)}{2y^{5}z} + \frac{P_{11}(w)}{2y^{5}z^{2}} + \frac{P_{12}(w)}{4y^{6}z} − \frac{(w − 2)^{4}}{4y^{8}} ,

where Ψ_{2}(n) and Ψ_{3}(n) are given as in (5.8) and (5.9), respectively. Applying the defining formulas of P_{10}, P_{11}, P_{12}, and G_{1} to the last inequality, we find

\frac{d(n)}{2y^{2}} ≤ −\frac{a_{1}(n)}{2y^{3}} + \frac{w^{2} − w + 1}{y^{2}} · Ψ_{1}(n) + \frac{P_{9}(w)}{2y^{3}} · Ψ_{1}(n) + 12.85Ψ_{3}(n) + \frac{(w − 1)^{2}}{2y^{2}}

− \frac{w^{2} − 6w}{2y^{3}} − \frac{71.3}{z^{4}} − \frac{463.2275}{z^{5}} − \frac{4585}{z^{6}} + 3.15 (Ψ_{2}(n) + \frac{1}{y^{2}} + \frac{w^{2} − w + 1}{y^{3}z})

− ∑_{k=2}^{4} \frac{(−1)^{k}}{k} (\frac{w − 1}{y} + \frac{w − 2}{y^{2}})^{k} + ∑_{k=4}^{6} \frac{P_{k+5}(w)}{2y^{k}z},

where Ψ_{1}(n) is given as in (5.7). Note that w^{2} −w+1 and P_{9}(w) are nonnegative. Therefore, we can apply (5.7) and (5.9) to the last inequality and get

\frac{d(n)}{2y^{2}} ≤ −\frac{a_{1}(n)}{2y^{3}} − \frac{w^{2} − w + 1}{y^{2}z} − \frac{P_{9}(w)}{2y^{3}z} − \frac{12.85}{z^{3}} − \frac{71.3}{z^{4}} − \frac{463.2275}{z^{5}} − \frac{4585}{z^{6}}

+ \frac{(w − 1)^{2}}{2y^{2}} − \frac{w^{2} − 6w}{2y^{3}} − ∑_{k=2}^{4} \frac{(−1)^{k}}{k} (\frac{w − 1}{y} + \frac{w − 2}{y^{2}})^{k}

+ 3.15 (Ψ_{2}(n) + \frac{1}{y^{2}} + \frac{w^{2} − w + 1}{y^{3}z}) + ∑_{k=4}^{6} \frac{P_{k+5}(w)}{2y^{k}z}.

Since P_{9}(x) = P_{5}(x) + 2 · 3.15(x^{2} − x + 1) and d(n) = 11.3 − b_{1}(n), the last inequality is equivalent to

\frac{5 − b_{1}(n)}{2y^{2}} ≤ −\frac{a_{1}(n)}{2y^{3}} − \frac{w^{2} − w + 1}{y^{2}z} + \frac{(w − 1)^{2}}{2y^{2}} − \frac{w^{2} − 6w}{2y^{3}} − ∑_{k=2}^{4} \frac{(−1)^{k}}{k} (\frac{w − 1}{y} + \frac{w − 2}{y^{2}})^{k}

+ 3.15Ψ_{2}(n) − \frac{12.85}{z^{3}} − \frac{71.3}{z^{4}} − \frac{463.2275}{z^{5}} − \frac{4585}{z^{6}} − \frac{P_{5}(w)}{2y^{3}z} + ∑_{k=4}^{6} \frac{P_{k+5}(w)}{2y^{k}z}.

Using (5.8) and Proposition 10, we get the inequality

\frac{5 − b_{1}(n)}{2y^{2}} ≤ −\frac{1}{z} − \frac{3.15}{z^{2}} − \frac{12.85}{z^{3}} − \frac{71.3}{z^{4}} − \frac{463.2275}{z^{5}} − \frac{4585}{z^{6}} + \frac{1}{y} − \frac{w}{y^{2}} + \frac{(w − 1)^{2}}{2y^{2}}

− \frac{w^{2} − 6w}{2y^{3}} − ∑_{k=2}^{4} \frac{(−1)^{k}}{k} (\frac{w − 1}{y} + \frac{w − 2}{y^{2}})^{k} − \frac{a_{1}(n)}{2y^{3}}

which is equivalent to

\frac{w − 2}{y} ≤ \frac{w − 1}{y} + \frac{w − 2}{y^{2}} − \frac{w^{2} − 6w + a_{1}(n)}{2y^{3}} − ∑_{k=2}^{4} \frac{(−1)^{k}}{k} (\frac{w − 1}{y} + \frac{w − 2}{y^{2}})^{k}

+ \frac{w^{2} − 6w + b_{1}(n)}{2y^{2}} − \frac{1}{z} − \frac{3.15}{z^{2}} − \frac{12.85}{z^{3}} − \frac{71.3}{z^{4}} − \frac{463.2275}{z^{5}} − \frac{4585}{z^{6}}. (5.11)

Since log(1 + t) ≥ ∑_{k=1}^{4}(−1)^{k+1}t^{k}/k for every t > −1 and both g_{1}(x) = −x^{2}/2 + x^{3}/3 and g_{2}(x) = −x^{4}/4 are decreasing on the interval [0, 1], we can use (5.2) and (5.3) to see that the inequality (5.11) implies

\frac{w − 2}{y} − \frac{w^{2} − 6w + b_{1}(n)}{2y^{2}} ≤ log (1 + \frac{w − 1}{y} + \frac{w − 2}{y^{2}} − \frac{w^{2} − 6w + a_{1}(n)}{2y^{3}}) − \frac{1}{z} − \frac{3.15}{z^{2}}

− \frac{12.85}{z^{3}} − \frac{71.3}{z^{4}} − \frac{463.2275}{z^{5}} − \frac{4585}{z^{6}}.

Now we add y + w − 1 to both sides of the last inequality und use (5.5) to get

y + w − 1 + \frac{w − 2}{y} − \frac{w^{2} − 6w + b_{1}(n)}{y} ≤ z − 1 − \frac{1}{z} − \frac{3.15}{z^{2}} − \frac{12.85}{z^{3}} − \frac{71.3}{z^{4}} − \frac{463.2275}{z^{5}} − \frac{4585}{z^{6}}.

Finally, we multiply the last inequality by n and apply (5.6) to complete the proof.

Now, we give a proof of Theorem 4.

Proof of Theorem 4. Clearly, Theorem 3 implies the validity of the inequality (1.14) for every integer n satisfying 2 ≤ n ≤ π(10^{19}). Next, we prove the inequality (1.14) for every n ≥ M_{0}, where M_{0} = π(10^{19})+1 = 234 057 667 276 344 608. In order to do this, let A_{0} = 0.914. Then, similar to the proof of Lemma 15, we get log n ≥ 0.914 log p_{n} for every integer n ≥ M_{0}. So can chose N_{0} = M_{0}. In the following table we give explicit values for B_{i}:

\begin{matrix}i & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ B_{i} & 0.132 & 3.021 & 1.11 & 0.023 & 1.993 & 0.055 & 0.0006 & 0.0199 & 0.055 & 0.0125 \end{matrix}

Then H_{i}(n) ≥ 0 for every n ≥ M_{0} and each integer i satisfying 1 ≤ i ≤ 10. So we can set K_{1} = M_{0}. The proof that H_{i}(n) ≥ 0 for every n ≥ M_{0} and each integer i with 1 ≤ i ≤ 10 can be found in Section 7. Furthermore, the above table indicates

3.15 − (B_{6} + B_{7} + B_{8} + B_{9} + B_{10}) = 3.007.  (5.12)

Step 1. We set a_{1}(n) = 0.2y − w^{2} + 6w. Then, by (1.11) and (5.2)–(5.5), we can choose K_{2} = 33. Using (5.5) and (5.12), we obtain

b_{1}(n) = 11.5 − \frac{2w^{3} − 18w^{2} + 65.390388w − 97.1}{3y} + ρ(n),

where

ρ(n) = \frac{w^{4} − 12w^{3} + 46.6w^{2} − 112w + 40}{2y^{2}} + \frac{2w^{4} − 21.3w^{3} + 40.3w^{2} − 41.5w + 12}{y^{3}}
+ \frac{9w^{4} − 56w^{3} + 129w^{2} − 132w + 52}{3y^{4}} + \frac{2w^{4} − 14w^{3} + 36w^{2} − 40w + 16}{y^{5}}. (5.13)

In this step, we show that b_{1}(n) ≤ 11.5 for every n ≥ M_{0}. For this purpose, we set

α(x, t) = 2(2x^{3} + 65.390388x)e^{4x} − 2(18t^{2} + 97.1)e^{4t}
+ 3(12x^{3} + 112x)e^{3x} − 3(t^{4} + 46.6t^{2} + 40)e^{3t}
+ 6(21.3x^{3} + 41.5x)e^{2x} − 6(2t^{4} + 40.3t^{2} + 12)e^{2t}
+ 2(56x^{3} + 132x)e^{x} − 2(9t^{4} + 129t^{2} + 52)e^{t}
+ 6(14x^{3} + 40x) − 6(2t^{4} + 36t^{2} + 16).

Note that this function satisfies the identity

α(w, w) = 6(11.5 − b_{1}(n))y^{5}.  (5.14)

If t_{0} ≤ x ≤ t_{1}, then α(x, x) ≥ α(t_{0}, t_{1}). We check with a computer that α(3.6 + i · 10^{−3}, 3.6 + (i + 1) · 10^{−3}) ≥ 0 for every integer i satisfying 0 ≤ i ≤ 5399. Hence by (5.14),

b_{1}(n) ≤ 11.5 (3.6 ≤ w ≤ 9).  (5.15)

Next, we show that α(x, x) ≥ 0 for every x ≥ 9. Since 2(2x^{3}−18x^{2}+65.390388x−97.1) ≥ 982 for every x ≥ 9, we have

α(x, x) ≥ 982e^{4x} − 3(x^{4} − 12x^{3} + 46.6x^{2} − 112x + 40)e^{3x}
− 6(2x^{4} − 21.3x^{3} + 40.3x^{2} − 41.5x + 12)e^{2x}
− 2(9x^{4} − 56x^{3} + 129x^{2} − 132x + 52)e^{x}
− 6(2x^{4} − 14x^{3} + 36x^{2} − 40x + 16).

Note that 982e^{x} − 3(x^{4} − 12x^{3} + 46.6x^{2} − 112x + 40) ≥ 7 955 369 for every x ≥ 9. Therefore, α(x, x) ≥ 0 for every x ≥ 9. Combined with (5.14) and (5.15), it gives b_{1}(n) ≤ 11.5 for every n ≥ M_{0} > exp(exp(3.6)). Applying this to Proposition 17, we get

p_{n} > n (y + w − 1 + \frac{w − 2}{y} − \frac{w^{2} − 6w + 11.5}{2y^{2}})

for every n ≥ M_{0}.

Step 2. We set a_{1}(n) = 11.5. Then K_{2} = 47 is a suitable choice for K_{2}. Combined with (5.5) and (5.12), it gives

b_{1}(n) = 11.3 − \frac{2w^{3} − 21w^{2} + 83.390388w − 131.6}{3y} + ρ(n),

where ρ(n) is defined as in (5.13). We set

β(x, t) = 0.15e^{5x} + 2(2x^{3} + 83.390388x)e^{4x} − 2(21t^{2} + 131.6)e^{4t}
+ 3(12x^{3} + 112x)e^{3x} − 3(t^{4} + 46.6t^{2} + 40)e^{3t}
+ 6(21.3x^{3} + 41.5x)e^{2x} − 6(2t^{4} + 40.3t^{2} + 12)e^{2t}
+ 2(56x^{3} + 132x)e^{x} − 2(9t^{4} + 129t^{2} + 52)e^{t}
+ 6(14x^{3} + 40x) − 6(2t^{4} + 36t^{2} + 16).

Then β(w, w) = 6(11.325 − b_{1}(n))y^{5}. Similarly to the first step, we get

b_{1}(n) ≤ 11.325 (3.686 ≤ w ≤ 7).

Therefore, it suffices to verify that β(x, x) ≥ 0 for every x ≥ 7. Notice that 0.15e^{x} + 2(2x^{3} − 21x^{2} + 83.390388x − 131.6) ≥ 382 for every x ≥ 7. Thus we get

β(x, x) ≥ 382e^{4x} − 3(x^{4} − 12x^{3} + 46.6x^{2} − 112x + 40)e^{3x}
− 6(2x^{4} − 21.3x^{3} + 40.3x^{2} − 41.5x + 12)e^{2x}
− 2(9x^{4} − 56x^{3} + 129x^{2} − 132x + 52)e^{x}
− 6(2x^{4} − 14x^{3} + 36x^{2} − 40x + 16).

Since 382e^{x} − 3(x^{4} − 12x^{3} + 46.6x^{2} − 112x + 40) ≥ 419 440 for every x ≥ 7, we conclude that β(x, x) ≥ 0 for every x ≥ 7. Hence b_{1}(n) ≤ 11.325 for every n ≥ M_{0} > exp(exp(3.686)). So, by Proposition 17,

p_{n} > n (y + w − 1 + \frac{w − 2}{y} − \frac{w^{2} − 6w + 11.325}{2y^{2}})

for every n ≥ M_{0}.

Step 3. Here we set a_{1}(n) = 11.325. Then we can choose K_{2} = 47. By (5.5) and (5.12),

b_{1}(n) = 11.3 − \frac{2w^{3} − 21w^{2} + 83.390388w − 131.075}{3y} + ρ(n),

where ρ(n) is defined as in (5.13). To show that b_{1}(n) ≤ 11.321 for every n ≥ M_{0}, we set

γ(x, t) = 0.126e^{5x} + 2(2x^{3} + 83.390388x)e^{4x} − 2(21t^{2} + 131.075)e^{4t}
+ 3(12x^{3} + 112x)e^{3x} − 3(t^{4} + 46.6t^{2} + 40)e^{3t} + 6(21.3x^{3} + 41.5x)e^{2x}
− 6(2t^{4} + 40.3t^{2} + 12)e^{2t} + 2(56x^{3} + 132x)e^{x} − 2(9t^{4} + 129t^{2} + 52)e^{t}
+ 6(14x^{3} + 40x) − 6(2t^{4} + 36t^{2} + 16).

Notice that γ(w, w) = 6(11.321 − b_{1}(n))y^{5}. Analogously to the first step, we obtain b_{1}(n) ≤ 11.321 for w satisfying 3.68 ≤ w ≤ 7. Next we find b_{1}(n) ≤ 11.321 for w ≥ 7. Note that 0.126e^{x} + 2(2x^{3} − 21x^{2} + 83.390388x − 131.075) ≥ 357.491 for every x ≥ 7. Therefore,

γ(x, x) ≥ 357e^{4x} − 3(x^{4} − 12x^{3} + 46.6x^{2} − 112x + 40)e^{3x}
− 6(2x^{4} − 21.3x^{3} + 40.3x^{2} − 41.5x + 12)e^{2x}
− 2(9x^{4} − 56x^{3} + 129x^{2} − 132x + 52)e^{x}
− 6(2x^{4} − 14x^{3} + 36x^{2} − 40x + 16).

Since 357e^{x} −3(x^{4} −12x^{3} +46.6x^{2} −112x+40) ≥ 392 024 for every x ≥ 7, we get γ(x, x) ≥ 0 for every x ≥ 7. So b_{1}(n) ≤ 11.321 for every n ≥ M_{0} > exp(exp(3.68)). Now Proposition 17 implies the required inequality for every n ≥ M_{0} which completes the proof.

Denoting the right-hand side of (1.11) by D_{low}(n) and the right-hand side of (1.14) by A_{low}(n), we use \underline{A006988} to compare the error term of the approximation from Theorem 4 with the approximation from (1.11) for the 10^{n}th prime number:

\begin{matrix}n & p_{n} & ⌈p_{n} − D_{low}(n)⌉ & ⌈p_{n} − A_{low}(n)⌉ \\ 10^{10} & 252 097 800 623 & 22 918 665 & 1 553 620 \\ 10^{11} & 2 760 727 302 517 & 221 928 766 & 12 203 725 \\ 10^{12} & 29 996 224 275 833 & 2 149 187 973 & 116 712 205 \\ 10^{13} & 323 780 508 946 331 & 20 674 500 003 & 1 107 237 510 \\ 10^{14} & 3 475 385 758 524 527 & 198 184 329 536 & 10 418 290 134 \\ 10^{15} & 37 124 508 045 065 437 & 1 896 434 754 032 & 97 120 372 631 \\ 10^{16} & 394 906 913 903 735 329 & 18 139 062 711 550 & 901 415 873 097 \\ 10^{17} & 4 185 296 581 467 695 669 & 173 543 282 219 005 & 8 342 526 771 836 \\ 10^{18} & 44 211 790 234 832 169 331 & 1 661 592 139 340 947 & 77 153 499 580 018 \\ 10^{19} & 465 675 465 116 607 065 549 & 15 924 846 933 652 812 & 713 638 559 773 813 \\ 10^{20} & 4 892 055 594 575 155 744 537 & 152 800 345 036 619 338 & 6 606 690 561 425 196 \end{matrix}

Remark 19. Compared to Theorem 4, the asymptotic expansion (1.2) implies a better lower bound for the nth prime number, which corresponds to the first five terms, namely that

p_{n} > n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2} − 6 log log n + 11}{2 log^{2} n}) (5.16)

for all sufficiently large values of n. Let r_{3} denote the smallest positive integer such that the inequality (5.16) holds for every n ≥ r_{3}. Under the assumption that the Riemann hypothesis is true, Arias de Reyna and Toulisse [1, Theorem 6.4] proved that 3.9·10^{30} < r_{3} ≤ 3.958·10^{30}.

6 New estimates for ϑ(p_{n})

Chebyshev’s ϑ-function is defined by

ϑ(x) = ∑_{p≤x} log p,

where p runs over primes not exceeding x. Notice that the prime number theorem is equivalent to

ϑ(x) ∼ x (x → ∞).  (6.1)

By proving the existence of a zero-free region for the Riemann zeta-function ζ(s) to the left of the line Re(s) = 1, de la Vallée-Poussin [20] found an estimate for the error term in (6.1) by proving ϑ(x) = x + O(xe^{−c\sqrt{log x}}), where c is a positive absolute constant. Applying (1.2) to the last asymptotic formula, we see that

ϑ(p_{n}) = n (log n + log_{2} n − 1 + \frac{log_{2} n − 2}{log n} − \frac{(log_{2} n)^{2} − 6 log_{2} n + 11}{2 log^{2} n} + O (\frac{(log_{2} n)^{3}}{log^{3} n} )) ,

where log_{2} n = log log n. In this direction, many estimates for ϑ(p_{n}) were obtained (see for example Massias and Robin [11, Théorème B]). The current best ones are due to Dusart [8, Propositions 5.11 and 5.12]. He found that

ϑ(p_{n}) ≥ n (log n + log log n − 1 + \frac{log log n − 2.04}{log n})

for every n ≥ π(10^{15}) + 1 = 29 844 570 422 670, and that the inequality

ϑ(p_{n}) ≤ n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{0.782}{log^{2} n})

holds for every n ≥ 781. Using Theorems 1 and 4, we find the following estimates for ϑ(p_{n}), which improve the estimates given by Dusart.

ϑ(p_{n}) > n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2} − 6 log log n + 11.621}{2 log^{2} n}) ,

and for every integer n ≥ 2581, we have

ϑ(p_{n}) < n (log n + log log n − 1 + \frac{log log n − 2}{log n} − \frac{(log log n)^{2} − 6 log log n + 10.367}{2 log^{2} n}) .

Proof. From [3, Theorem 1], it follows that

p_{n} − \frac{0.15p_{n}}{log^{3} p_{n}} < ϑ(p_{n}) < p_{n} + \frac{0.15p_{n}}{log^{3} p_{n}} ,  (6.2)

where the left-hand side inequality is valid for every integer n ≥ 841 508 302 and the righthand side inequality holds for every positive integer n. By Rosser and Schoenfeld [16, Corollary 1], we have n > p_{n}/ log p_{n} for every n ≥ 7. Applying the last inequality to the left-hand side inequality of (6.2), we get ϑ(p_{n}) > p_{n}−0.15n/ log^{2} n for every n ≥ 841 508 302. Now we apply Theorem 4 to get the desired lower bound for ϑ(p_{n}) for every n ≥ 841 508 302. By Büthe [4, Theorem 2], we have

ϑ(x) ≥ x − \frac{\sqrt{x}}{8π} log^{2} x (599 < x ≤ 1.89 × 10^{21}).  (6.3)

Now we apply Theorem 3 to (6.3) and get the required lower bound for ϑ(p_{n}) for every integer n with 200 000 ≤ n ≤ 841 508 301. We check the remaining cases for n with a computer.

Similarly to the first part of the proof, we apply the inequality n > p_{n}/ log p_{n} to the right-hand side inequality of (6.2) to get ϑ(p_{n}) < p_{n} + 0.15n/ log^{2} n for every n ≥ 7. Now we use Theorem 1 to get the required upper bound for ϑ(p_{n}) for every n ≥ 46 254 381. For smaller values of n, we use a computer.

7 Appendix

Let M_{0} = π(10^{19}) + 1 = 234 057 667 276 344 608. In the proof of Theorem 4, we note a table in which we give explicit values of B_{i}. In this appendix, we show that the H_{i} defined at the start of paragraph 5 are non-negative for every integer n ≥ M_{0} for the given values of B_{i}. We start with the claim concerning H_{1}.

Proposition 21. If B_{1} = 0.132, then H_{1}(n) ≥ 0 for every integer n ≥ M_{0}.

Proof. Let x_{0} = log log M_{0}. We have P_{11}(x) ≥ 0 for every x ≥ 0.6 and P_{6}(x) ≥ 0 for every x ≥ 0.6. Using Lemma 15, we get

H_{1}(n) ≥ \frac{f_{1}(log log n)}{4 log^{6} n log p_{n}}  (7.1)

for every integer n ≥ M_{0}, where f_{1}(x) = 0.528xe^{3x} − 2P_{10}(x)e^{x} + 1.74P_{11}(x) + P_{12}(x) + 19.45233P_{6}(x). We show that f(x) ≥ 0 for every x ≥ x_{0}. For this, we set g(x) = (57.024 + 42.768x)e^{2x} + (−24.6x^{4} − 322.1x^{3} − 1137.1x^{2} − 1265.98x − 512.24). It is easy to show that g(x) ≥ 3 · 10^{5} for every x ≥ x_{0}. So, f_{1}^{(4)}(x) = g(x)e^{x} + 240x − 1005.6 ≥ 0 for every x ≥ x_{0}. Now it is easy to see that f(x) ≥ 0 for every x ≥ x_{0}. Applying this to (7.1), we get H_{1}(n) ≥ 0 for every integer n ≥ M_{0}.

Let B_{2} = 3.021. Before we check that H_{2}(n) ≥ 0 for every integer n ≥ M_{0}, we introduce the following

Definition 22. For x ≥ 1, let

Φ(x) = e^{x} + x + log (1 + \frac{x − 1}{e^{x}} + \frac{x − 2.1}{e^{2x}}) .

We note the following three properties of the function Φ(x).

Lemma 23. For every x ≥ 0.179, we have Φ′(x) ≥ e^{x} + 3/4.

Proof. We have Φ′(x) ≥ e^{x} + 3/4 if and only if g(x) = e^{2x} − 3xe^{x} + 7e^{x} − 7x + 18.7 ≥ 0. Since g′′(x) = 4e^{2x} − (3x − 1)e^{x} ≥ 0 for every x ≥ 0 and g′(0.179) ≥ 0, we obtain g′(x) ≥ 0 for every x ≥ 0.179. If we combine this with g(0.179) ≥ 26.6, we get g(x) ≥ 0 for every x ≥ 0.179.

Lemma 24. For every x ≥ 1.246, we have Φ(x) ≥ e^{x} + x.

Proof. The desired inequality holds if and only if (x − 1)e^{x} + x − 2.1 ≥ 0. Since the last inequality holds for every x ≥ 1.246, we arrived at the end of the proof.

Lemma 25. For every integer n ≥ 3, we have Φ(log log n) ≤ log p_{n}.

Proof. The claim follows directly from (1.11).

Next, we use these properties to see that H_{2}(n) ≥ 0 for every integer n ≥ M_{0}.

Proposition 26. Let B_{2} = 3.021. Then H_{2}(n) ≥ 0 for every integer n ≥ M_{0}.

Proof. Let x_{0} = log log M_{0}. We set f_{2}(x) = 3.021xΦ^{3}(x) + 12.85xe^{x}Φ^{2}(x) − 71.3e^{3x} and use Lemmata 23 and 24 to obtain

f′_{2}(x) ≥ 3.021(e^{x} + x)^{3} + 21.913xe^{x}(e^{x} + x)^{2} + 12.85e^{x}(e^{x} + x)^{2}
+ 25.7xe^{2x}(e^{x} + x) − 213.9e^{3x}  (7.2)

for every x ≥ 1.25. We denote the right-hand side of the last inequality by g_{2}(x). A straightforward calculation gives g_{2}^{(3)}(x) ≥ (1285.551x − 4061.232)e^{3x} ≥ 0 for every x ≥ x_{0}. Now it is easy to see that g_{2}(x) ≥ 0 for every x ≥ x_{0}. Applying this to (7.2), we see that f′_{2}(x) ≥ 0 for every x ≥ x_{0}. Since f_{2}(x_{0}) ≥ 268.5, we obtain f_{2}(log log n) ≥ 0 for every integer n ≥ M_{0}. Finally, we apply Lemma 25.

Proposition 27. If B_{3} = 1.11, then H_{3}(n) ≥ 0 for every integer n ≥ M_{0}.

Proof. Let x_{0} = log log M_{0} and let f_{3}(x) = 2.22xΦ(x) − 35.15x^{2} + 44.6x − 42.08. Using Lemmata 23 and 24, we get f′_{3}(x) ≥ (2.22e^{x} −65.86)x ≥ 0 holds for every x ≥ x_{0}. Combined with f_{3}(x_{0}) ≥ 2.42 and Lemma 25, we get that H_{3}(n) ≥ 0 for every integer n ≥ M_{0}.

Proposition 28. Let B_{4} = 0.023. Then H_{4}(n) ≥ 0 for every integer n ≥ M_{0}.

Proof. Let x_{0} = log log M_{0}. We set f_{4}(x) = 0.046xe^{x}Φ(x)+3.15x^{3}−57.45x^{2}+113.01x−80.05 and have f_{4}(M_{0}) ≥ 10.103. By Lemmata 23 and 24, we get f′_{4}(x) ≥ (0.046(e^{x}(e^{x} +x)+e^{2x})+ 9.45x − 114.9)x ≥ 0 for every x ≥ x_{0}. Hence f_{4}(log log n) ≥ 0 for every integer n ≥ M_{0} and we can apply Lemma 25.

Proposition 29. Let B_{5} = 1.993. Then we have H_{5}(n) ≥ 0 for every integer n ≥ M_{0}.

Proof. Let x_{0} = log log M_{0}. To proof the claim, we define f_{5}(x) = 3.986xe^{x} − 2x^{4} + 5x^{3} − 47.15x^{2} +60x−48.28. Since f′′′_{5}(x) ≥ 0 for every x ≥ x_{0} and f′′_{5}(x_{0}) ≥ 0, we obtain f′′_{5}(x) ≥ 0 for every x ≥ x_{0}. Combined with f′_{5}(x_{0}) ≥ 0, it turns out that f′_{5}(x) ≥ 0 for every x ≥ x_{0}. Together with f_{5}(x_{0}) ≥ 0.203, we conclude that f_{5}(log log n) ≥ 0, and thus H_{5}(n) ≥ 0, for every integer n ≥ M_{0}.

Adding the constants B_{1}, …, B_{5} given in Proposition 21 and Propositions 26-29, we get 12.85 − B_{1} − B_{2} − B_{3} − B_{4} − B_{5} = 6.571. Now we set B_{6} = 0.055 to obtain the following result.

Proposition 30. Let B_{6} = 0.055. Then H_{6}(n) ≥ 0 for every integer n ≥ M_{0}.

Proof. Let x_{0} = log log M_{0}. Furthermore, let r(x, t) = (0.118e^{x} + 4.116)xΦ(x) + 3.15xe^{x} − 3.15(t^{2} + 1)e^{t} and let f_{6}(x) = r(x, x). If t_{0} ≤ x ≤ t_{1}, then f_{6}(x) ≥ r(t_{0}, t_{1}). We check with a computer that r(3.6 + i · 10^{−3}, 3.6 + (i + 1) · 10^{−3}) ≥ 0 for every integer i such that 0 ≤ i ≤ 599. Hence f_{6}(x) ≥ 0 for every x such that 3.6 ≤ x ≤ 4.2. To show that f_{6}(x) ≥ 0 for every x ≥ 4.2, we set

g(x) = (0.055(xe^{x} + e^{x}) + 6.571)(e^{x} + x) + (0.055e^{x} + 6.571)xe^{x} − 3.15xe^{x}(1 + x).

Then g′(x) = h(x)e^{x} + 6.571 where h(x) = 0.22(1 + x)e^{x} − 3.095x^{2} − 2.714x + 10.047. Since h(x) ≥ 0 for every x ≥ 4.2, we get g′(x) ≥ 0 for every x ≥ 4.2. Together with g(4.2) ≥ 0, we see that g(x) ≥ 0 for every x ≥ 4.2. Using Lemmata 23 and 24, we obtain f′_{6}(x) ≥ g(x) ≥ 0 for every x ≥ 4.2. Combined with f_{6}(4.2) ≥ 17.047, we have f_{6}(x) ≥ 0 for every x ≥ 4.2. Hence f_{6}(x) ≥ 0 for every x ≥ x_{0} ≥ 3.6. Now we apply Lemma 25 to get H_{6}(n) ≥ 0 for every integer n ≥ M_{0}.

Proposition 31. If B_{7} = 0.0006, then we have H_{7}(n) ≥ 0 for every integer n ≥ M_{0}.

Proof. Let x_{0} = log log M_{0}. Substituting the definition of P_{5}(x), we get

H_{7}(n) = \frac{0.0006w}{y^{2}z} − \frac{38.55w^{2} − 77.1w + 66.82}{2y^{3}z^{3}}.

To show that H_{7}(n) ≥ 0 for every integer n ≥ M_{0}, we first consider the function f_{7}(x) = 0.0012xe^{x}Φ^{2}(x) − 38.55x^{2} + 77.1x − 66.82. We have f_{7}(x_{0}) ≥ 31.88. Additionally, we use Lemmata 23 and 24 to get f′_{7}(x) ≥ (0.0012(e^{x} + x)^{2}(1 + e^{x}) + 0.0024e^{2x}(e^{x} + x) − 77.1)x ≥ 0 for every x ≥ x_{0}. Hence, f_{7}(log log n) ≥ 0 for every integer n ≥ M_{0}. Finally, it suffices to apply Lemma 25.

Proposition 32. Let B_{8} = 0.0199. Then H_{8}(n) ≥ 0 for every integer n ≥ M_{0}.

Proof. Let x_{0} = log log M_{0}. We set f_{8}(x) = 0.0199x Φ^{2}(x) − 12.85(x^{2} − x + 1). We have f_{8}(x_{0}) ≥ 0.906. By Lemmata 23 and 24, we obtain f′_{8}(x) ≥ (0.0199(e^{x} + x) + 0.0398(e^{x} + x)e^{x} − 25.7)x ≥ 0 for every x ≥ x_{0}. Hence f_{8}(log log n) ≥ 0 for every integer n ≥ M_{0}. Finally, we use Lemma 25.

Proposition 33. If B_{9} = 0.055, then H_{9}(n) ≥ 0 for every integer n ≥ M_{0}.

Proof. Let x_{0} = log log M_{0}. We define f_{9}(x) = 0.055x Φ^{4}(x) − 463.2275e^{2x}. By Lemmata 23 and 24, we have f′_{9}(x) ≥ ((0.055 + 0.22x)(e^{x} + x)^{2} − 926.455)e^{2x} ≥ 0 for every x ≥ x_{0}. Combined with f_{9}(x_{0}) ≥ 2263.343, we get f_{9}(x) ≥ 0 for every x ≥ x_{0}. Substituting x = log log n in f_{9}(x), we apply Lemma 25 to see that H_{9}(n) ≥ 0 for every integer n ≥ M_{0}.

Finally, we set B_{10} = 0.0125 and check that H_{10}(n) ≥ 0 for every integer n ≥ M_{0}.

Proposition 34. Let B_{10} = 0.0125. Then we have H_{10}(n) ≥ 0 for every integer n ≥ M_{0}.

Proof. Let x_{0} = log log M_{0} and let f_{10}(x) = 0.0125x Φ^{5}(x) − 4585e^{2x}. Applying Lemmata 23 and 24, we get f′_{10}(x) ≥ (0.4x(e^{x} + x)^{3} − 9170)e^{2x} ≥ 0 for every x ≥ x_{0}. Together with f_{10}(x_{0}) ≥ 55867.822, we see that f_{10}(log log n) ≥ 0 for every integer n ≥ M_{0}. Now, we use Lemma 25 to conclude that H_{10}(n) ≥ 0 for every integer n ≥ M_{0}.

8 Acknowledgment

I would like to thank the anonymous referees for useful comments to improve the quality of this paper. Furthermore, I would like to thank R. for being a never-ending inspiration.

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2010 Mathematics Subject Classification: Primary 11N05; Secondary 11A41. Keywords: Chebyshev’s ϑ-function, prime counting function, prime number.

(Concerned with sequences \underline{A000040} and \underline{A006988}.)

Received December 12 2017; revised versions received March 15 2018; February 18 2019; February 25 2019; May 21 2019. Published in Journal of Integer Sequences, May 23 2019.

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