On a Sequence Involving Prime Numbers

Christian Axler Departament of Mathematics Heinrich-Heine-Universität 40225 Düsseldorf Germany

                    Christian.Axler@hhu.de

Abstract We study a particular sequence C_{n} = np_{n} − ∑_{k≤n} p_{k}, n ∈ ℕ, involving prime numbers by deriving two asymptotic formulae, and we find a new lower bound for C_{n} that improves the currently known estimates. Furthermore, for the first time we determine an upper bound for C_{n}.

1 Introduction

In this paper, we study the sequence (C_{n})_{n∈ℕ} with

C_{n} = np_{n} − ∑_{k≤n} p_{k},

where p_{n} is the nth prime number. The motivation for considering this special sequence is an inequality conjectured by Mandl [7, p. 1] that asserts that

\frac{np_{n}}{2} − ∑_{k≤n} p_{k} ≥ 0  (1)

for every n ≥ 9. This inequality originally appeared without proof. In his 1998 thesis [4], Dusart used the equality

C_{n} = ∫_{2}^{p_{n}} π(x) dx,

where π(x) denotes the number of primes ≤ x, and explicit estimates for the prime counting function π(x) to prove that

C_{n} ≥ \frac{np_{n}}{2} ,

which is equivalent to Mandl’s inequality (1), for every n ≥ 9. At the same time, Dusart [4] showed that

C_{n} ≥ c + \frac{p_{n}^{2}}{2 log p_{n}} + \frac{3p_{n}^{2}}{4 log^{2} p_{n}}  (2)

for every n ≥ 109, where c ≐ −47.1. The first goal of this article is to study the asymptotic behaviour of the sequence C_{n}. This is done in the following two theorems.

Theorem 1 (Corollary 7). For each s ∈ ℕ there is a unique monic polynomial U_{s} of degree s with rational coefficients, so that for every m ∈ ℕ

C_{n} = \frac{n^{2}}{2} (log n + log log n − \frac{1}{2} + ∑_{s=1}^{m} \frac{(−1)^{s+1}U_{s}(log log n)}{s log^{s} n}) + O (\frac{n^{2}(log log n)^{m+1}}{log^{m+1} n} ) .

Theorem 2 (Theorem 10). For each m ∈ ℕ we have

C_{n} = ∑_{k=1}^{m−1}(k − 1)! (1 − \frac{1}{2^{k}} ) \frac{p_{n}^{2}}{log^{k} p_{n}} + O (\frac{p_{n}^{2}}{log^{m} p_{n}} ) .  (3)

By setting m = 9 in (3), we get

C_{n} = \frac{p_{n}^{2}}{2 log p_{n}} + \frac{3p_{n}^{2}}{4 log^{2} p_{n}} + \frac{7p_{n}^{2}}{4 log^{3} p_{n}} + χ(n) + O (\frac{p_{n}^{2}}{log^{9} p_{n}} ) ,  (4)

where

χ(n) = \frac{45p_{n}^{2}}{8 log^{4} p_{n}} + \frac{93p_{n}^{2}}{4 log^{5} p_{n}} + \frac{945p_{n}^{2}}{8 log^{6} p_{n}} + \frac{5715p_{n}^{2}}{8 log^{7} p_{n}} + \frac{80325p_{n}^{2}}{16 log^{8} p_{n}} .

In view of (4), we improve the inequality (2) by finding the following lower bound for C_{n}.

Theorem 3 (Proposition 18). If n ≥ 52703656, then

C_{n} ≥ \frac{p_{n}^{2}}{2 log p_{n}} + \frac{3p_{n}^{2}}{4 log^{2} p_{n}} + \frac{7p_{n}^{2}}{4 log^{3} p_{n}} + Θ(n),

where

Θ(n) = \frac{43.6p_{n}^{2}}{8 log^{4} p_{n}} + \frac{90.9p_{n}^{2}}{4 log^{5} p_{n}} + \frac{927.5p_{n}^{2}}{8 log^{6} p_{n}} + \frac{5620.5p_{n}^{2}}{8 log^{7} p_{n}} + \frac{79075.5p_{n}^{2}}{16 log^{8} p_{n}} .

Finally, for the first time we give an upper bound for C_{n}, by proving the following theorem.

Theorem 4 (Proposition 21). For every n ∈ ℕ,

C_{n} ≤ \frac{p_{n}^{2}}{2 log p_{n}} + \frac{3p_{n}^{2}}{4 log^{2} p_{n}} + \frac{7p_{n}^{2}}{4 log^{3} p_{n}} + Ω(n),

where

Ω(n) = \frac{46.4p_{n}^{2}}{8 log^{4} p_{n}} + \frac{95.1p_{n}^{2}}{4 log^{5} p_{n}} + \frac{962.5p_{n}^{2}}{8 log^{6} p_{n}} + \frac{5809.5p_{n}^{2}}{8 log^{7} p_{n}} + \frac{118848p_{n}^{2}}{16 log^{8} p_{n}} .

2 Two asymptotic formulae for C_{n}

From here on, we use the following notation. Cipolla [3] showed that for each s ∈ ℕ and each 0 ≤ i ≤ s there exist unique rational numbers a_{is}, where a_{ss} = 1, such that for every m ∈ ℕ

p_{n} = n (log n + log log n − 1 + ∑_{s=1}^{m} \frac{(−1)^{s+1}}{s log^{s} n} ∑_{i=0}^{s} a_{is}(log log n)^{i}) + O(c_{m}(n)),  (5)

where

c_{m}(n) = \frac{n(log log n)^{m+1}}{log^{m+1} n} .

We set

h_{m}(n) = ∑_{j=1}^{m} \frac{(j − 1)!}{2^{j} log^{j} n}.

Further, we recall the following definition from [2].

Definition 5. Let s, i, j, r ∈ ℕ_{0} with j ≥ r. We define the integers b_{s,i,j,r} ∈ ℤ as follows:

• If j = r = 0, then

b_{s,i,0,0} = 1.  (6)

• If j ≥ 1, then

b_{s,i,j,j} = b_{s,i,j−1,j−1} · (−i + j − 1).  (7)

• If j ≥ 1, then

b_{s,i,j,0} = b_{s,i,j−1,0} · (s + j − 1).  (8)

• If j > r ≥ 1, then

b_{s,i,j,r} = b_{s,i,j−1,r} · (s + j − 1) + b_{s,i,j−1,r−1} · (−i + r − 1).  (9)

Using (5) and [2, Thm. 2.5], we obtain the first asymptotic formula for C_{n}.

Theorem 6. For each m ∈ ℕ we have

C_{n} = \frac{n^{2}}{2} (log n + log log n − \frac{1}{2} + h_{m}(n))

+ \frac{n^{2}}{2} ∑_{s=1}^{m} \frac{(−1)^{s+1}}{s log^{s} n} ∑_{i=0}^{s} a_{is} (2(log log n)^{i} − ∑_{j=0}^{m−s} ∑_{r=0}^{min\{i,j\}} \frac{b_{s,i,j,r}(log log n)^{i−r}}{2^{j} log^{j} n} )

+ O(nc_{m}(n)).

Proof. From [2, Thm. 2.5] we know that

∑_{k≤n} p_{k} = \frac{n^{2}}{2} (g(n) − h_{m}(n) + ∑_{s=1}^{m} \frac{(−1)^{s+1}}{s log^{s} n} ∑_{i=0}^{s} a_{is} ∑_{j=0}^{m−s} ∑_{r=0}^{min\{i,j\}} \frac{b_{s,i,j,r}(log log n)^{i−r}}{2^{j} log^{j} n} )
+ O(nc_{m}(n)),  (10)

where g(n) = log n + log log n − 3/2. Now we multiply (5) by n and substract (10) to get the result.

Corollary 7. For each s ∈ ℕ there is a unique monic polynomial U_{s} of degree s with rational coefficients, so that for every m ∈ ℕ

C_{n} = \frac{n^{2}}{2} (log n + log log n − \frac{1}{2} + ∑_{s=1}^{m} \frac{(−1)^{s+1}U_{s}(log log n)}{s log^{s} n}) + O(nc_{m}(n)).  (11)

In particular, U_{1}(x) = x − 3/2 and U_{2}(x) = x^{2} − 5x + 15/2.

Proof. Since a_{ss} = 1 and b_{s,s,0,0} = 1, the formula (11) follows from Theorem 6. Now let m = 2. Cipolla [3] showed that a_{01} = −2, a_{11} = 1, a_{02} = 11, a_{12} = −6 and a_{22} = 1. Further, we use formulae (6)–(9) to compute the integers b_{s,i,j,r}. Then, using Theorem 6, we find the polynomials U_{1} and U_{2}.

To find another asymptotic formula for C_{n}, we use the identity (see Dusart [4, p. 50] or Hassani [5, p. 3])

C_{n} = ∫_{2}^{p_{n}} π(x) dx,  (12)

which allows us to estimate C_{n} by using explicit bounds for π(x). Further, we use the following integration rules (see Lemma 8), where the logarithmic integral li(x) is defined for every real x ≥ 2 by

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}\} ≈ ∫_{2}^{x} \frac{dt}{log t} + 1.04516 ….

Lemma 8. Let r, s ∈ ℝ with s ≥ r > 1.

(i) ∫_{r}^{s} \frac{x dx}{log x} = li(s^{2}) − li(r^{2}).

(ii) ∫_{r}^{s} \frac{x dx}{log^{2} x} = 2 li(s^{2}) − 2 li(r^{2}) − \frac{s^{2}}{log s} + \frac{r^{2}}{log r}.

(iii) If n ∈ ℕ, then

∫_{r}^{s} \frac{x dx}{log^{n+1} x} = \frac{r^{2}}{n log^{n} r} − \frac{s^{2}}{n log^{n} s} + \frac{2}{n} ∫_{r}^{s} \frac{x}{log^{n} x} dx.

(iv) For every m ∈ ℕ with m ≥ 2 we have

∫_{r}^{s} \frac{x dx}{log^{m} x} = \frac{2^{m−2}}{(m − 1)!} ∫_{r}^{s} \frac{x dx}{log^{2} x} − ∑_{k=2}^{m−1} \frac{2^{m−1−k}(k − 1)!}{(m − 1)!} (\frac{s^{2}}{log^{k} s} − \frac{r^{2}}{log^{k} r}) .

Proof. The rules (i) and (ii) are from Dusart [4, Lemma 1.6]. Now, (iii) follows by integration by parts and (iv) can be shown by induction on m.

The next proposition plays an important role for the proof of the second asymptotic formula (Theorem 2, see Introduction) for C_{n}.

Proposition 9. Let m ∈ ℕ with m ≥ 2. Let a_{2}, …, a_{m} ∈ ℝ and let r, s ∈ ℝ with s ≥ r > 1. Then

∑_{k=2}^{m} a_{k} ∫_{r}^{s} \frac{x dx}{log^{k} x} = t_{m−1,1} ∫_{r}^{s} \frac{x dx}{log^{2} x} − ∑_{k=2}^{m−1} t_{m−1,k} (\frac{s^{2}}{log^{k} s} − \frac{r^{2}}{log^{k} r}) ,

where

t_{i,j} = (j − 1)! ∑_{l=j}^{i} \frac{2^{l−j}a_{l+1}}{l!} .  (13)

Proof. If m = 2, the claim is obviously true. By induction hypothesis, we have

∑_{k=2}^{m+1} a_{k} ∫_{r}^{s} \frac{x dx}{log^{k} x} = t_{m−1,1} ∫_{r}^{s} \frac{x dx}{log^{2} x} − ∑_{k=2}^{m−1} t_{m−1,k} (\frac{s^{2}}{log^{k} s} − \frac{r^{2}}{log^{k} r}) + a_{m+1} ∫_{r}^{s} \frac{x dx}{log^{m+1} x}.

By Lemma 8(iii), we get

∑_{k=2}^{m+1} a_{k} ∫_{r}^{s} \frac{x dx}{log^{k} x} = t_{m−1,1} ∫_{r}^{s} \frac{x dx}{log^{2} x} − ∑_{k=2}^{m−1} t_{m−1,k} (\frac{s^{2}}{log^{k} s} − \frac{r^{2}}{log^{k} r}) + \frac{2a_{m+1}}{m} ∫_{r}^{s} \frac{x dx}{log^{m} x}

− \frac{a_{m+1}s^{2}}{m log^{m} s} + \frac{a_{m+1}r^{2}}{m log^{m} r}.

Now we use Lemma 8(iv) and the equality t_{m−1,1} + 2^{m−1}a_{m+1}/m! = t_{m,1} to obtain

∑_{k=2}^{m+1} a_{k} ∫_{r}^{s} \frac{x dx}{log^{k} x} = t_{m,1} ∫_{r}^{s} \frac{x dx}{log^{2} x} − ∑_{k=2}^{m−1} (\frac{2^{m−k}a_{m+1}(k − 1)!}{m!} + t_{m−1,k}) (\frac{s^{2}}{log^{k} s} − \frac{r^{2}}{log^{k} r})

− \frac{a_{m+1}(m − 1)!}{m!} (\frac{s^{2}}{log^{m} s} − \frac{r^{2}}{log^{m} r}) .

Since we have

\frac{2^{m−k}a_{m+1}(k − 1)!}{m!} + t_{m−1,k} = t_{m,k}

and t_{m,m} = a_{m+1}(m − 1)!/(m!), the proposition is proved.

Now we are able to prove Theorem 2. Theorem 10. For each m ∈ ℕ we have

C_{n} = ∑_{k=1}^{m−1}(k − 1)! (1 − \frac{1}{2^{k}} ) \frac{p_{n}^{2}}{log^{k} p_{n}} + O (\frac{p_{n}^{2}}{log^{m} p_{n}} ) .

Proof. A well-known asymptotic formula for the prime counting function π(x) is given by

π(x) = \frac{x}{log x} + \frac{x}{log^{2} x} + \frac{2x}{log^{3} x} + \frac{6x}{log^{4} x} + … + \frac{(m − 1)!x}{log^{m} x} + O (\frac{x}{log^{m+1} x}) .  (14)

Using (14) and (12), we get

C_{n} = ∑_{k=1}^{m}(k − 1)! ∫_{2}^{p_{n}} \frac{x dx}{log^{k} x} + O (∫_{2}^{p_{n}} \frac{x dx}{log^{m+1} x}) .

Integration by parts gives

C_{n} = ∑_{k=1}^{m}(k − 1)! ∫_{2}^{p_{n}} \frac{x dx}{log^{k} x} + O (\frac{p_{n}^{2}}{log^{m} p_{n}} ) .

We now apply Proposition 9 to get

C_{n} = ∫_{2}^{p_{n}} \frac{x dx}{log x} + (2^{m−1} − 1) ∫_{2}^{p_{n}} \frac{x dx}{log^{2} x} − ∑_{k=2}^{m−1} (\frac{(k − 1)!(2^{m−k} − 1)p_{n}^{2}}{log^{k} p_{n}} ) + O (\frac{p_{n}^{2}}{log^{m} p_{n}} ) .

It follows from Lemma 8(i) and Lemma 8(ii) that

C_{n} = (2^{m} − 1) li(p_{n}^{2}) − ∑_{k=1}^{m−1} (\frac{(k − 1)!(2^{m−k} − 1)p_{n}^{2}}{log^{k} p_{n}} ) + O (\frac{p_{n}^{2}}{log^{m} p_{n}} ) .

Now we use the well-known asymptotic formula

li(x) = \frac{x}{log x} + \frac{x}{log^{2} x} + \frac{2x}{log^{3} x} + \frac{6x}{log^{4} x} + … + \frac{(m − 1)!x}{log^{m} x} + O (\frac{x}{log^{m+1} x}) (15)

to obtain

C_{n} = (2^{m} − 1) ∑_{k=1}^{m−1} \frac{(k − 1)! p_{n}^{2}}{2^{k} log^{k} p_{n}} − ∑_{k=1}^{m−1} (\frac{(k − 1)!(2^{m−k} − 1)p_{n}^{2}}{log^{k} p_{n}} ) + O (\frac{p_{n}^{2}}{log^{m} p_{n}} )

and the theorem is proved.

Using (14), we get the following corollary. Corollary 11. For each m ∈ ℕ we have

∑_{k≤n} p_{k} = π(p_{n}^{2}) + O (\frac{p_{n}^{2}}{log^{m} p_{n}} ) .

Proof. From Theorem 10 and the definition of C_{n} it follows that

∑_{k≤n} p_{k} = np_{n} − ∑_{k=1}^{m−1} \frac{(k − 1)! p_{n}^{2}}{log^{k} p_{n}} + ∑_{k=1}^{m−1} \frac{(k − 1)! p_{n}^{2}}{2^{k} log^{k} p_{n}} + O (\frac{p_{n}^{2}}{log^{m} p_{n}} ) .

Since n = π(p_{n}), we obtain

∑_{k≤n} p_{k} = π(p_{n})p_{n} − ∑_{k=1}^{m−1} \frac{(k − 1)! p_{n}^{2}}{log^{k} p_{n}} + ∑_{k=1}^{m−1} \frac{(k − 1)! p_{n}^{2}}{2^{k} log^{k} p_{n}} + O (\frac{p_{n}^{2}}{log^{m} p_{n}} ) .

Using (14), we get the asymptotic formula

∑_{k≤n} p_{k} = ∑_{k=1}^{m−1} \frac{(k − 1)! p_{n}^{2}}{2^{k} log^{k} p_{n}} + O (\frac{p_{n}^{2}}{log^{m} p_{n}} ) = π(p_{n}^{2}) + O (\frac{p_{n}^{2}}{log^{m} p_{n}} )

and the corollary is proved.

Using (14), (15) and Corollary 11, we obtain the following result concerning the sum of the first n prime numbers. Corollary 12. For each m ∈ ℕ we have

∑_{k≤n} p_{k} = li(p_{n}^{2}) + O (\frac{p_{n}^{2}}{log^{m} p_{n}} ) .

3 A lower bound for C_{n}

Let m ∈ ℕ with m ≥ 2 and let a_{2}, …, a_{m}, x_{0}, y_{0} ∈ ℝ, so that

π(x) ≥ \frac{x}{log x} + ∑_{k=2}^{m} \frac{a_{k}x}{log^{k} x}  (16)

for every x ≥ x_{0} and

li(x) ≥ ∑_{j=1}^{m−1} \frac{(j − 1)!x}{log^{j} x}  (17)

for every x ≥ y_{0}. Then, we obtain the following lower bound for C_{n}. Theorem 13. If n ≥ max\{π(x_{0}) + 1, π(\sqrt{y_{0}}) + 1\}, then

C_{n} ≥ d_{0} + ∑_{k=1}^{m−1} (\frac{(k − 1)!}{2^{k}}(1 + 2t_{k−1,1})) \frac{p_{n}^{2}}{log^{k} p_{n}} ,

where t_{i,j} is defined as in (13) and d_{0} is given by

d_{0} = d_{0}(m, a_{2}, …, a_{m}, x_{0}) = ∫_{2}^{x_{0}} π(x) dx − (1 + 2t_{m−1,1}) li(x_{0}^{2}) + ∑_{k=1}^{m−1} t_{m−1,k} \frac{x_{0}^{2}}{log^{k} x_{0}} .

Proof. Since p_{n} ≥ x_{0}, we use (12) and (16) to obtain

C_{n} ≥ ∫_{2}^{x_{0}} π(x) dx + ∫_{x_{0}}^{p_{n}} \frac{x dx}{log x} + ∑_{k=2}^{m} a_{k} ∫_{x_{0}}^{p_{n}} \frac{x dx}{log^{k} x}.

Now we apply Lemma 8(i) and Proposition 9 to get

C_{n} ≥ ∫_{2}^{x_{0}} π(x) dx − li(x_{0}^{2}) + li(p_{n}^{2}) + t_{m−1,1} ∫_{x_{0}}^{p_{n}} \frac{x dx}{log^{2} x} − ∑_{k=2}^{m−1} t_{m−1,k} (\frac{p_{n}^{2}}{log^{k} p_{n}} − \frac{x_{0}^{2}}{log^{k} x_{0}} ) .

Using Lemma 8(ii), we obtain

C_{n} ≥ d_{0} + (1 + 2t_{m−1,1}) li(p_{n}^{2}) − ∑_{k=1}^{m−1} t_{m−1,k} \frac{p_{n}^{2}}{log^{k} p_{n}} .

Since p_{n}^{2} ≥ y_{0}, we use (17) to conclude

C_{n} ≥ d_{0} + ∑_{k=1}^{m−1} (\frac{(k − 1)!}{2^{k}} + \frac{(k − 1)!}{2^{k−1}} t_{m−1,1} − t_{m−1,k}) \frac{p_{n}^{2}}{log^{k} p_{n}}

and it remains to apply the definition of t_{ij}.

4 An upper bound for C_{n}

Next, we derive for the first time an upper bound for C_{n}. Let m ∈ ℕ with m ≥ 2 and let a_{2}, …, a_{m}, x_{1} ∈ ℝ so that

π(x) ≤ \frac{x}{log x} + ∑_{k=2}^{m} \frac{a_{k}x}{log^{k} x}  (18)

for every x ≥ x_{1} and let λ, y_{1} ∈ ℝ so that

li(x) ≤ ∑_{j=1}^{m−2} \frac{(j − 1)!x}{log^{j} x} + \frac{λx}{log^{m−1} x}  (19)

for every x ≥ y_{1}. Setting

d_{1} = d_{1}(m, a_{2}, …, a_{m}, x_{1}) = ∫_{2}^{x_{1}} π(x) dx − (1 + 2t_{m−1,1}) li(x_{1}^{2}) + ∑_{k=1}^{m−1} t_{m−1,k} \frac{x_{1}^{2}}{log^{k} x_{1}} ,

where t_{m−1,k} is defined by (13), we obtain the following Theorem 14. If n ≥ max\{π(x_{1}) + 1, π(\sqrt{y_{1}}) + 1\}, then

C_{n} ≤ d_{1} + ∑_{k=1}^{m−2} (\frac{(k − 1)!}{2^{k}}(1 + 2t_{k−1,1})) \frac{p_{n}^{2}}{log^{k} p_{n}} + (\frac{(1 + 2t_{m−1,1})λ}{2^{m−1}} − \frac{a_{m}}{m − 1}) \frac{p_{n}^{2}}{log^{m−1} p_{n}} .

Proof. Since p_{n} ≥ x_{1}, we use (12) and (18) to get

C_{n} ≤ ∫_{2}^{x_{1}} π(x) dx + ∫_{x_{1}}^{p_{n}} \frac{x dx}{log x} + ∑_{k=2}^{m} a_{k} ∫_{x_{1}}^{p_{n}} \frac{x dx}{log^{k} x}.

We apply Lemma 8(i) and Proposition 9 to obtain

C_{n} ≤ ∫_{2}^{x_{1}} π(x) dx − li(x_{1}^{2}) + li(p_{n}^{2}) + t_{m−1,1} ∫_{x_{1}}^{p_{n}} \frac{x dx}{log^{2} x} − ∑_{k=2}^{m−1} t_{m−1,k} (\frac{p_{n}^{2}}{log^{k} p_{n}} − \frac{x_{1}^{2}}{log^{k} x_{1}} ) .

Using Lemma 8(ii), we get

C_{n} ≤ d_{1} + (1 + 2t_{m−1,1}) li(p_{n}^{2}) − ∑_{k=1}^{m−1} t_{m−1,k} \frac{p_{n}^{2}}{log^{k} p_{n}} .

Now we use the inequality (19) to obtain

C_{n} ≤ d_{1} + ∑_{k=1}^{m−2} (\frac{(k − 1)!}{2^{k}} + \frac{t_{m−1,1}(k − 1)!}{2^{k−1}} − t_{m−1,k}) \frac{p_{n}^{2}}{log^{k} p_{n}}

+ (\frac{(1 + 2t_{m−1,1})λ}{2^{m−1}} − t_{m−1,m−1}) \frac{p_{n}^{2}}{log^{m−1} p_{n}}

and it remains to apply the definition of t_{ij}.

5 Numerical results

5.1 An explicit lower bound for C_{n}

The goal of this subsection is to improve the inequality (2) in view of (4). In order to do this, we first give two lemmata concerning explicit estimates for li(x) and π(x), respectively.

Lemma 15. If x ≥ 4171, then

li(x) ≥ \frac{x}{log x} + \frac{x}{log^{2} x} + \frac{2x}{log^{3} x} + \frac{6x}{log^{4} x} + \frac{24x}{log^{5} x} + \frac{120x}{log^{6} x} + \frac{720x}{log^{7} x} + \frac{5040x}{log^{8} x}.

Proof. We denote the right hand side of the above inequality by α(x) and let f(x) = li(x) − α(x). Then, f(4171) ≥ 0.00019 and f′(x) = 40320/ log^{9} x, and the lemma is proved.

Lemma 16. If x ≥ 10^{16}, then

li(x) ≤ \frac{x}{log x} + \frac{x}{log^{2} x} + \frac{2x}{log^{3} x} + \frac{6x}{log^{4} x} + \frac{24x}{log^{5} x} + \frac{120x}{log^{6} x} + \frac{900x}{log^{7} x}.

Proof. Similarly to the proof of Lemma 15.

Lemma 17. If x ≥ 1332450001, then

π(x) > \frac{x}{log x} + \frac{x}{log^{2} x} + \frac{2x}{log^{3} x} + \frac{5.65x}{log^{4} x} + \frac{23.65x}{log^{5} x} + \frac{118.25x}{log^{6} x} + \frac{709.5x}{log^{7} x} + \frac{4966.5x}{log^{8} x}.

Proof. See [1, Thm. 1.2].

Setting

Θ(n) = \frac{43.6p_{n}^{2}}{8 log^{4} p_{n}} + \frac{90.9p_{n}^{2}}{4 log^{5} p_{n}} + \frac{927.5p_{n}^{2}}{8 log^{6} p_{n}} + \frac{5620.5p_{n}^{2}}{8 log^{7} p_{n}} + \frac{79075.5p_{n}^{2}}{16 log^{8} p_{n}} .

we get the following improvement of (2).

Proposition 18. If n ≥ 52703656, then

C_{n} ≥ \frac{p_{n}^{2}}{2 log p_{n}} + \frac{3p_{n}^{2}}{4 log^{2} p_{n}} + \frac{7p_{n}^{2}}{4 log^{3} p_{n}} + Θ(n).

Proof. We choose m = 9, a_{2} = 1, a_{3} = 2, a_{4} = 5.65, a_{5} = 23.65, a_{6} = 118.25, a_{7} = 709.5, a_{8} = 4966.5, a_{9} = 0, x_{0} = 1332450001 and y_{0} = 4171. By Lemma 17, we obtain the inequality (16) for every x ≥ x_{0} and (17) holds for every x ≥ y_{0} by Lemma 15. Substituting these values in Theorem 13, we get

C_{n} ≥ d_{0} + \frac{p_{n}^{2}}{2 log p_{n}} + \frac{3p_{n}^{2}}{4 log^{2} p_{n}} + \frac{7p_{n}^{2}}{4 log^{3} p_{n}} + Θ(n)

for every n ≥ 66773605, where d_{0} = d_{0}(9, 1, 2, 5.65, 23.65, 118.25, 709.5, 4966.5, 0, x_{0}) is given by

d_{0} = ∫_{2}^{x_{0}} π(x) dx − \frac{753.1}{3} li(x_{0}^{2}) + \frac{375.05x_{0}^{2}}{3 log x_{0}} + \frac{186.025x_{0}^{2}}{3 log^{2} x_{0}} + \frac{183.025x_{0}^{2}}{3 log^{3} x_{0}} + \frac{88.6875x_{0}^{2}}{log^{4} x_{0}}

+ \frac{165.55x_{0}^{2}}{log^{5} x_{0}} + \frac{354.75x_{0}^{2}}{log^{6} x_{0}} + \frac{709.5x_{0}^{2}}{log^{7} x_{0}} .

Since x_{0}^{2} ≥ 10^{16}, it follows from Lemma 16 that

d_{0} ≥ ∫_{2}^{x_{0}} π(x) dx − \frac{x_{0}^{2}}{2 log x_{0}} − \frac{3x_{0}^{2}}{4 log^{2} x_{0}} − \frac{7x_{0}^{2}}{4 log^{3} x_{0}} − \frac{5.45x_{0}^{2}}{log^{4} x_{0}} − \frac{22.725x_{0}^{2}}{log^{5} x_{0}}

− \frac{115.9375x_{0}^{2}}{log^{6} x_{0}} − \frac{1055.578125x_{0}^{2}}{log^{7} x_{0}} .

Using log x_{0} ≥ 21.01027, we get

d_{0} ≥ ∫_{2}^{x_{0}} π(x) dx − 4.22512933 · 10^{16} − 0.30164729 · 10^{16} − 0.03349997 · 10^{16}
− 0.0049656 · 10^{16} − 0.00098548 · 10^{16} − 0.0002393 · 10^{16} − 0.0001037 · 10^{16}
= ∫_{2}^{x_{0}} π(x) dx − 4.56657067 · 10^{16}.  (20)

Since x_{0} = p_{66773604}, we use (12) a computer to obtain

∫_{2}^{x_{0}} π(x) dx = C_{66773604} = 45665745738169817.

Hence, by (20), we get d_{0} ≥ 3.9·10^{10} > 0. So we obtain the asserted inequality for every n ≥ 66773605. For every 52703656 ≤ n ≤ 66773604 we check the inequality with a computer.

5.2 An explicit upper bound for C_{n}

We begin with the following two lemmata.

Lemma 19. If x ≥ 10^{18}, then

li(x) ≤ \frac{x}{log x} + \frac{x}{log^{2} x} + \frac{2x}{log^{3} x} + \frac{6x}{log^{4} x} + \frac{24x}{log^{5} x} + \frac{120x}{log^{6} x} + \frac{720x}{log^{7} x} + \frac{6300x}{log^{8} x}.

Proof. Similarly to the proof of Lemma 15.

Lemma 20. If x > 1, then

π(x) < \frac{x}{log x} + \frac{x}{log^{2} x} + \frac{2x}{log^{3} x} + \frac{6.35x}{log^{4} x} + \frac{24.35x}{log^{5} x} + \frac{121.75x}{log^{6} x} + \frac{730.5x}{log^{7} x} + \frac{6801.4x}{log^{8} x}.

Proof. See [1, Thm. 1.1].

Using these upper bounds, we obtain the following explicit upper bound for C_{n}, where

Ω(n) = \frac{46.4p_{n}^{2}}{8 log^{4} p_{n}} + \frac{95.1p_{n}^{2}}{4 log^{5} p_{n}} + \frac{962.5p_{n}^{2}}{8 log^{6} p_{n}} + \frac{5809.5p_{n}^{2}}{8 log^{7} p_{n}} + \frac{118848p_{n}^{2}}{16 log^{8} p_{n}} .

Proposition 21. For every n ∈ ℕ,

C_{n} ≤ \frac{p_{n}^{2}}{2 log p_{n}} + \frac{3p_{n}^{2}}{4 log^{2} p_{n}} + \frac{7p_{n}^{2}}{4 log^{3} p_{n}} + Ω(n).

Proof. We choose a_{2} = 1, a_{3} = 2, a_{4} = 6.35, a_{5} = 24.35, a_{6} = 121.75, a_{7} = 730.5, a_{8} = 6801.4, λ = 6300, x_{1} = 11 and y_{1} = 10^{18}. By Lemma 20, we get that the inequality (18) holds for every x ≥ x_{1} and by Lemma 19, that (19) holds for all y ≥ y_{1}. By substituting these values in Theorem 14, we get

C_{n} ≤ d_{1} + \frac{p_{n}^{2}}{2 log p_{n}} + \frac{3p_{n}^{2}}{4 log^{2} p_{n}} + \frac{7p_{n}^{2}}{4 log^{3} p_{n}} + Ω(n) − \frac{0.875p_{n}^{2}}{16 log^{8} p_{n}}  (21)

for every n ≥ 50847535, where d_{1} = d_{1}(9, 1, 2, 6.35, 24.35, 121.75, 730.5, 6801.4, 0, x_{1}) is given by

d_{1} = ∫_{2}^{x_{1}} π(x) dx − \frac{950777}{3150} li(x_{0}^{2}) + \frac{947627x_{0}^{2}}{6300 log x_{0}} + \frac{941327x_{0}^{2}}{12600 log^{2} x_{0}} + \frac{928727x_{0}^{2}}{12600 log^{3} x_{0}}

+ \frac{902057x_{0}^{2}}{8400 log^{4} x_{0}} + \frac{425461x_{0}^{2}}{2100 log^{5} x_{0}} + \frac{187163x_{0}^{2}}{420 log^{6} x_{0}} + \frac{34007x_{0}^{2}}{35 log^{7} x_{0}} .

Since li(x_{1}^{2}) ≥ 34.59 and log x_{1} ≥ 2.39, we obtain d_{1} ≤ 450. We define

f(x) = \frac{0.875x^{2}}{16 log^{8} x} − 450.

Since f(6 · 10^{6}) ≥ 109 and f′(x) ≥ 0 for every x ≥ e^{4}, we get f(p_{n}) ≥ 0 for every n ≥ π(6 · 10^{6}) + 1 = 412850. Now we can use (21) to obtain the desired inequality for every n ≥ 50847535. For every 1 ≤ n ≤ 50847534 a computer makes the rest of work.

References

[1] C. Axler, New bounds for the prime counting function π(x), preprint, 2015. Available at

   http://arxiv.org/abs/1409.1780.

[2] C. Axler, On the sum of the first n prime numbers, preprint, 2014. Available at

   http://arxiv.org/abs/1409.1777.

[3] M. Cipolla, La determinazione assintotica dell’ n^{imo} numero primo, Rend. Accad. Sci. Fis. Mat. Napoli 8 (1902), 132–166.

[4] P. Dusart, Autour de la fonction qui compte le nombre de nombres premiers, Dissertation, Université de Limoges, 1998.

[5] M. Hassani, A remark on the Mandl’s inequality, preprint, 2006. Available at

   http://arxiv.org/abs/math/0606765.

[6] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, http://oeis.org.

[7] J. B. Rosser and L. Schoenfeld, Sharper bounds for the Chebyshev functions θ(x) and ψ(x), Math. Comp. 29 (1975), 243–269.

2010 Mathematics Subject Classification: Primary 11A41; Secondary 11B83, 11N05. Keywords: asymptotic formula, Mandl’s inequality, prime number.

(Concerned with sequences \underline{A000040}, \underline{A007504}, \underline{A124478}, and \underline{A152535}.)

Received March 19 2015; revised versions received March 23 2015; June 19 2015; July 10 2015; July 13 2015. Published in Journal of Integer Sequences, July 16 2015.

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