On a Family of Functions Defined Over Sums of Primes

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

                christian.axler@hhu.de

Abstract

Let r and m be real numbers so that the sum S_{r,m}(x) = ∑_{p≤x} p^{r} log^{m} p diverges as x → ∞. Here p runs over all primes not exceeding x. In this paper, we give an asymptotic formula for each S_{r,m}(x) as x → ∞. The case where x is the nth prime number is of particular interest. Here we use a method developed by Salvy to give an asymptotic formula for S_{r,m}(p_{n}) as n → ∞, which generalizes, for instance, the previously known one for S_{1,0}(p_{n}), the sum of the first n prime numbers.

1 Introduction

Let m and r be real numbers and let S_{r,m} : [0, ∞) → ℝ be defined by

S_{r,m}(x) = ∑_{p≤x} p^{r} log^{m} p,

where p runs over all primes not exceeding x. The sum S_{r,m}(x) diverges as x → ∞ if and only if

(i) r > −1 and m ∈ ℝ  or  (ii) r = −1 and m ≥ 0.

The aim of this paper is to find the asymptotic behaviour of the sum S_{r,m}(x) in the case where m and r satisfy conditions (i) or (ii). First, we study the case (i). Let π(x) denote the number of primes not exceeding x. A well-known result (see [10]) concerning this function is the prime number theorem, which states that

π(x) = li(x) + O(xe^{−a\sqrt{log x}})  (1)

as x → ∞, where a is a positive absolute constant, and the logarithmic integral li(x) is defined for every real x ≥ 0 as follows:

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)

Denoting the sum of the first prime numbers not exceeding x by S(x), Szalay [9, Lemma 1] used (1) and Stieltjes integration to find

S(x) = li(x^{2}) + O(x^{2}e^{−a\sqrt{log x}})  (3)

as x → ∞. The case x = p_{n}, where p_{n} denotes the nth prime number, is of particular interest. Here, S(x) = ∑_{k≤n} p_{k} is equal to the sum of the first n prime numbers. Massias and Robin [4, p. 217] found that

∑_{k=1}^{n} p_{k} = li((li^{−1}(n))^{2}) + O(n^{2}e^{−c\sqrt{log n}})  (4)

as n → ∞, where c is a positive absolute constant and li^{−1}(x) is the inverse function of li(x). Then they [4, p. 217] used (4) and a result of Robin [7] to derive the asymptotic expansion

∑_{k=1}^{n} p_{k} = \frac{n^{2}}{2} (log n + ∑_{i=0}^{N} \frac{A_{i+1}(log log n)}{log^{i} n}) + O_{N} (\frac{n^{2}(log log n)^{N+1}}{log^{N+1} n} )  (5)

as n → ∞, where N is a nonnegative integer and the polynomials A_{k} satisfy the formulas A_{0}(x) = 1 and

A′_{i+1} = A′_{i} − (i − 1)A_{i}.  (6)

It follows that deg(A_{0}) = 0, deg(A_{1}) = 1, and deg(A_{i}) = i − 1 for every integer i ≥ 2. Unfortunately, the recursive formula (6) for the derivatives does not yield a description of the polynomials A_{i}, since the constant coefficient of the polynomials A_{i} remains undetermined by this equation. This problem was fixed in [3, Theorem 1.4] by applying a method developed by Salvy [8, Theorem 2]. We use the same method to give the following result concerning the sum S_{r,m}(p_{n}) in the case where r > −1 and m ∈ ℝ. Here, we use the notation

\binom{δ}{0} = 1 and \binom{δ}{k} = \frac{δ(δ − 1) ⋯ (δ − k + 1)}{k!}

for a real number δ and a positive integer k.

Theorem 1. Let r and m be real numbers with r > −1 and let N be a nonnegative integer. As n → ∞, we have

∑_{k=1}^{n} p_{k}^{r} log^{m} p_{k} = \frac{n^{r+1} log^{r+m} n}{r + 1} (∑_{i=0}^{N} \frac{A_{r,m,i}(log log n)}{log^{i} n} + O_{r,m,N} (\frac{(log log n)^{N+1}}{log^{N+1} n} )) , (7)

where the polynomials A_{r,m,i} ∈ ℝ[x] are defined by

A_{r,m,0} = 1, A′_{r,m,i+1} = A′_{r,m,i} + (m + r − i)A_{r,m,i}.  (8)

The polynomials A_{r,m,i} can be computed explicitly. In particular,

A_{r,m,1}(x) = (r + m)x − \frac{m − 1}{r + 1} − r − 1

and

A_{r,m,2}(x) = \binom{r + m}{2}x^{2} + \frac{(−r^{3} − mr^{2} + (−2m + 3)r − m^{2} + 2m)x}{r + 1} + λ_{r,m},

where

λ_{r,m} = \frac{(m − 1)(m − 2)}{(r + 1)^{2}} + \frac{r(r − 3)}{2} − 2.

Remark 2. The polynomials A_{i} given in (6) and A_{r,m,i} are connected by the formula A_{i} = A_{1,0,i}.

Remark 3. Since S_{1,0}(p_{n}) = S(p_{n}), Theorem 1 yields a generalization of (5).

Remark 4. For some alternative asymptotic formulae for S(p_{n}), see [2, Theorems 1 and 2].

In the second part of this paper, we study the case (ii); i.e. r = −1 and m ≥ 0. If m = 0, we see that S_{r,m}(x) is equal to the sum of the reciprocals of all prime numbers not exceeding x. Mertens [5, p. 52] proved that log log x is the right order of magnitude for this sum by showing

∑_{p≤x} \frac{1}{p} = log log x + B + O (\frac{1}{log x})

as x → ∞. Here B denotes the Mertens’ constant and is defined by

B = γ + ∑_{p} (log (1 − \frac{1}{p}) + \frac{1}{p}) = 0.261 …,

where γ = 0.577 … denotes the Euler–Mascheroni constant. So it suffices to consider the case where r = −1 and m > 0. Here, we find the following result.

Theorem 5. Let m be a positive real number and let N be a nonnegative integer. As n → ∞, we have

∑_{k=1}^{n} \frac{log^{m} p_{k}}{p_{k}} = \frac{log^{m} n}{m} (∑_{i=0}^{N} \frac{B_{m,i}(log log n)}{log^{i} n} + O_{N,m} (\frac{(log log n)^{N+1}}{log^{N+1} n} )) ,  (9)

where the polynomials B_{m,i} ∈ ℝ[x] are defined by

B_{m,0} = 1, B′_{m,i+1} = B′_{m,i} + (m − i)B_{m,i}.

The polynomials B_{m,i} can be computed explicitly. For example, we have

B_{m,1}(x) = mx  and  B_{m,2}(x) = \binom{m}{2}x^{2} + mx − m.

2 Proof of Theorem 1

In order to prove Theorem 1, we first note Abel’s summation formula, which can be found in [1].

Lemma 6 (Abel’s summation formula). Let a : ℕ → ℂ be a function, and let A(x) = ∑_{n≤x} a(n), where A(x) = 0 if x < 1. If g has a continuous derivative on the interval [y, x], where 0 < y < x, then

∑_{y<n≤x} a(n)g(n) = A(x)g(x) − A(y)g(y) − ∫_{y}^{x} A(t)g′(t) dt.

Proof. See [1, Theorem 4.2].

Using this Lemma, we get the following result.

Proposition 7. Let r and m be real numbers with r > −1 and let N be a nonnegative integer. For a nonnegative integer j, we set

χ_{r,m,j} = \frac{(−1)^{j}j!}{(r + 1)^{j}} \binom{m − 1}{j}.  (10)

As x → ∞, we have

∫_{2}^{x} t^{r} log^{m−1} t dt = \frac{x^{r+1} log^{m−1} x}{r + 1} ∑_{j=0}^{N} \frac{χ_{r,m,j}}{log^{j} x} + χ_{r,m,N+1} ∫_{2}^{x} t^{r} log^{m−2−N} t dt + O(1).

Proof. Integration by parts and induction.

Remark 8. In the case where m is a positive integer and N ≥ m − 1, we get χ_{r,m,N+1} = 0. Here, Proposition 7 gives

∫_{2}^{x} t^{r} log^{m−1} t dt = \frac{x^{r+1} log^{m−1} x}{r + 1} ∑_{j=0}^{N} \frac{χ_{r,m,j}}{log^{j} x} + O(1)

as x → ∞.

Proposition 7 implies the following asymptotic formula.

Corollary 9. Let r and m be real numbers with r > −1 and let N be a nonnegative integer. As x → ∞, we have

∫_{2}^{x} t^{r} log^{m−1} t dt = \frac{x^{r+1} log^{m−1} x}{r + 1} (∑_{j=0}^{N} \frac{χ_{r,m,j}}{log^{j} x} + O_{r,m,N} (\frac{1}{log^{N+1} x})) .

Proof. Using L’Hospital’s rule, we see that

∫_{2}^{x} t^{r} log^{m−2−N} t dt = O_{r,m,N} (\frac{x^{r+1}}{log^{N+2−m} x})

as x → ∞, and it suffices to apply this equation to Proposition 7.

Now we apply Corollary 9 to find the following result.

Proposition 10. Let r and m be real numbers with r > −1 and let N be a nonnegative integer. Then

∑_{p≤x} p^{r} log^{m} p = \frac{x^{r+1} log^{m−1} x}{r + 1} (∑_{j=0}^{N} \frac{χ_{r,m,j}}{log^{j} x} + O_{r,m,N} (\frac{1}{log^{N+1} x}))

as x → ∞, where χ_{r,m,j} is defined by (10).

Proof. For x ≥ 2, let R(x) = π(x) − li(x). As x → ∞, the asymptotic fomula (1) implies

R(x) = O(xe^{−a\sqrt{log x}})  (11)

for some positive absolute constant a. Furthermore, let y = 3/2, g(x) = x^{r} log^{m} x, and

a(n) = \begin{cases}1, & if n is prime; \\ 0, & otherwise. \end{cases}

We use Lemma 6 to get

∑_{p≤x} p^{r} log^{m} p = (li(x) + R(x))x^{r} log^{m} x − ∫_{2}^{x}(li(t) + R(t))t^{r−1}(r log^{m} t + m log^{m−1} t) dt.

If we combine this with (11) and

∫_{2}^{x} t^{r} log^{m−1} t dt = li(x)x^{r} log^{m} x − ∫_{2}^{x} li(t)t^{r−1}(r log^{m} t + m log^{m−1} t) dt + O(1)

as x → ∞, we see that

∑_{p≤x} p^{r} log^{m} p = ∫_{2}^{x} t^{r} log^{m−1} t dt + O(x^{r+1}e^{−b\sqrt{log x}}) + O (∫_{2}^{x} t^{r}e^{−b\sqrt{log t}} dt) (12)

as x → ∞, where the real number b satisfies 0 < b < a. Using L’Hospital’s rule, we get

∫_{2}^{x} t^{r}e^{−a\sqrt{log t}} dt = O(x^{r+1}e^{−a\sqrt{log x}})

as x → ∞. So we can rewrite (12) as

∑_{p≤x} p^{r} log^{m} p = ∫_{2}^{x} t^{r} log^{m−1} t dt + O(x^{r+1}e^{−b\sqrt{log x}})  (13)

as x → ∞. Finally, we apply Corollary 9 to complete the proof.

In the following corollary, we give a generalization of (3).

Corollary 11. Let r be a real number with r > −1 and let m be a nonnegative integer. As x → ∞, we have

∑_{p≤x} \frac{p^{r}}{log^{m} p} = \frac{(r + 1)^{m}}{m!} (li(x^{r+1}) − ∑_{j=0}^{m−1} \frac{j! x^{r+1}}{(r + 1)^{j+1} log^{j+1} x}) + O(x^{r+1}e^{−a\sqrt{log x}}).

Proof. Induction over m and integration by parts gives

∫_{2}^{x} \frac{t^{r}}{log^{m+1} t} dt = \frac{(r + 1)^{m}}{m!} (li(x^{r+1}) − ∑_{j=0}^{m−1} \frac{j! x^{r+1}}{(r + 1)^{j+1} log^{j+1} x}) + O(1)

as x → ∞. Now it suffices to apply the equation (13).

To give a proof of Theorem 1, we also need the following result of Salvy [8]. Here we use the notation from Robin [7].

Proposition 12 (Salvy). Let y = y(x) satisfies e^{y}y^{−α}D(1/y) ≈ x as x → ∞, with D(u) = ∑_{n≥0} d_{n}u^{n} a formal power series, α ≠ 0, and D(0) ≠ 0. Then for any formal power series G with nonzero constant term the following asymptotic expansion hold:

e^{βy}y^{γ}G(1/y) ≈ (\frac{x}{d_{0}} )^{β} (log x)^{αβ+γ} ∑_{n≥0} \frac{Q_{n}(log log x)}{log^{n} x} (x → ∞).

Here Q_{n} are polynomials with Q_{0} = G(0) and Q′_{n+1}/α = Q′_{n} + (αβ + γ − n)Q_{n}.

Proof. See [8, Theorem 2].

Now we can give a proof of Theorem 1.

Proof of Theorem 1. Let N be a nonnegative integer and let D_{N}(u) = ∑_{j=0}^{N} j! u^{j}. We define the formal power series

D(u) = ∑_{j=0}^{∞} j! u^{j}.

Then D(0) = 1. First, we note that repeated integration by parts in (2) gives

li(x) ≈ \frac{x}{log x} D (\frac{1}{log x})  (14)

as x → ∞. 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.  (15)

We combine (14) and (15) to obtain

e^{y}y^{−1}D(1/y) ≈ n  (16)

as n → ∞, where y = log li^{−1}(n). Next, we define δ(n) by p_{n} = li^{−1}(n) + δ(n). By Massias and Robin [4, p. 217], we have

δ(n) = O(ne^{−c\sqrt{log n}})  (17)

as n → ∞, where c is a positive absolute constant. Since p_{n} ∼ n log n as n → ∞ (see, for example, [8]), we see that li^{−1}(n) ∼ n log n as n → ∞. Substituting x = p_{n} in Proposition 10, we get

∑_{k=1}^{n} p_{k}^{r} log^{m} p_{k} = \frac{p_{n}^{r+1} log^{m−1} p_{n}}{r + 1} (∑_{j=0}^{N} \frac{χ_{r,m,j}}{log^{j} p_{n}} + O_{r,m,N} (\frac{1}{log^{N+1} p_{n}} ))  (18)

as n → ∞. Using the mean value theorem and (17), we deduce

log^{m−1}(p_{n}) = y^{m−1} + O(e^{−d\sqrt{log n}})  (19)

as n → ∞, where d is a real number satisfying 0 < d < c. Combined with (18), this gives

∑_{k=1}^{n} p_{k}^{r} log^{m} p_{k} = \frac{p_{n}^{r+1}y^{m−1}}{r + 1} (∑_{j=0}^{N} \frac{χ_{r,m,j}}{log^{j} p_{n}} + O_{r,m,N} (\frac{1}{y^{N+1}} ))  (20)

as n → ∞. By the binomial theorem, we have

p_{n}^{r+1} = (li^{−1}(n) + δ(n))^{r+1} = li^{−1}(n)^{r+1} + O_{r}(n^{r+1}e^{−c\sqrt{log n}})

as n → ∞. Applying this to (20), we see that

∑_{k=1}^{n} p_{k}^{r} log^{m} p_{k} = \frac{e^{(r+1)y}y^{m−1}}{r + 1} (∑_{j=0}^{N} \frac{χ_{r,m,j}}{log^{j} p_{n}} + O_{r,m,N} (\frac{1}{y^{N+1}} ))

as n → ∞. Let f(x) = 1/ log^{k} x, where k is an integer with 0 ≤ k ≤ m − 1. Again, by the mean value theorem there exists a real ξ ∈ (min{p_{n}, li^{−1}(n)}, max{p_{n}, li^{−1}(n)}) such that f(p_{n}) = f(li^{−1}(n)) + δ(n)f′(ξ). Since f′(x) = O(1/(x log^{k+1} x)) as x → ∞, we get f(p_{n}) = f(li^{−1}(n)) + O(e^{−c\sqrt{log n}}) as n → ∞. Hence

∑_{k=1}^{n} p_{k}^{r} log^{m} p_{k} = \frac{e^{(r+1)y}y^{m−1}}{r + 1} (G(1/y) + O_{r,m,N} (\frac{1}{y^{N+1}} ))  (21)

as n → ∞, where

G(u) = G_{r,m,N}(u) = ∑_{j=0}^{N} χ_{r,m,j} u^{j}.

Since (16) holds, we can apply Proposition 12 with α = 1, β = r + 1, and γ = m − 1 to see that

\frac{e^{(r+1)y}y^{m−1}G(1/y)}{r + 1} = \frac{n^{r+1} log^{r+m} n}{r + 1} (∑_{i=0}^{N} \frac{A_{r,m,i}(log_{2} n)}{log^{i} n} + O_{r,m,N} (\frac{(log_{2} n)^{N+1}}{log^{N+1} n} )) (22)

as n → ∞, where log_{2} x = log log x and the polynomials A_{r,m,i} ∈ ℝ[x] are defined by (8). If we combine (21) and (22), we arrive at the end of the proof of (7). Again, we apply the symbolic algebra system Maple to compute the polynomials A_{r,m,1}, …, A_{r,m,N} from the appendices of [8]. It suffices to write, in Maple,

  'sum'(p^r*log(p)^m) = 1/(r+1)*theorem2 part2(1,r+1,m-1,D N,G {r,m,N},n,N);

with N = 2 to get the polynomials A_{r,m,1} and A_{r,m,2}. This completes the proof.

Remark 13. We define lc(P) to be the leading coefficient of a polynomial P ∈ ℝ[x]. If r + m ∈ ℕ, we can use (8) to see that the polynomials A_{r,m,0}, …, A_{r,m,N} ∈ ℝ[x] satisfy

deg(A_{r,m,i}) = \begin{cases}i, & if i ≤ r + m; \\ i − 1, & otherwise, \end{cases}

and

lc(A_{r,m,i}) = \begin{cases}\binom{r + m}{i}, & if i ≤ r + m; \\ (−1)^{i−1−r−m}\binom{i − 1}{r + m}^{−1}, & otherwise. \end{cases}

In the case where r + m ∈ ℝ \ (ℕ ∪ {0}), we deduce from (8) that

deg(A_{r,m,i}) = i and lc(A_{r,m,i}) = \binom{r + m}{i}.

If r + m = 0, the equation in (8) implies that A′_{r,m,1} = 0 and we are not able to say anything in general about the degree or the leading coefficient of the polynomials A_{r,m,i}, where i ≥ 1.

Example 14. Let log_{2} x = log log x. We write

  'sum'(p^r*log(p)^m) = 1/(r+1)*theorem2 part2(1,r+1,m-1,D N,G {r,m,N},n,N);

with (r, m, N) = (1, −1, 2), (r, m, N) = (1, −2, 2), and (r, m, N) = (1, −3, 2) respectively. As n → ∞, this gives the asymptotic formulae

∑_{k=1}^{n} \frac{p_{k}}{log p_{k}} = \frac{n^{2}}{2} (1 − \frac{1}{log n} + \frac{log_{2} n − 3/2}{log^{2} n} + O (\frac{(log_{2} n)^{2}}{log^{3} n} )) ,

∑_{k=1}^{n} \frac{p_{k}}{log^{2} p_{k}} = \frac{n^{2}}{2 log n} (1 − \frac{log_{2} n + 1/2}{log n} + \frac{(log_{2} n)^{2}}{log^{2} n} + O (\frac{(log_{2} n)^{3}}{log^{3} n} )) ,

∑_{k=1}^{n} \frac{p_{k}}{log^{3} p_{k}} = \frac{n^{2}}{2 log^{2} n} (1 − \frac{2 log_{2} n}{log n} + \frac{3(log_{2} n)^{2} − 2 log_{2} n + 2}{log^{2} n} + O (\frac{(log_{2} n)^{3}}{log^{3} n} )) ,

respectively.

Remark 15. We have A_{1,1,0} = 1. The last example implies that lc(A_{1,1,1}) = 1. Now we can use (8) and induction to get lc(A_{1,1,i}) = 1 for all integers i with i ≥ 0.

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 (1) is equivalent to

ϑ(x) = x + O(xe^{−c_{1}\sqrt{log x}})

as x → ∞, where c_{1} is a positive absolute constant. Applying a well-known asymptotic expansion for the nth prime number (see [8]), 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} ))

as n → ∞, where log_{2} x = log log x, which gives the asymptotic expansion for S_{r,m}(p_{n}) in the case r = 0 and m = 1. For some other cases, we apply again the method developed by Salvy [8, Theorem 2] to get the following further results.

Example 16. Let log_{2} x = log log x. Again, it suffices to write

  'sum'(p^r*log(p)^m) = 1/(r+1)*theorem2 part2(1,r+1,m-1,D N,G {N,r,m},n,N);

with (r, m, N) = (0, 2, 2), (r, m, N) = (1, 1, 2), and (r, m, N) = (1, 2, 1), respectively. As n → ∞, this gives the asymptotic formulae

∑_{k=1}^{n} log^{2} p_{k} = n log^{2} n (1 + \frac{2 log_{2} n − 2}{log n} + \frac{(log_{2} n)^{2} − 2}{log^{2} n} + O (\frac{(log_{2} n)^{2}}{log^{3} n} )) ,

∑_{k=1}^{n} p_{k} log p_{k} = \frac{n^{2} log^{2} n}{2} (1 + \frac{2 log_{2} n − 2}{log n} + \frac{(log_{2} n)^{2} − 3}{log^{2} n} + +O (\frac{(log_{2} n)^{2}}{log^{3} n} )) ,

∑_{k=1}^{n} p_{k} log^{2} p_{k} = \frac{n^{2} log^{3} n}{2} (1 + \frac{3 log_{2} n − 5/2}{log n} + O (\frac{(log_{2} n)^{2}}{log^{2} n} )) ,

respectively.

3 Proof of Theorem 5

In 1857, de Polignac [6, part 3] stated without proof that log x is the right asymptotic behaviour for ∑_{p≤x} log p/p as x → ∞, where p runs over primes not exceeding x. A rigorous proof of this statement was given by Mertens [5]. He showed that

∑_{p≤x} \frac{log p}{p} = log x + O(1)

as x → ∞. We find the following generalization of this asymptotic formula.

Proposition 17. Let m be a positive real number. As x → ∞, we have

∑_{p≤x} \frac{log^{m} p}{p} = \frac{log^{m} x}{m} + O(1).

Proof. Similar to Proposition 10.

Finally, we use Proposition 17 to find the following proof of Theorem 5.

Proof of Theorem 5. We apply Proposition 17 with x = p_{n} and use (19) to see that

∑_{k=1}^{n} \frac{log^{m} p_{k}}{p_{k}} = \frac{y^{m}}{m} + O(1)

as n → ∞, where y = log li^{−1}(n). Since (16) holds, we can apply Proposition 12 with α = 1, β = 0, and γ = m to get the asymptotic expansion (9). In order to compute the polynomials B_{m,1}, …, B_{m,N}, we use the Maple code given in the appendices of [8]. It suffices to write

        'sum'(log(p)^m/p) = 1/(r+1)*theorem2 part2(1,0,m,D N,1,n,N);.

In particular, we get the desired polynomials B_{m,1} and B_{m,2}.

4 Acknowledgement

I would like to express my sincere thanks to Jean-Louis Nicolas for his support. Without his helpful comments this paper would not have appeared in this form. I would also like to thank R. for being a never-ending inspiration. Further, I would like to thank the anonymous referees for useful suggestions to improve the quality of this paper.

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2010 Mathematics Subject Classification: Primary 11N37; Secondary 11A41. Keywords: asymptotic formula, logarithmic integral, sum of primes.

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

Received December 17 2018; revised versions received December 19 2018; March 29 2019; July 8 2019. Published in Journal of Integer Sequences, August 23 2019.

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