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NNTDM 17 (2011), 2, 1-3 On a limit involving the product of prime numbers József Sándor1 and Antoine Verroken2 1 Babeş–Bolyai University of Cluj, Romania e-mail: [email protected] 2 Univ. of Gent, Gent, Belgium e-mail: [email protected] Abstract. Let pk denote the kth prime number. The aim of this note is to prove that the √ limit of the sequence (pn / n p1 · · · pn ) is e. Keywords and phrases: Arithmetic functions, Estimates, Primes AMS Subject Classification: 11A25, 11N37 1 Introduction Let pn denote the nth prime number. The famous prime number theorem asserts that pn ∼ n log n as n → ∞, (1.1) pn = 1. There are many consequences of (1.1). As immediate applications, n→∞ n log n we can deduce pn+1 → 1, (1.2) pn log pn → 1, (1.3) log n i.e. lim as n → ∞. From (1.2) or (1.3) easily follows √ n pn → 1. (1.4) Various limits, including e.g. nlog pn+1 → 1, (n + 1) log pn (1.5) are induced in [4] (see pp. 247–254), where the unsolved conjecture of the first author, i.e. pn+1 − pn → 0, √ pn 1 (1.6) is also stated. √ The aim of this paper is to study the limit of pn / n p1 · · · pn , and to show in fact that √ n pn → e. p1 · · · pn (1.7) One of the main ingredients will be the use of a certain inequality involving Chebyshev’s function X θ(x) = log p, p≤x where p runs through the primes ≤ x. 2 Main results We need also the following limit relation: Lemma 2.1. log pn − pn → 1 as n → ∞. n (2.1) Proof. By a result of P. Dusart [1] one has pn = n(log n + log log n − 1) + n · θ(n), (2.2) where θ(n) > 0 and θ(n) → 0 as n → ∞. Thus log pn − pn = log nf (n) + nθ(n) − f (n) − θ(n) = n log log n − 1 + θ(n) = log 1 + + 1 − θ(n) = log n log log n − 1 + θ(n) = log 1 + + 1 − θ(n) → 1, log n log log n − 1 + θ(n) → 0. Here f (n) = log n + log log n − 1. For details see also the log n first author’s paper [3]. since The following result is due to Rosser and Schoenfeld (see [2]). Lemma 2.2. There exists a positive constant c > 0 such that for all x > 1 one has |θ(x) − x| < c · x . log2 x The main result of this paper is contained in the following: Theorem 2.1. The relation (1.7) holds true. 2 (2.3) √ Proof. Put An = pn / n p1 · · · pn . Then, by definition of function θ(x), one can write log An = log pn − 1 pn 1 θ(pn ) = log pn − + pn − θ(pn ) . n n n By (2.1) it will be sufficient to show that pn − θ(pn ) → 0. n (2.4) |θ(pn ) − pn | c · pn . By (1.1) and (1.3) one < n n log2 pn pn 1 |θ(pn ) − pn | has ∼ , so clearly → 0. This implies (2.4), and the proof of 2 log pn n n log pn Theorem 2.1 is finished, as log An → 1 implies An → e, (i.e.) relation (1.7) holds true. Now, by (2.3) of Lemma 2.2 one gets References [1] P. Dusart, The k th prime is greater than k(ln k + ln ln k − 1) for k ≥ 2, Math. Comp., 68(1999), no. 225, 411–415. [2] J. B. Rosser and L. Schoenfeld, Approximate formulas for some functions of prime numbers, Illinois J. Math., 6(1962), 64–94. [3] J. Sándor, On a limit for the sequence of primes, Octogon Math. Mag., 8(2000), no. 1, 180–181. [4] J. Sándor, Geometric theorems, diophantine equations, and arithmetic functions, American Research Press, 2002, USA. 3