. INEQUALITIES FOR THE EULER-MASCHERONI CONSTANT CHAO-PING CHEN Pn 1 1 2 Abstract. Let Rn = k=1 k −log n + 2 ,H(n) = n (Rn −γ), n = 1, 2,..., where γ is the Euler-Mascheroni constant. We prove that for all integers n ≥ 1, H(n) and [(n + 1/2)/n]2H(n) are strictly increasing, while [(n + 1)/n]2H(n) is strictly decreasing. For all integers n ≥ 1, 1 1 ≤ Rn − γ < 24(n + a)2 24(n + b)2 with the best possible constants 1 1 a = − 1 = 0.55106 ... and b = . p24[−γ + 1 − log(3/2)] 2 This refines result of D. W. DeTemple, who proved that the double inequality holds with a = 1 and b = 0. 1. Introduction The Euler-Mascheroni constant γ = 0.57721 ... is defined by n X 1 γ = lim Dn, where Dn = − log n. n→∞ k k=1 The speed of convergence of the sequence Dn has been studied by different authors. In 1971, S. R. Tims and J. A. Tyrrell [7] proved that 1 1 < D − γ < , n = 2, 3,.... 2(n + 1) n 2(n − 1) In 1991, R. M. Young [8] presented an elegant geometrical proof for the double inequality 1 1 < D − γ < , n = 1, 2,.... 2(n + 1) n 2n In 1997, G. D. Anderson et al. [3] established the the double inequality 1 − γ 1 ≤ D − γ < , n = 1, 2,.... n n 2n In 1998, H. Alzer [2] proved that for all integers n ≥ 1, the inequality 1 1 ≤ D − γ < 2(n + a) n 2(n + b) holds with the best possible constants 1 1 a = − 1 = 0.1826 ... and b = . 2(1 − γ) 6 2000 Mathematics Subject Classification. Primary 11B83; Secondary 11B73; 26D15. Key words and phrases. Euler-Mascheroni constant; inequality; psi function; asymptotic expansion. This work was supported by SF for pure research of Natural Science of the Education Depart- ment of Henan Province (2007110011). 1 2 CH.-P. CHEN The convergence of the sequence Dn to γ is very slow. In 1993, D. W. DeTemple [5] studied a modified sequence which converges faster and proved 1 1 < R − γ < , (1) 24(n + 1)2 n 24n2 where n X 1 1 R = − log n + . n k 2 k=1 Now let 2 H(n) = n (Rn − γ), n ≥ 1. Since n X 1 ψ(n + 1) = −γ + , k k=1 we see that 1 H(n) = (R − γ)n2 = ψ(n + 1) − log n + n2, (2) n 2 where ψ = Γ0/Γ is the psi function. Some computer experiments led M. Vuorinen to conjecture that H(n) increases on the interval [1, ∞) from H(1) = −γ + 1 − log(3/2) = 0.0173 ... to 1/24 = 0.0416 .... E. A. Karatsuba [6] proved that for all integers n ≥ 1, H(n) < H(n + 1), by clever use of Stirling formula and Fourier series. Some computer experiments also seem to indicate that [(n + 1)/n]2H(n) is a decreasing convex function [4]. The following Theorem 1 shows the monotonicity properties of H(n), [(n + 1/2)/n]2H(n) and [(n + 1)/n]2H(n). Theorem 1. Let H(n)(n = 1, 2,...) be defined by (2). Then for all integers n ≥ 1, H(n) and [(n+1/2)/n]2H(n) are both strictly increasing, while [(n+1)/n]2H(n) is strictly decreasing. Remark 1. By the asymptotic formula [1, p. 550] 1 1 ψ(x) = log x − + + O(x−4) as x → ∞, 2 24(x − 1/2)2 we conclude that 1 1 lim H(n) = and lim [(n + 1)/n]2H(n) = . (3) n→∞ 24 n→∞ 24 From the monotonicity of H(n), [(n+1)/n]2H(n) and (3), we obtain the inequality (1). In view of the inequality (1) it is natural to ask: What is the smallest number a and What is the largest number b such that the inequality 1 1 ≤ R − γ ≤ 24(n + a)2 n 24(n + b)2 holds for all integers n ≥ 1 ? The following Theorem 2 answers this question. Theorem 2. For all integers n ≥ 1, then 1 1 ≤ R − γ < (4) 24(n + a)2 n 24(n + b)2 INEQUALITIES FOR THE EULER-MASCHERONI CONSTANT 3 with the best possible constants 1 1 a = − 1 = 0.55106 ... and b = . p24[−γ + 1 − log(3/2)] 2 Remark 2. From the monotonicity of [(n + 1/2)/n]2H(n) and the limit relation 2 1 1 limn→∞[(n + 1/2)/n] H(n) = 24 , we obtain the right inequality of (4) with b = 2 . 2. Proofs of theorems In order prove our Theorem 1 and Theorem 2 we need to the following results 1 [1]: For x > − 2 ,N = 0, 1, 2,..., 2N 1 X B2k(1/2) log x − − < ψ(x) 1 2k 2 2k(x − 2 ) k=1 (5) 2N+1 1 X B2k(1/2) < log x − − 2 2k(x − 1 )2k k=1 2 and 2N+1 (n − 1)! X B2k(1/2) (n + 2k − 1)! + < (−1)n+1ψ(n)(x) 1 n 1 n+2k (x − 2 ) (2k)! (x − 2 ) k=1 (6) 2N (n − 1)! X B2k(1/2) (n + 2k − 1)! < + , n = 1, 2,..., (x − 1 )n (2k)! (x − 1 )n+2k 2 k=1 2 where 1 B (1/2) = − 1 − B , k = 0, 1, 2,..., k 2k−1 k Bk are Bernoulli numbers defined by ∞ t X tk = B . et − 1 k k! k=0 First four Bernoulli numbers with even indices are 1 1 1 1 B = ,B = − ,B = ,B = − , 2 6 4 30 6 42 8 30 this yields 1 7 31 127 B (1/2) = − ,B (1/2) = ,B (1/2) = − ,B (1/2) = . 2 12 4 240 6 1344 8 3840 From (5), we get 1 1 7 log x − + − < ψ(x) 2 24(x − 1 )2 960(x − 1 )4 2 2 (7) 1 1 < log x − + . 1 2 2 24(x − 2 ) From (6), we obtain 1 1 1 1 7 − < ψ0(x) < − + , 1 1 3 1 1 3 1 5 (8) x − 2 12(x − 2 ) x − 2 12(x − 2 ) 240(x − 2 ) 4 CH.-P. CHEN 1 1 7 31 − + − < ψ0(x) x − 1 12(x − 1 )3 240(x − 1 )5 1344(x − 1 )7 2 2 2 2 (9) 1 1 7 < − + . 1 1 3 1 5 x − 2 12(x − 2 ) 240(x − 2 ) Now we are in position to prove our Theorem 1 and Theorem 2. Proof of Theorem 1. Let a ≥ 0 be a real number and fa(x) be defined by 1 1 f (x) = (x + a)2 ψ(x + 1) − log x + , x > − . (10) a 2 2 differentiation yields 1 0 1 0 1 fa(x) = 2 ψ(x + 1) − log x + + (x + a) ψ (x + 1) − 1 . x + a 2 x + 2 By (7) and (8), we obtain 1 0 1 0 1 f0(x) = 2 ψ(x + 1) − log x + + x ψ (x + 1) − 1 x 2 x + 2 1 7 x > 2 − − 1 2 1 4 1 3 24(x + 2 ) 960(x + 2 ) 12(x + 2 ) 20x + 3 3 = > 0 for x > − . 1 4 480(x + 2 ) 20 This means that the sequence H(n) is strictly increasing for all integers n ≥ 1. By (7) and (9), we have 1 0 1 1 0 1 1 f1/2(x) = 2 ψ(x + 1) − log x + + x + ψ (x + 1) − 1 x + 2 2 2 x + 2 1 7 > 2 − 1 2 1 4 24(x + 2 ) 960(x + 2 ) 1 1 7 31 + x + − + − 1 3 1 5 1 7 2 12(x + 2 ) 240(x + 2 ) 1344(x + 2 ) 98x2 + 98x − 131.5 = > 0 for x > 0.76168 .... 1 6 6720(x + 2 ) This means that the sequence [(n+1/2)/n]2H(n) is strictly increasing for all integers n ≥ 1. By (7) and (8), we obtain 1 0 1 0 1 f1(x) = 2 ψ(x + 1) − log x + + (x + 1) ψ (x + 1) − 1 x + 1 2 x + 2 1 1 7 < + (x + 1) − + 12(x + 1 )2 12(x + 1 )3 240(x + 1 )5 2 2 √ 2 10x2 + 3x − 9 −3 + 189 = − 2 < 0 for x > = 0.537 .... 1 5 240(x + 2 ) 20 This means that the sequence [(n+1)/n]2H(n) is strictly decreasing for all integers n ≥ 1. The proof of Theorem 1 is complete. INEQUALITIES FOR THE EULER-MASCHERONI CONSTANT 5 Proof of Theorem 2. The inequality (4) can be written as 1 a < − n ≤ b. p24[ψ(n + 1) − log(n + 1/2)] In order to prove (4) we define 1 f(x) = − x. p24[ψ(x + 1) − log(x + 1/2)] Differentiation yields 13/2 24 ψ(x + 1) − log x + f 0(x) 2 3/2 1 0 1 = 12 1 − ψ (x + 1) − 24 ψ(x + 1) − log x + . x + 2 2 Now we show that there exists a positive real number x0 such that the function f is strictly decreasing on (x0, ∞). In order to find x0, we consider 3/2 1 0 1 12 1 − ψ (x + 1) < 24 ψ(x + 1) − log x + . x + 2 2 By (7) and (9), it is sufficient to consider 1 7 31 12 − + 12(x + 1 )3 240(x + 1 )5 1344(x + 1 )7 2 2 2 (11) 1 7 3/2 < 24 − . 1 2 1 4 24(x + 2 ) 960(x + 2 ) 1 Set u = x + 2 ,(11) become 7 31 7 3/2 1 − + < 1 − .