Some Completely Monotonic Functions Involving Polygamma Functions and an Application ✩

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J. Math. Anal. Appl. 310 (2005) 303–308 provided by Elsevier - Publisher Connector www.elsevier.com/locate/jmaa

Some completely monotonic functions involving polygamma functions and an application ✩

Feng Qi ∗, Run-Qing Cui, Chao-Ping Chen, Bai-Ni Guo

Department of Applied Mathematics and Informatics, Research Institute of Applied Mathematics, Henan Polytechnic University, Jiaozuo City, Henan 454010, China Received 4 February 2004 Available online 16 March 2005 Submitted by S. Kaijser

Abstract By using the ﬁrst Binet’s formula the strictly completely monotonic properties of functions involving the psi and polygamma functions are obtained. As direct consequences, two inequalities are proved. As an application, the best lower and upper bounds of the nth harmonic number are estab- lished.  2005 Elsevier Inc. All rights reserved.

Keywords: Completely monotonic function; Gamma function; Psi function; Polygamma function; Harmonic number; Harmonic sequence

✩ The authors were supported in part by the National Natural Science Foundation of China grant 10001016, Science Foundation for the Prominent Youth of Henan Province grant 0112000200, Science Foundation of Fostering Project for Innovation Talents at Universities of Henan Province, Doctor Foundation of Henan Polytechnic University, China. * Corresponding author. E-mail addresses: [email protected], [email protected] (F. Qi), [email protected], [email protected] (C.-P. Chen), [email protected] (B.-N. Guo). URLs: http://rgmia.vu.edu.au/qi.html, http://dami.hpu.edu.cn/staff/qifeng.html (F. Qi).

1. Introduction

It is well known that the Euler gamma function Γ(z)is deﬁned for Re z>0asΓ(z)= ∞ z−1 −t = Γ (x) 0 t e dt. The function ψ(x) Γ(x), the logarithmic derivative of the gamma function, and ψ(i)(x) for i ∈ N are called the psi or digamma function and the polygamma functions. A function f is said to be completely monotonic on an interval I if f has derivatives of all orders on I which alternate successively in sign, that is (−1)nf (n)(x) 0(1) for x ∈ I and n 0. If inequality (1) is strict for all x ∈ I and for all n 0, then f is said to be strictly completely monotonic. ∞ 1 The series i=1 i is called the harmonic series. The sum of the ﬁrst few terms of the harmonic series is given in [1], [3, p. 593] and [9, p. 104] analytically by the nth harmonic number n 1 H(n)= = γ + ψ(n+ 1), (2) i i=1 where γ = 0.57721566 ...is the Euler–Mascheroni constant. L. Tóth in [8, p. 264] posed the following problems:

(1) Prove that for every positive integer n we have 1 1

In this article, by exploiting the ﬁrst Binet’s formula in [1] and [9, p. 106], we obtain the following strictly completely monotonic properties of some functions involving the psi and polygamma functions.

Theorem 1. The following functions 1 1 ψ(x)− ln x + + , (4) 2x 12x2 1 ln x − − ψ(x), (5) 2x 1 1 1 1 ψ (x) − − − + , (6) x 2x2 6x3 30x5 1 1 1 + + − ψ (x) (7) x 2x2 6x3 are strictly completely monotonic in (0, ∞). F. Qi et al. / J. Math. Anal. Appl. 310 (2005) 303–308 305

Remark 1. Recently, some completely monotonic properties of functions involving the gamma, psi, and polygamma functions are veriﬁed in [4–7].

As direct consequences of Theorem 1, we have the following two inequalities.

Corollary 1. For x>0, we have 1 1 1 − <ψ(x+ 1) − ln x< , (8) 2x 12x2 2x 1 1 1 1 1 1 − < − ψ (x + 1)< − + . (9) 2x2 6x3 x 2x2 6x3 30x5 As an application of Corollary 1, we establish the best lower and upper bounds of the nth harmonic number H(n)as follows.

Theorem 2. For arbitrary natural number n ∈ N, we have 1 1 H(n)− ln n − γ< . (10) + 1 − + 1 2n 1−γ 2 2n 3 1 − 1 The constants 1−γ 2 and 3 are the best possible.

2. Proofs of theorems

Proof of Theorem 1. Clearly for x>0 and any nonnegative integer k, ∞ 1 1 − = tke xt dt. (11) xk+1 k! 0 The ﬁrst Binet’s formula ([1] and [9, p. 106]) states that for x>0, ∞ 1 √ 1 1 1 e−xt ln Γ(x)= x − ln x − x + ln 2π − + − dt. (12) 2 2 t 1 − e−t t 0 Differentiating (12), integrating by part and using formula (11), it is deduced that ∞ 1 1 1 − ψ(x)− ln x + = − e xt dt. (13) x t et − 1 0 Using formulas (11) and (13) yields 1 1 q(x) ψ(x)− ln x + + 2x 12x2 ∞ 1 1 1 t − = − − + e xt dt t et − 1 2 12 0 306 F. Qi et al. / J. Math. Anal. Appl. 310 (2005) 303–308

∞ 2 t 2 (12 − 6t + t )e − (12 + 6t + t ) − = e xt dt 12t(et − 1) 0 > 0 (14) and 1 p(x) ψ(x)− ln x + 2x ∞ ∞ t 1 1 1 − (2 − t)e − (t + 2) − = − − e xt dt = e xt dt t et − 1 2 2t(et − 1) 0 0 < 0, (15) since it is easy to see that (12−6t +t2)et −(12+6t +t2)>0 and (2−t)et −(t +2)<0in (0, ∞). Hence, (−1)kq(k)(x) > 0 and (−1)k+1p(k)(x) > 0in(0, ∞) for any nonnegative integer k. Formula (13) implies ∞ 1 1 t − ψ (x) − − = − 1 e xt dt. (16) x x2 et − 1 0 Employing formulas (11) and (16) leads to

1 1 1 r(x) + + − ψ (x) x 2x2 6x3 ∞ 2 t t t − = 1 − − + e xt dt et − 1 2 12 0 ∞ 2 t 2 (12 − 6t + t )e − (12 + 6t + t ) − = e xt dt 12(et − 1) 0 > 0 (17) and

1 1 1 1 s(x) + + − − ψ (x) x 2x2 6x3 30x5 ∞ 2 4 t t t t − = 1 − − + − e xt dt et − 1 2 12 720 0 ∞ 4 2 4 2 t (t − 60t − 360t − 720) − (t − 60t + 360t − 720)e − = e xt dt 720(et − 1) 0 < 0, (18) F. Qi et al. / J. Math. Anal. Appl. 310 (2005) 303–308 307 since it is not difﬁcult to obtain (t4 − 60t2 − 360t − 720) − (t4 − 60t2 + 360t − 720)et < 0 in (0, ∞). Therefore, (−1)kr(k)(x) > 0 and (−1)k+1s(k)(x) > 0in(0, ∞) for any nonnegative integer k. The proof is complete. 2

+ = + 1 2 Proof of Corollary 1. These follow from Theorem 1 and using ψ(x 1) ψ(x) x .

The ﬁrst proof of Theorem 2. From formula (2), it is deduced that inequality (10) can be rearranged as 1 1 1 < − 2n − 2. (19) 3 ψ(n+ 1) − ln n 1 − γ Deﬁne for x>0, 1 φ(x)= − 2x. (20) ψ(x + 1) − ln x 12 Straightforward computation and utilizing (8) and (9) reveal that for x> 5 , 1 ψ(x + 1) − ln x 2φ (x) = − ψ (x + 1) − 2 ψ(x + 1) − ln x 2 x 1 1 1 1 1 2 < − + − 2 − 2x2 6x3 30x5 2x 12x2 12 − 5x = < 0, (21) 360x5 12 and φ(x) decreases with x> 5 . Straightforward calculation produces 1 φ(1) = − 2 = 0.36527211862544155 ..., (22) 1 − γ 1 φ(2) = − 4 = 0.35469600731465752 ..., (23) 3 − − 2 γ ln 2 1 φ(3) = − 6 = 0.34898948531361115 .... (24) 11 − − 6 γ ln 3 Therefore, the sequence 1 φ(n)= − 2n, n ∈ N, (25) ψ(n+ 1) − ln n is decreasing strictly, and for n ∈ N, 1 lim φ(n)<φ(n) φ(1) = − 2. (26) n→∞ 1 − γ Making use of approximating expansion of ψ in [1], [3, p. 594], or [9, p. 108] gives

1 1 − ψ(x)= ln x − − + O(x 4)(x→∞), (27) 2x 12x2 308 F. Qi et al. / J. Math. Anal. Appl. 310 (2005) 303–308 and then − 1 + O(x 2) 1 lim φ(n)= lim φ(x)= lim 3 = . (28) n→∞ x→∞ x→∞ 1 + O(x−1) 3 The proof is complete. 2

The second proof of Theorem 2. Deﬁne the sequence {xn}n∈N by 1 n 1 = − ln x − γ. (29) 2n + xn k k=1 Then, using that (see [2, p. 466] and [8]) n 1 1 1 ε = ln n + γ + − + n , (30) k 2n 12n2 120n4 k=1 where εn ∈ (0, 1) for n ∈ N, it is easy to show that the sequence {xn}n∈N is strictly decreas- = 1 2 ing and that limn→∞ xn 3 .

References

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