MONOTONICITY RESULTS FOR THE GAMMA FUNCTION

CHAO-PING CHEN AND FENG QI

Abstract. The function f(x) = \frac{[Γ(x+1)]^{1/x}}{x+1} is strictly decreasing on [1, ∞), the function g(x) = \frac{[Γ(x+1)]^{1/x}}{\sqrt{x}} is strictly increasing on [2, ∞), and the function h(x) = \frac{[Γ(x+1)]^{1/x}}{\sqrt{x+1}} is strictly increasing on [1, ∞), respectively. From these, some inequalities, for example, the Minc-Sathre inequality, are deduced, and two open problems posed by the second author are partially solved.

1. Introduction

In [12], H. Minc and L. Sathre proved that, if r is a positive integer and ϕ(r) = (r!)^{1/r}, then

1 < \frac{ϕ(r + 1)}{ϕ(r)} < \frac{r + 1}{r}.  (1)

In [1, 11], H. Alzer and J. S. Martins refined the right inequality in (1) and showed that, if n is a positive integer, then, for all positive real numbers r, we have

\frac{n}{n + 1} < (\frac{1}{n} ∑_{i=1}^{n} i^{r}/\frac{1}{n + 1} ∑_{i=1}^{n+1} i^{r})^{1/r} < \begin{matrix}^{n} & \sqrt{n!} \\ ^{n+1} & \sqrt{(n + 1)!} \end{matrix}.  (2)

Both bounds in (2) are the best possible.

Many extensions and generalizations of the inequalities in (2) have been found, please see [3, 4, 10, 13, 14, 20, 22, 27] and the references therein.

The inequalities in (1) were refined and generalized in [15, 21, 23, 24, 25] and the following inequalities were obtained:

\frac{n + k + 1}{n + m + k + 1} < ( ∏_{i=k+1}^{n+k} i)^{1/n}/( ∏_{i=k+1}^{n+m+k} i)^{1/(n+m)} ≤ \sqrt{\frac{n + k}{n + m + k}} , (3)

2000 Mathematics Subject Classification. Primary 33B15; Secondary 26D07. Key words and phrases. Gamma function, monotonicity, inequality. The authors were supported in part by NNSF (#10001016) of China, SF for the Prominent

Youth of Henan Province, SF of Henan Innovation Talents at Universities, Doctor Fund of Jiaozuo Institute of Technology, China.

This paper was typeset using AMS-L^{A}TEX.

2 CH.-P. CHEN AND F. QI

where k is a nonnegative integer, n and m are natural numbers. For n = m = 1, the equality in (3) is valid.

In [16], the inequalities in (3) were generalized to obtain the following inequalities on the ratio for the geometric means of a positive arithmetic sequence with unit difference for any nonnegative integer k and natural numbers n and m:

\frac{n + k + 1 + α}{n + m + k + 1 + α} ≤ \begin{matrix}[∏_{i=k+1}^{n+k}(i + α)] & 1/n \\ [∏_{i=k+1}^{n+m+k}(i + α)] & 1/(n+m) \end{matrix} ≤ \sqrt{\frac{n + k + α}{n + m + k + α}} , (4)

where α ∈ [0, 1] is a constant.

Furthermore, for nonnegative integer k and natural numbers n and m, we have

\frac{a(n + k + 1) + b}{a(n + m + k + 1) + b} ≤ \begin{matrix}[∏_{i=k+1}^{n+k}(ai + b)] & \frac{1}{n} \\ [∏_{i=k+1}^{n+m+k}(ai + b)] & \frac{1}{n+m} \end{matrix} ≤ \sqrt{\frac{a(n + k) + b}{a(n + m + k) + b}} , (5)

where a and b are positive constants. See [8].

It is clear that the inequalities in (5) extend those in (4).

In [21], the following monotonicity result was obtained: The function

\frac{[Γ(x + y + 1)/Γ(y + 1)]^{1/x}}{x + y + 1}  (6)

is decreasing in x ≥ 1 for fixed y ≥ 0. Then, for positive real numbers x and y, we have

\frac{x + y + 1}{x + y + 2} ≤ \frac{[Γ(x + y + 1)/Γ(y + 1)]^{1/x}}{[Γ(x + y + 2)/Γ(y + 1)]^{1/(x+1)}} .  (7)

Inequality (7) extends and generalizes inequality (3), since Γ(n + 1) = n!.

In [8, 21], the second author, F. Qi, posed the following Open Problem 1. For positive real numbers x and y, we have

\frac{[Γ(x + y + 1)/Γ(y + 1)]^{1/x}}{[Γ(x + y + 2)/Γ(y + 1)]^{1/(x+1)}} ≤ \sqrt{\frac{x + y}{x + y + 1}} ,  (8)

where Γ denotes the gamma function. Open Problem 2. For any positive real number z, define z! = z(z − 1) ⋯ {z}, where {z} = z − [z − 1], and [z] denotes Gauss function whose value is the largest integer not more than z. Let x > 0 and y ≥ 0 be real numbers, then

\frac{x + 1}{x + y + 1} ≤ \begin{matrix}^{x} & \sqrt{x!} \\ ^{x+y} & \sqrt{(x + y)!} \end{matrix} ≤ \sqrt{\frac{x}{x + y}} .  (9)

MONOTONICITY RESULTS FOR THE GAMMA FUNCTION 3

It is well known that Γ(z + 1) = z!, see [28, pp. 90–91]. Hence the inequalities in (8) and (9) are equivalent to the following monotonicity results in some sense for x ≥ 1, which are the main results of this paper.

Theorem 1. The function f(x) = \frac{[Γ(x+1)]^{1/x}}{x+1} is strictly decreasing on [1, ∞), the function g(x) = \frac{[Γ(x+1)]^{1/x}}{\sqrt{x}} is strictly increasing on [2, ∞), and the function h(x) = \frac{[Γ(x+1)]^{1/x}}{\sqrt{x+1}} is strictly increasing on [1, ∞), respectively.

Remark 1. Note that the function f(x) is a special case of the function (6). In this paper, we will give a new and simple proof for the monotonicity of f(x). Theorem 1 partially solves the two open problems above. Remark 2. In recent years, many monotonicity results and inequalities involving the gamma function and incomplete gamma functions have been established, please see [5, 6, 7, 17, 18, 19, 24, 26] and some of the references therein.

2. Proof of Theorem 1

For x > 1, the following double inequalities are stated in [9, p. 431]:

0 < ln Γ(x) − [(x − \frac{1}{2}) ln x − x + \frac{1}{2} ln(2π)] < \frac{1}{x},  (10)
\frac{1}{2x} < ln x − \frac{Γ′(x)}{Γ(x)} < \frac{1}{x},  (11)
\frac{1}{x} < \frac{d^{2}}{dx^{2}} ln Γ(x) < \frac{1}{x − 1}.  (12)

In [28, pp. 103–105], the following formula was given:

\frac{Γ′(z)}{Γ(z)} + γ = ∫_{0}^{∞} \frac{e^{−t} − e^{−zt}}{1 − e^{−t}} dt = ∫_{0}^{1} \frac{1 − t^{z−1}}{1 − t} dt,  (13)

where γ denotes Euler constant and γ = 0.57721566490153286060651 · · ·. See [28, p. 94]. Formula (13) can be used to calculate Γ′(k) for k ∈ ℕ. We call ψ(z) = \frac{Γ′(z)}{Γ(z)} the digamma or psi function. See [2, p. 71].

Taking the logarithm yields

ln f(x) = \frac{1}{x} ln Γ(x + 1) − ln(x + 1).  (14)

Differentiating with respect to x on both sides of (14) and using the double inequalities (10) and (11) gives us

x^{2} \frac{f′(x)}{f(x)} = − ln Γ(x + 1) + x\frac{Γ′(x + 1)}{Γ(x + 1)} − \frac{x^{2}}{x + 1}

< − [(x + \frac{1}{2}) ln(x + 1) − (x + 1) + \frac{1}{2} ln(2π)]

+ x [ln(x + 1) − \frac{1}{2(x + 1)}] − \frac{x^{2}}{x + 1}

= −\frac{1}{2} ln(x + 1) − \frac{1}{2(x + 1)} + \frac{1}{2}[3 − ln(2π)]

≜ ϕ(x),

(15)

By direct computation, we have

ϕ′(x) = −\frac{x}{2(x + 1)^{2}} < 0.

Thus, the function ϕ(x) is strictly decreasing, and then ϕ(x) ≤ ϕ(1) = \frac{5}{4} − \frac{1}{2} ln(4π) < 0. Therefore f′(x) < 0 and f(x) is strictly decreasing on [1, ∞).

Straightforward calculation and utilizing the inequalities in (12) for x > 1 produces

ln g(x) = \frac{1}{x} ln Γ(x + 1) − \frac{1}{2} ln x,  (16)
x^{2} \frac{g′(x)}{g(x)} = − ln Γ(x + 1) + x \frac{d}{dx} ln Γ(x + 1) − \frac{1}{2}x ≜ φ(x),  (17)

\begin{aligned}φ′(x) & = x \frac{d^{2}}{dx^{2}} ln Γ(x + 1) − \frac{1}{2} \\ & > \frac{x}{x + 1} − \frac{1}{2} \\ & = \frac{x − 1}{2(x + 1)} \\ & > 0. \end{aligned}  (18)

Therefore, the function φ(x) is strictly increasing, and φ(x) ≥ φ(2) = Γ′(3) − 1 − ln 2 > 0 by (13). Thus g′(x) > 0 and then g(x) is strictly increasing on [2, ∞).

Direct computation and using the inequalities in (12) for x > 1 produces

ln h(x) = \frac{1}{x} ln Γ(x + 1) − \frac{1}{2} ln(x + 1),  (19)
x^{2} \frac{h′(x)}{h(x)} = − ln Γ(x + 1) + x \frac{d}{dx} ln Γ(x + 1) − \frac{x^{2}}{2(x + 1)} ≜ τ(x),  (20)

\begin{aligned}τ′(x) & = x \frac{d^{2}}{dx^{2}} ln Γ(x + 1) − \frac{x(2 + x)}{2(1 + x)^{2}} \\ & > \frac{x}{x + 1} − \frac{x(2 + x)}{2(1 + x)^{2}} \\ & = \frac{x^{2}}{2(x + 1)^{2}} \\ & > 0. \end{aligned}  (21)

Therefore, function τ(x) is strictly increasing, and τ(x) ≥ τ(1) = Γ′(2) − \frac{1}{4} > 0. Thus h′(x) > 0 and then h(x) is strictly increasing on [1, ∞).

The proof is complete.

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(Chen) Department of Applied Mathematics and Informatics, Jiaozuo Institute of Technology, Jiaozuo City, Henan 454000, China

(Qi) Department of Applied Mathematics and Informatics, Jiaozuo Institute of Technology, Jiaozuo City, Henan 454000, China

E-mail address: qifeng@jzit.edu.cn
URL: http://rgmia.vu.edu.au/qi.html