INEQUALITIES AND MONOTONICITY OF THE RATIO FOR THE GEOMETRIC MEANS OF A POSITIVE ARITHMETIC

SEQUENCE WITH UNIT DIFFERENCE

FENG QI

Abstract. For any nonnegative integer k and natural numbers n and m, we have the following inequalities on the ratio for the geometric means of a positive arithmetic sequence with unit difference:

\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}} ,

where α ∈ [0, 1] is a constant. The equality above is valid for n = 1 and m = 1. Moreover, some monotonicity results for the sequences involving \sqrt[n]{∏_{i=k+1}^{n+k}(i + α)} are obtained, and the related inequalities are generalized.

1. Introduction

It is known that, for n ∈ ℕ, the following inequalities were given in [6]:

\frac{n}{n + 1} < \begin{matrix}^{n} & \sqrt{n!} \\ ^{n+1} & \sqrt{(n + 1)!} \end{matrix} < 1.  (1)

In [1], the left inequality in (1) was refined by

\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)

for all positive real numbers r. Both bounds are best possible.

There is a rich literature on refinements, extensions, and generalizations of the inequalities in (2), for examples, [2, 7, 8, 10, 14] and references therein.

2000 Mathematics Subject Classification. Primary 26D15; Secondary 26A48. Key words and phrases. Arithmetic sequence, geometric mean, inequality, monotonicity, ratio. The author was supported in part by NNSF (#10001016) of China, SF for the Prominent

Youth of Henan Province (#0112000200), SF of Henan Innovation Talents at Universities, China. This paper was typeset using AMS-L^{A}TEX.

Using analytic method and Stirling’s formula, in [9, 4, 12, 13], for n, m ∈ ℕ and k being a nonnegative integer, the author and others proved the following

\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)

the equality in (3) holds for n = 1 and m = 1. These extend and refine inequalities in (1).

Meanwhile, an inequality involving the ratio of gamma function was obtained in [4, 12]: 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)}} .  (4)

In this paper, using the ideas and method in [3, 5, 11] and the mathematical induction, we further generalize the inequalities in (3) and obtain the following inequalities of the ratio for the geometric means of a positive arithmetic sequence with unit difference and monotonicity results. Theorem 1. Let k be a nonnegative integer, n and m positive integers, and α ∈ [0, 1] a constant. Then

\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}} . (5)

If n = 1 and m = 1, then equality in the right hand side inequality of (5) hold. Theorem 2. The sequence

\frac{\sqrt[n]{∏_{i=k+1}^{n+k}(i + α)}}{n + k + 1 + α}  (6)

is strictly decreasing with n ∈ ℕ, and strictly increasing with nonnegative integer k and α ∈ [0, 1].

The sequence

\begin{matrix}^{n} & \sqrt{∏_{i=k+1}^{n+k}(i + α)} \\ √ & \overline{n + k + α} \end{matrix}  (7)

is strictly increasing with n ∈ ℕ, nonnegative integer k, and α ∈ [0, 1]. The sequence

\begin{matrix}[∏_{i=k+1}^{n+k}(i + α)] & 1/n \\ [∏_{i=k+1}^{n+m+k}(i + α)] & 1/(n+m) \end{matrix}  (8)

INEQUALITIES AND MONOTONICITY OF THE RATIO FOR THE GEOMETRIC MEANS 3

is strictly increasing with nonnegative integer k and α ∈ [0, 1] for fixed numbers n, m ∈ ℕ.

2. Proofs of Theorem 1 and Theorem 2

The left hand side inequality of (5) can be rearranged as

\frac{[∏_{i=k+1}^{n+k}(i + α)]^{\frac{1}{n}}}{n + k + 1 + α} ≥ \frac{[∏_{i=k+1}^{n+m+k}(i + α)]^{\frac{1}{n+m}}}{n + m + k + 1 + α} ,
[ ∏_{i=k+1}^{n+k} \frac{i + α}{n + k + 1 + α}]^{\frac{1}{n}} ≥ [ ∏_{i=k+1}^{n+m+k} \frac{i + α}{n + m + k + 1 + α}]^{\frac{1}{n+m}} ,
\frac{1}{n} ∑_{i=k+1}^{n+k} ln \frac{i + α}{n + k + 1 + α} ≥ \frac{1}{n + m} ∑_{i=k+1}^{n+m+k} ln \frac{i + α}{n + m + k + 1 + α},

which is equivalent to

\frac{1}{n} ∑_{i=k+1}^{n+k} ln \frac{i + α}{n + k + 1 + α} ≥ \frac{1}{n + 1} ∑_{i=k+1}^{n+k+1} ln \frac{i + α}{n + k + 2 + α}.  (9)

The same argument gives us that the right hand side inequality of (5) is equivalent to

\frac{1}{n} ∑_{i=k+1}^{n+k} ln \frac{i + α}{\sqrt{n + k + α}} ≤ \frac{1}{n + 1} ∑_{i=k+1}^{n+k+1} ln \frac{i + α}{\sqrt{n + k + 1 + α}}.  (10)

Proof of the left hand side inequality in (5). Since ln x is increasing and concave on (0, 1], from definition of the concave function and monotonicity of the function ln x, it follows that

\frac{i − k}{n + 1} ln \frac{i + 1 + α}{n + k + 2 + α} + \frac{n + k − i + 1}{n + 1} ln \frac{i + α}{n + k + 2 + α}

≤ ln (\frac{i − k}{n + 1} · \frac{i + 1 + α}{n + k + 2 + α} + \frac{n + k − i + 1}{n + 1} · \frac{i + α}{n + k + 2 + α})

≤ ln \frac{i + α}{n + k + 1 + α}.

(11)

Summing up on both sides of (11) leads to

∑_{i=k+1}^{n+k} [\frac{i − k}{n + 1} ln \frac{i + 1 + α}{n + k + 2 + α} + \frac{n + k − i + 1}{n + 1} ln \frac{i + α}{n + k + 2 + α}]

= \frac{n}{n + 1} ∑_{i=k+1}^{n+k} ln \frac{i + α}{n + k + 2 + α} + \frac{n}{n + 1} ln \frac{n + k + 1 + α}{n + k + 2 + α}

≤ ∑_{i=k+1}^{n+k} ln \frac{i + α}{n + k + 1 + α},

which is equivalent to

\frac{n}{n + 1} ∑_{i=k+1}^{n+k+1} ln \frac{i + α}{n + k + 2 + α} ≤ ∑_{i=k+1}^{n+k} ln \frac{i + α}{n + k + 1 + α}.

The proof of the left hand side of inequality (5) is complete. □

Proof of the right hand side inequality in (5). By standard argument, the inequality (10) can be rearranged as

\frac{1}{n} ∑_{i=k+1}^{n+k} ln(i + α) ≤ \frac{n + 1}{2} ln(n + k + α) − \frac{n − 1}{2} ln(n + k + 1 + α).  (12)

If n = 1, inequality (12) holds with equality.

If n = 2, it is easy to verify by a standard argument that inequality (12) is valid. Assume that inequality (12) holds for some positive integer n ∈ ℕ. Straightfor-

ward computation yields

\frac{1}{n + 1} ∑_{k+1}^{n+k+1} ln(i + α)

= \frac{n}{n + 1} [\frac{1}{n} ∑_{k+1}^{n+k} ln(i + α) + \frac{1}{n} ln(n + k + 1 + α)]

≤ \frac{n}{n + 1} [\frac{n + 1}{2} ln(n + k + α) − \frac{n − 1}{2} ln(n + k + 1 + α)]

+ \frac{ln(n + k + 1 + α)}{n + 1}

= \frac{n}{2} ln(n + k + α) − \frac{n − 2}{2} ln(n + k + 1 + α).

INEQUALITIES AND MONOTONICITY OF THE RATIO FOR THE GEOMETRIC MEANS 5

Therefore, by induction on n, to prove inequality (12), it is sufficient to verify the following

n ln(n + k + α) − (n − 2) ln(n + k + 1 + α) ≤ (n + 1) ln(n + k + 1 + α) − (n − 1) ln(n + k + 2 + α),

which is equivalent to

n[ln(n + k + α) − ln(n + k + 1 + α)] ≤ (n − 1)[ln(n + k + 1 + α) − ln(n + k + 2 + α)],

which can be further rewritten as

\frac{n}{n − 1} ≥ \frac{ln(n + k + α) − ln(n + k + 1 + α)}{ln(n + k + 1 + α) − ln(n + k + 2 + α)}.  (13)

Using Cauchy’s mean-value theorem for derivative, it follows that, there exists a number ξ ∈ (0, 1) satisfying

\frac{ln(n + k + α) − ln(n + k + 1 + α)}{ln(n + k + 1 + α) − ln(n + k + 2 + α)} = \frac{n + k + 1 + α + ξ}{n + k + α + ξ} .

Since k + 1 + α + ξ > 1 and n ≥ 2 and the function \frac{x}{x−1} is decreasing for x > 1, we have

\frac{n + k + 1 + α + ξ}{n + k + α + ξ} ≤ \frac{n}{n − 1},

which implies the inequality (13) holds. The right hand side of inequality (5) follows. □

Proof of Theorem 2. The monotonicities with n of the sequences (6) and (7) follow from Theorem 1 easily.

Direct calculating yields

\frac{\sqrt[n]{∏_{i=k+1}^{n+k}(i + α)}}{n + k + 1 + α} = \frac{n + k + 2 + α}{n + k + 1 + α} \sqrt[n]{\frac{k + 1 + α}{n + k + 1 + α}} · \frac{\sqrt[n]{∏_{i=k+2}^{n+k+1}(i + α)}}{n + k + 2 + α}.

Let

ψ(t) = ln(n + 2 + α + t) + \frac{1}{n} ln(1 + α + t) − (1 + \frac{1}{n}) ln(n + 1 + α + t)

for t ≥ 0. Direct computing and simplifying yields

ψ′(t) = \frac{n + 1}{(1 + α + t)(n + 1 + α + t)(n + 2 + α + t)} > 0,

and then ψ(t) is increasing. Therefore, exp(ψ(t)) is increasing, and

lim_{t→∞} exp(ψ(t)) = lim_{t→∞} (\frac{n + t + 2 + α}{n + t + 1 + α} \sqrt[n]{\frac{t + 1 + α}{n + t + 1 + α}}) = 1.

Thus

\frac{n + k + 2 + α}{n + k + 1 + α} \sqrt[n]{\frac{k + 1 + α}{n + k + 1 + α}} < 1.

The sequence (6) increases strictly with k.

The fact that

\frac{1}{n} ∑_{i=k+1}^{n+k} \frac{1}{i + α} − \frac{1}{n + k + 1 + α} > 0

implies the following sequence

\frac{1}{n} ∑_{i=k+1}^{n+k} ln(i + α) − ln(n + k + 1 + α)

is strictly increasing with α, and then the sequence (6) is also strictly increasing with α.

Easy computing yields

\begin{matrix}^{n} & \sqrt{∏_{i=k+1}^{n+k}(i + α)} \\ √ & \overline{n + k + α} \end{matrix} = \sqrt{\frac{n + k + 1 + α}{n + k + α}} \sqrt[n]{\frac{k + 1 + α}{n + k + 1 + α}} · \frac{\sqrt[n]{∏_{i=k+2}^{n+k+1}(i + α)}}{n + k + 1 + α}.

Let

ϕ(t) = \frac{1}{n} ln(1 + α + t) + (\frac{1}{2} − \frac{1}{n}) ln(n + 1 + α + t) − \frac{1}{2} ln(n + α + t)

for t ≥ 0. Computing directly and simplifying yields

ϕ′(t) = \frac{2n + t + α − 1}{2(1 + α + t)(n + α + t)(n + 1 + α + t)} > 0,

and then ϕ(t) is increasing. Therefore, exp(ϕ(t)) is increasing, and

lim_{t→∞} exp(ϕ(t)) = lim_{t→∞} (\sqrt{\frac{n + t + 1 + α}{n + t + α}} \sqrt[n]{\frac{t + 1 + α}{n + t + 1 + α}}) = 1.

Thus

\sqrt{\frac{n + k + 1 + α}{n + k + α}} \sqrt[n]{\frac{k + 1 + α}{n + k + 1 + α}} < 1.

The sequence (7) increases strictly with k.

The fact that

\frac{1}{n} ∑_{i=k+1}^{n+k} \frac{1}{i + α} − \frac{1}{2} · \frac{1}{n + k + α} > 0

INEQUALITIES AND MONOTONICITY OF THE RATIO FOR THE GEOMETRIC MEANS 7

implies the following sequence

\frac{1}{n} ∑_{i=k+1}^{n+k} ln(i + α) − \frac{1}{2} ln(n + k + α)

is strictly increasing, and then the sequence (7) is strictly increasing with α. Straightforward calculation gives us

\begin{matrix}[∏_{i=k+2}^{n+k+1}(i + α)] & \frac{1}{n} \\ [∏_{i=k+2}^{n+m+k+1}(i + α)] & \frac{1}{n+m} \end{matrix}

= [\frac{(n + k + 1 + α)^{n+m}}{(n + m + k + 1 + α)^{n}(k + 1 + α)^{m}} ]^{\frac{1}{n(n+m)}} \begin{matrix}[∏_{i=k+1}^{n+k}(i + α)] & \frac{1}{n} \\ [∏_{i=k+1}^{n+m+k}(i + α)] & \frac{1}{n+m} \end{matrix} .

Let

τ(t) = (m + t) ln(k + 1 + α + t) − t ln(m + k + 1 + α + t) − m ln(k + 1 + α)

for t ≥ 0. Then

τ(0) = 0,
τ′(t) = \frac{m + t}{k + α + 1 + t} − \frac{t}{k + m + α + 1 + t} − ln \frac{k + m + α + 1 + t}{k + α + 1 + t}

≥ \frac{m + t}{k + α + 1 + t} − \frac{t}{k + m + α + 1 + t} − \frac{m}{k + α + 1 + t}

≥ 0,

and τ(t) is increasing and nonnegative, which implies

\frac{(n + k + 1 + α)^{n+m}}{(n + m + k + 1 + α)^{n}(k + 1 + α)^{m}} > 1.

Therefore, the sequence (8) is strictly increasing with k.

The inequality

\frac{1}{n} ∑_{i=k+1}^{n+k} \frac{1}{i + α} > \frac{1}{n + m} ∑_{i=k+1}^{n+m+k} \frac{1}{i + α},  (14)

which is equivalent to

∑_{i=k+1}^{n+k} \frac{1}{i + α} > \frac{n}{n + k + 1 + α}

being valid clearly, implies that the sequence (8) is strictly increasing with α. □

3. Open problem

It is clear that it is natural to pose the following open problem.

Open Problem. For all nonnegative integers 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)] & 1/n \\ [∏_{i=k+1}^{n+m+k}(ai + b)] & 1/(n+m) \end{matrix} ≤ \sqrt{\frac{a(n + k) + b}{a(n + m + k) + b}} , (15)

where a and b are positive constants.

We will discuss this open problem in a subsequent paper.

Acknowledgements. This paper was done while the author’s visit to the RGMIA between November 1, 2001 and January 31, 2002, as a Visiting Professor with grants from the Victoria University of Technology and Jiaozuo Institute of Technology.

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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