Then van der Corput’s inequality [5] states that (1.1

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http://jipam.vu.edu.au/

Volume 7, Issue 4, Article 127, 2006

A REFINEMENT OF VAN DER CORPUT’S INEQUALITY

DA-WEI NIU, JIAN CAO, AND FENG QI SCHOOL OFMATHEMATICS ANDINFORMATICS

HENANPOLYTECHNICUNIVERSITY

JIAOZUOCITY, HENANPROVINCE

454010, CHINA

nnddww@hotmail.com 21caojian@163.com

RESEARCHINSTITUTE OFMATHEMATICALINEQUALITYTHEORY

HENANPOLYTECHNICUNIVERSITY

JIAOZUOCITY, HENANPROVINCE

454010, CHINA

qifeng@hpu.edu.cn

URL:http://rgmia.vu.edu.au/qi.html Received 08 May, 2006; accepted 10 August, 2006

Communicated by B. Yang

ABSTRACT. In this note, a refinement of van der Corput’s inequality is given.

Key words and phrases: Refinement, van der Corput’s inequality, Harmonic number.

2000 Mathematics Subject Classification. Primary 26D15.

1. INTRODUCTION

Leta_n ≥0forn∈Nsuch that0<P∞

n=1(n+ 1)a_n <∞, andS_n=Pn m=1

1

m, the harmonic number. Then van der Corput’s inequality [5] states that

(1.1)

∞

X

n=1 n

Y

k=1

a^1/k_k

!_Sn¹

< e^1+γ

∞

X

n=1

(n+ 1)an,

whereγ = 0.57721566. . . stands for Euler-Mascheroni’s constant. The factore^1+γ in (1.1) is the best possible.

ISSN (electronic): 1443-5756 c

The authors were supported in part by the Science Foundation of the Project for Fostering Innovation Talents at Universities of Henan Province, China.

136-06

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In 2003, Hu in [3] gave a strengthened version of (1.1) by (1.2)

∞

X

n=1 n

Y

k=1

a^1/k_k

!_Sn¹

< e^1+γ

∞

X

n=1

n−lnn 4

a_n.

Recently, Yang in [7] obtained a better result than Hu’s inequality (1.2) as (1.3)

∞

X

n=1 n

Y

k=1

a^1/k_k

!_Sn¹

< e^1+γ

∞

X

n=1

n−lnn 3

a_n.

Moreover, he also extended (1.1) in [7] as follows (1.4)

∞

X

n=1 n

Y

k=1

a^1/(k+β)_k

!_Sn(β)¹

< e^1+γ(β)

∞

X

n=1

n+ 1

2+β

a_n, whereβ ∈(−1,∞),S_n(β) = Pn

k=1 1 k+β, and γ(β) = lim

n→∞

" _n X

k=1

1

k+β −ln(n+β)

# .

Applyingβ = 0in (1.4) leads to (1.5)

∞

X

n=1 n

Y

k=1

a^1/k_k

!_Sn¹

< e^1+γ

∞

X

n=1

n+ 1

2

a_n,

which improved inequality (1.1) clearly, but is not more accurate than (1.2) and (1.3).

In [1], among other things, the authors established a sharper inequality than (1.1), (1.2), (1.3) and (1.5) as follows

(1.6)

∞

X

n=1 n

Y

k=1

a^1/k_k

!_Sn¹

< e^1+γ

∞

X

n=1

n

1− lnn 3n−1/4

a_n.

The purpose of this note is to refine further inequality (1.6). Our main result is the following.

Theorem 1.1. Forn∈N, letS_n =Pn m=1

1

m, the harmonic number. Ifa_n ≥0forn ∈Nand 0<

∞

X

n=1

n

1− lnn 2n+ lnn+ 11/6

a_n <∞, then

(1.7)

∞

X

n=1 n

Y

k=1

a^1/k_k

!_Sn¹

< e^1+γ

∞

X

n=1

e⁻

6(6n+1)γ−9 (6n+1)(12n+11)n

1− lnn 2n+ lnn+ 11/6

a_n,

whereγ = 0.57721566. . . is Euler-Mascheroni’s constant.

Remark 1.2. Let

A_n=e^1+γe⁻

6(6n+1)γ−9 (6n+1)(12n+11)n

1− lnn 2n+ lnn+ 11/6

forn ∈ N. Numerical computation showsA₁ = 4.40. . . < e^1+γ

1−_3−1/4^{ln 1}

= 4.84. . . and A₂ = 7.99. . . < e^1+γ

2− _6−1/4^{2 ln 2}

= 8.51. . .. and A₃ = 11.95. . . < e^1+γ

3−_9−1/4^{3 ln 2}

=

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12.70. . .. Whenn ≥4, inequality2n+¹¹₆ + lnn <3n− ¹₄ is valid, which can be rearranged as

1− lnn

2n+ lnn+ 11/6 ≤1− lnn 3n−1/4. This implies that inequality (1.7) is a refinement of (1.6).

2. LEMMAS

In order to prove our main result, some lemmas are necessary.

Lemma 2.1 ([4]). Forn ∈N,

(2.1) 1

2n+ 1/(1−γ)−2 < S_n−lnn−γ < 1 2n+ 1/3. The constants _1−γ¹ −2and ¹₃ in (2.1) are the best possible.

Lemma 2.2 ([2, 6]). Ifx >0, then (2.2)

1 + 1

x x

< e

1− 1

2x+ 11/6

. Lemma 2.3. Forn ∈N,

(2.3) B_n ,

(n+ 1)S_n+1 nS_n

nSn

< e^1+γe⁻

6(6n+1)γ−9 (6n+1)(12n+11)n

1− lnn 2n+ lnn+ 11/6

.

Proof. By virtue of Lemma 2.2, it follows that

(n+ 1)S_n+1 nSn

_Sn+1^nSn

=

1 + S_n+ 1 nSn

_Sn+1^nSn

< e

1− S_n+ 1

2nS_n+ 11(S_n+ 1)/6

< e

1− 1

2n+ 11/6

.

Applying Lemma 2.1 yields Bn<

e

1− 1

2n+ 11/6

Sn+1

(2.4)

< e¹⁺^2n+1/3¹ ^+γ+lnⁿ

1− 1

2n+ 11/6

1+_2n+1/3¹ +γ+lnn

. Taking advantage of inequalities 1− ¹_x−x

> eforx >1ande^−x ≤ _1+x¹ forx >−1leads to (2.5)

1− 1

2n+ 11/6 lnn

≤exp

− lnn 2n+ 11/6

≤ 2n+ 11/6 2n+ 11/6 + lnn and

1− 1

2n+ 11/6

_2n+1/3¹ +1+γ

exp

1 2n+ 1/3

(2.6)

<exp

1

2n+ 1/3− 1 +γ

2n+ 11/6 − 1

(2n+ 11/6)(2n+ 1/3)

= exp

− 6(6n+ 1)γ−9 (6n+ 1)(12n+ 11)

.

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Combination of (2.4), (2.5), (2.6) gives

(2.7) B_n< e^1+γ−

6(6n+1)γ−9 (6n+1)(12n+11)

n− nlnn

2n+ lnn+ 11/6

.

Lemma 2.3 is proved.

3. PROOF OFTHEOREM1.1 Forn ∈Nand1≤k ≤n, let

(3.1) c_k = [(k+ 1)S_k+1]^kS^k

(kS_k)^kS^k−1 with assumptionS₀ = 0, then

(3.2)

n

Y

k=1

c^1/k_k

!⁻_Sn¹

= 1

(n+ 1)S_n+1.

By using the discrete weighted arithmetic-geometric mean inequality and interchanging the order of summations,

∞

X

n=1 n

Y

k=1

a^1/k_k

!_Sn¹

=

∞

X

n=1

" _n Y

k=1

(c_ka_k)^1/k

#_Sn¹ _n Y

k=1

c^1/k_k

!−_Sn¹

(3.3)

≤

∞

X

n=1 n

Y

k=1

c^1/k_k

!−_Sn¹

1 S_n

n

X

k=1

ckak

k

≤

∞

X

n=1

1 (n+ 1)S_n+1S_n

n

X

k=1

ckak

k

=

∞

X

k=1

ckak

k

∞

X

n=k

1 (n+ 1)S_nS_n+1

=

∞

X

k=1

ckak

k

∞

X

n=k

1

S_n − 1 S_n+1

=

∞

X

k=1

c_ka_k kS_k =

∞

X

k=1

(k+ 1)S_k+1 kS_k

kSk

a_k =

∞

X

n=1

B_na_n.

Substituting (2.7) into (3.3) leads to (1.7). The proof of Theorem 1.1 is complete.

REFERENCES

[1] J. CAO, D.-W. NIUANDF. QI, An extension and a refinement of van der Corput’s inequality, Int. J.

Math. Math. Sci., 2006 (2006), Article ID 70786, 10 pages.

[2] CH.-P. CHENANDF. QI, On further sharpening of Carleman’s inequality, Dàxué Shùxué (College Mathematics), 21(2) (2005), 88–90. (Chinese)

[3] K. HU, On van der Corput’s inequality, J. Math. (Wuhan), 23(1) (2003), 126–128. (Chinese) [4] F. QI, R.-Q. CUI, CH.-P. CHEN,ANDB.-N. GUO, Some completely monotonic functions involving

polygamma functions and an application, J. Math. Anal. Appl., 310(1) (2005), 303–308.

[5] J.G. VAN DER CORPUT, Generalization of Carleman’s inequality, Proc. Akad. Wet. Amsterdam, 39 (1936), 906–911.

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[6] Z.-T. XIEANDY.-B. ZHONG, A best approximation for constanteand an improvment to Hardy’s inequality, J. Math. Anal. Appl., 252 (2000), 994–998.

[7] B.-CH. YANG, On an extension and a refinement of van der Corput’s inequality, Chinese Quart. J.

Math., 22 (2007), in press.