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
[email protected] [email protected]
RESEARCHINSTITUTE OFMATHEMATICALINEQUALITYTHEORY
HENANPOLYTECHNICUNIVERSITY
JIAOZUOCITY, HENANPROVINCE
454010, CHINA
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
Letan ≥0forn∈Nsuch that0<P∞
n=1(n+ 1)an <∞, andSn=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
a1/kk
!Sn1
< e1+γ
∞
X
n=1
(n+ 1)an,
whereγ = 0.57721566. . . stands for Euler-Mascheroni’s constant. The factore1+γ in (1.1) is the best possible.
ISSN (electronic): 1443-5756 c
2006 Victoria University. All rights reserved.
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
In 2003, Hu in [3] gave a strengthened version of (1.1) by (1.2)
∞
X
n=1 n
Y
k=1
a1/kk
!Sn1
< e1+γ
∞
X
n=1
n−lnn 4
an.
Recently, Yang in [7] obtained a better result than Hu’s inequality (1.2) as (1.3)
∞
X
n=1 n
Y
k=1
a1/kk
!Sn1
< e1+γ
∞
X
n=1
n−lnn 3
an.
Moreover, he also extended (1.1) in [7] as follows (1.4)
∞
X
n=1 n
Y
k=1
a1/(k+β)k
!Sn(β)1
< e1+γ(β)
∞
X
n=1
n+ 1
2+β
an, whereβ ∈(−1,∞),Sn(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
a1/kk
!Sn1
< e1+γ
∞
X
n=1
n+ 1
2
an,
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
a1/kk
!Sn1
< e1+γ
∞
X
n=1
n
1− lnn 3n−1/4
an.
The purpose of this note is to refine further inequality (1.6). Our main result is the following.
Theorem 1.1. Forn∈N, letSn =Pn m=1
1
m, the harmonic number. Ifan ≥0forn ∈Nand 0<
∞
X
n=1
n
1− lnn 2n+ lnn+ 11/6
an <∞, then
(1.7)
∞
X
n=1 n
Y
k=1
a1/kk
!Sn1
< e1+γ
∞
X
n=1
e−
6(6n+1)γ−9 (6n+1)(12n+11)n
1− lnn 2n+ lnn+ 11/6
an,
whereγ = 0.57721566. . . is Euler-Mascheroni’s constant.
Remark 1.2. Let
An=e1+γe−
6(6n+1)γ−9 (6n+1)(12n+11)n
1− lnn 2n+ lnn+ 11/6
forn ∈ N. Numerical computation showsA1 = 4.40. . . < e1+γ
1−3−1/4ln 1
= 4.84. . . and A2 = 7.99. . . < e1+γ
2− 6−1/42 ln 2
= 8.51. . .. and A3 = 11.95. . . < e1+γ
3−9−1/43 ln 2
=
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 < Sn−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) Bn ,
(n+ 1)Sn+1 nSn
nSn
< e1+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)Sn+1 nSn
Sn+1nSn
=
1 + Sn+ 1 nSn
Sn+1nSn
< e
1− Sn+ 1
2nSn+ 11(Sn+ 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− 1x−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)
.
Combination of (2.4), (2.5), (2.6) gives
(2.7) Bn< e1+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) ck = [(k+ 1)Sk+1]kSk
(kSk)kSk−1 with assumptionS0 = 0, then
(3.2)
n
Y
k=1
c1/kk
!−Sn1
= 1
(n+ 1)Sn+1.
By using the discrete weighted arithmetic-geometric mean inequality and interchanging the order of summations,
∞
X
n=1 n
Y
k=1
a1/kk
!Sn1
=
∞
X
n=1
" n Y
k=1
(ckak)1/k
#Sn1 n Y
k=1
c1/kk
!−Sn1
(3.3)
≤
∞
X
n=1 n
Y
k=1
c1/kk
!−Sn1
1 Sn
n
X
k=1
ckak
k
≤
∞
X
n=1
1 (n+ 1)Sn+1Sn
n
X
k=1
ckak
k
=
∞
X
k=1
ckak
k
∞
X
n=k
1 (n+ 1)SnSn+1
=
∞
X
k=1
ckak
k
∞
X
n=k
1
Sn − 1 Sn+1
=
∞
X
k=1
ckak kSk =
∞
X
k=1
(k+ 1)Sk+1 kSk
kSk
ak =
∞
X
n=1
Bnan.
Substituting (2.7) into (3.3) leads to (1.7). The proof of Theorem 1.1 is complete.
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