Journal of Inequalities in Pure and
Applied Mathematics
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
S CHOOL OF M ATHEMATICS AND I NFORMATICS
H ENAN P OLYTECHNIC U NIVERSITY
J IAOZUO C ITY, H ENAN P ROVINCE
454010, C HINA
nnddww@hotmail.com
21caojian@163.com
R ESEARCH I NSTITUTE OF M ATHEMATICAL I NEQUALITY T HEORY
H ENAN P OLYTECHNIC U NIVERSITY
J IAOZUO C ITY, H ENAN P ROVINCE
454010, C HINA
qifeng@hpu.edu.cn
URL: http://rgmia.vu.edu.au/qi.html
Received 08 May, 2006; accepted 10 August, 2006
Communicated by B. Yang
A BSTRACT. 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. I NTRODUCTION
P
Pn
Let an ≥ 0 for n ∈ N such that 0 < ∞
n=1 (n + 1)an < ∞, and Sn =
m=1
number. Then van der Corput’s inequality [5] states that
! S1
∞
n
∞
n
X
X
Y
1/k
1+γ
<e
(n + 1)an ,
(1.1)
ak
n=1
k=1
1
,
m
the harmonic
n=1
where γ = 0.57721566 . . . stands for Euler-Mascheroni’s constant. The factor e1+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
2
D.-W. N IU , J. C AO , AND F. Q I
In 2003, Hu in [3] gave a strengthened version of (1.1) by
! S1
n
∞
∞ n
Y
X
X
ln n
1/k
1+γ
ak
(1.2)
<e
n−
an .
4
n=1
n=1
k=1
Recently, Yang in [7] obtained a better result than Hu’s inequality (1.2) as
! S1
∞
n
∞ n
X
X
Y
ln n
1/k
1+γ
<e
n−
an .
(1.3)
ak
3
n=1
n=1
k=1
Moreover, he also extended (1.1) in [7] as follows
! S 1(β)
∞
n
∞ n
X
X
Y
1
1/(k+β)
1+γ(β)
<e
n + + β an ,
(1.4)
ak
2
n=1
n=1
k=1
P
1
where β ∈ (−1, ∞), Sn (β) = nk=1 k+β
, and
" n
#
X 1
γ(β) = lim
− ln(n + β) .
n→∞
k+β
k=1
Applying β = 0 in (1.4) leads to
∞
n
X
Y
(1.5)
n=1
! S1
n
1/k
ak
<e
1+γ
∞ X
n=1
k=1
1
n+
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
! S1
∞
n
∞
n
X
Y
X
ln n
1/k
1+γ
(1.6)
ak
an .
<e
n 1−
3n
−
1/4
n=1
n=1
k=1
The purpose of this note is to refine further inequality (1.6). Our main result is the following.
P
Theorem 1.1. For n ∈ N, let Sn = nm=1 m1 , the harmonic number. If an ≥ 0 for n ∈ N and
∞
X
ln n
0<
n 1−
an < ∞,
2n
+
ln
n
+
11/6
n=1
then
(1.7)
∞
n
X
Y
n=1
! S1
n
1/k
ak
<e
1+γ
∞
X
6(6n+1)γ−9
− (6n+1)(12n+11)
e
n=1
k=1
ln n
n 1−
2n + ln n + 11/6
an ,
where γ = 0.57721566 . . . is Euler-Mascheroni’s constant.
Remark 1.2. Let
ln n
n 1−
An = e e
2n + ln n + 11/6
ln 1
= 4.84 . . . and
for n ∈ N. Numerical computation shows A1 = 4.40 . . . < e1+γ 1 − 3−1/4
3 ln 2
2 ln 2
1+γ
1+γ
3 − 9−1/4 =
A2 = 7.99 . . . < e
2 − 6−1/4 = 8.51 . . . . and A3 = 11.95 . . . < e
6(6n+1)γ−9
1+γ − (6n+1)(12n+11)
J. Inequal. Pure and Appl. Math., 7(4) Art. 127, 2006
http://jipam.vu.edu.au/
A R EFINEMENT OF VAN DER C ORPUT ’ S I NEQUALITY
3
12.70 . . . . When n ≥ 4, inequality 2n + 11
+ ln n < 3n − 14 is valid, which can be rearranged
6
as
ln n
ln n
≤1−
.
1−
2n + ln n + 11/6
3n − 1/4
This implies that inequality (1.7) is a refinement of (1.6).
2. L EMMAS
In order to prove our main result, some lemmas are necessary.
Lemma 2.1 ([4]). For n ∈ N,
1
1
< Sn − ln n − γ <
.
2n + 1/(1 − γ) − 2
2n + 1/3
(2.1)
The constants
1
1−γ
− 2 and
1
3
in (2.1) are the best possible.
Lemma 2.2 ([2, 6]). If x > 0, then
x
1
1
(2.2)
1+
<e 1−
.
x
2x + 11/6
Lemma 2.3. For n ∈ N,
nSn
6(6n+1)γ−9
(n + 1)Sn+1
ln n
1+γ − (6n+1)(12n+11)
(2.3)
Bn ,
<e e
n 1−
.
nSn
2n + ln n + 11/6
Proof. By virtue of Lemma 2.2, it follows that
nSn
nSn
(n + 1)Sn+1 Sn +1
Sn + 1 Sn +1
= 1+
nSn
nSn
1
Sn + 1
<e 1−
.
<e 1−
2nSn + 11(Sn + 1)/6
2n + 11/6
Applying Lemma 2.1 yields
(2.4)
Bn < e 1 −
Sn +1
1
2n + 11/6
1
1+ 2n+1/3
+γ+ln n
1
1
1+ 2n+1/3
+γ+ln n
<e
1−
.
2n + 11/6
−x
1
Taking advantage of inequalities 1 − x1
> e for x > 1 and e−x ≤ 1+x
for x > −1 leads to
ln n
1
ln n
2n + 11/6
(2.5)
1−
≤ exp −
≤
2n + 11/6
2n + 11/6
2n + 11/6 + ln n
and
(2.6)
1
2n+1/3
+1+γ
1
exp
2n + 1/3
1
1+γ
1
< exp
−
−
2n + 1/3 2n + 11/6 (2n + 11/6)(2n + 1/3)
6(6n + 1)γ − 9
= exp −
.
(6n + 1)(12n + 11)
1
1−
2n + 11/6
J. Inequal. Pure and Appl. Math., 7(4) Art. 127, 2006
http://jipam.vu.edu.au/
4
D.-W. N IU , J. C AO , AND F. Q I
Combination of (2.4), (2.5), (2.6) gives
6(6n+1)γ−9
Bn < e
(2.7)
1+γ− (6n+1)(12n+11)
n ln n
n−
2n + ln n + 11/6
.
Lemma 2.3 is proved.
3. P ROOF OF T HEOREM 1.1
For n ∈ N and 1 ≤ k ≤ n, let
ck =
(3.1)
[(k + 1)Sk+1 ]kSk
(kSk )kSk−1
with assumption S0 = 0, then
n
Y
(3.2)
!− S1
n
1/k
ck
=
k=1
1
.
(n + 1)Sn+1
By using the discrete weighted arithmetic-geometric mean inequality and interchanging the
order of summations,
! S1
" n
# S1
!− S1
∞
n
∞
n
n
n
n
X
Y
X
Y
Y
1/k
1/k
ak
(3.3)
=
(ck ak )1/k
ck
n=1
n=1
k=1
≤
∞
n
X
Y
n=1
≤
∞
X
n=1
=
=
=
k=1
k=1
!− S1
n
1/k
ck
k=1
n
1 X c k ak
Sn k=1 k
n
X
1
c k ak
(n + 1)Sn+1 Sn k=1 k
∞
∞
X
c k ak X
k=1
∞
X
k=1
∞
X
k=1
k
c k ak
k
c k ak
kSk
1
(n + 1)Sn Sn+1
n=k
∞ X
1
1
−
Sn Sn+1
n=k
kS
∞ ∞
X
X
(k + 1)Sk+1 k
=
ak =
Bn an .
kSk
n=1
k=1
Substituting (2.7) into (3.3) leads to (1.7). The proof of Theorem 1.1 is complete.
R EFERENCES
[1] J. CAO, D.-W. NIU AND F. 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. CHEN AND F. 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, AND B.-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.
J. Inequal. Pure and Appl. Math., 7(4) Art. 127, 2006
http://jipam.vu.edu.au/
A R EFINEMENT OF VAN DER C ORPUT ’ S I NEQUALITY
5
[6] Z.-T. XIE AND Y.-B. ZHONG, A best approximation for constant e and 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.
J. Inequal. Pure and Appl. Math., 7(4) Art. 127, 2006
http://jipam.vu.edu.au/