Journal of Inequalities in Pure and
Applied Mathematics
NOTE ON THE CARLEMAN’S INEQUALITY FOR A NEGATIVE
POWER NUMBER
volume 4, issue 1, article 2,
2003.
THANH LONG NGUYEN, VU DUY LINH NGUYEN AND
THI THU VAN NGUYEN
Department of Mathematics and Computer Science,
College of Natural Science,
Vietnam National University HoChiMinh City,
227 Nguyen Van Cu Str.,
Dist.5, HoChiMinh City, Vietnam.
EMail: longnt@hcmc.netnam.vn
Received 30 October, 2002;
accepted 25 November, 2002.
Communicated by: H. Bor
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
113-02
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Abstract
By the method of indeterminate coefficients we prove the inequality
∞
X
1
n=1 a1
+ a12
∞
X
≤2
n=1
n
+ ... + a1n
!
∞
X
1
1
2
− 4n
an ,
1−
3n + 1
k(k + 1)2 (3k + 1)(3k + 4)
where an > 0, n = 1, 2, ...
Note on the Carleman’s
Inequality for a Negative Power
Number
k=n
P∞
n=1 an
< ∞.
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
2000 Mathematics Subject Classification: 26D15.
Key words: Carleman’s inequality.
Title Page
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
Contents
3
6
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1.
Introduction
The following Carleman inequality is well known (see [1, Chapter 9.12]).
 1
p
1
1
X
∞
∞
p
p
p
X
a1 + a2 + · · · + an 
p

(1.1)
≤
an ,
n
p − 1 n=1
n=1
P
where an ≥ 0, n = 1, 2, . . . , ∞
n=1 an < ∞, and p > 1.
Letting p → +∞, it follows from (1.1) that
(1.2)
∞
X
1
(a1 a2 · · · an ) n ≤ e
n=1
∞
X
an .
n=1
Note on the Carleman’s
Inequality for a Negative Power
Number
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
In practice, the inequality (1.2) is strict; i.e.,
Title Page
(1.3)
∞
X
1
(a1 a2 · · · an ) n < e
n=1
∞
X
an ,
n=1
P∞
if an ≥ 0, n = 1, 2, . . . , 0 < n=1 an < ∞.
The constant e is sharp in the sense that it cannot be replaced by a smaller
one.
Recently, the inequality (1.3) has also been improved by many authors, for
example: Yang Bicheng and L. Debnath [2] with
∞
∞ X
X
1
1
(1.4)
(a1 a2 · · · an ) n < e
1−
an ,
2n + 2
n=1
n=1
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in [3] by Yan Ping and Sun Guozheng with
∞
∞ X
X
1
n
(1.5)
(a1 a2 . . . an ) < e
1+
n=1
n=1
1
n + 1/5
−1/2
an ,
and in [4] by X. Yang with
∞
X
(1.6)
1
(a1 a2 . . . an ) n
n=1
<e
∞ X
n=1
1
1
1
1−
−
−
2
2(n + 1) 24(n + 1)
48(n + 1)3
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
and
∞
X
(1.7)
an ,
Note on the Carleman’s
Inequality for a Negative Power
Number
1
n
(a1 a2 . . . an ) < e
n=1
∞
X
1−
n=1
6
X
k=1
bk
(n + 1)k
!
an ,
where
1
1
1
73
11
1945
b1 = , b 2 = , b 3 = , b 4 =
, b5 =
, b6 =
,
2
24
48
5760
1280
580608
∞
X
an ≥ 0, n = 1, 2, . . . , 0 <
an < ∞.
n=1
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1
p
We rewrite the inequality (1.1) with r = as follows
1/r
∞
∞ r
X
X
a1 + ar2 + · · · + arn
−1/r
(1.8)
≤ (1 − r)
an ,
n
n=1
n=1
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P
where an ≥ 0, n = 1, 2, . . . , ∞
n=1 an < ∞ and 0 < r < 1.
In [5], we have improved Carleman’s inequality (1.8) for a negative power
number r < 0 as follows
(1.9)
1/r
∞ r
X
a + ar + · · · + ar
1
n=1
2
n
n
≤

∞
P

(1 − r)−1/r
an , if − 1 ≤ r < 1, r 6= 0,



n=1




r
2(r−1)/r
r−1
∞
P
an , if r < −1,
n=1
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
P∞
where an > 0, n = 1, 2, . . . , n=1 an < ∞.
In the case of r = −1, we obtain from (1.9) the inequality
(1.10)
∞
X
1
n=1 a1
+
where an > 0, n = 1, 2, . . . ,
1
a2
P∞
n
+ ··· +
n=1
1
an
an < ∞.
≤2
Note on the Carleman’s
Inequality for a Negative Power
Number
∞
X
n=1
an ,
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2.
Main Result
In this paper, we shall prove the following theorem.
P
Theorem 2.1. Let an > 0, n = 1, 2, . . . , and ∞
n=1 an < ∞. Then we have
(2.1)
∞
X
1
n=1 a1
+
1
a2
n
+ ··· +
1
an
∞
X
1
1
2
1−
− 4n
2
3n + 1
k(k + 1) (3k + 1)(3k + 4)
k=n
∞
X
!
an .
Note on the Carleman’s
Inequality for a Negative Power
Number
Remark 2.1. From the inequality (2.1), we obtain the following inequalities:
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
≤2
n=1
(2.2)
∞
X
1
n=1 a1
+
1
a2
n
+ ··· +
<2
∞ X
n=1
(2.3)
∞
X
1
n=1 a1
+
1
an
1
a2
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1
4n
1−
−
2
3n + 1 (n + 1) (3n + 1)(3n + 4)
n
+ ··· +
1
an
<2
∞ X
n=1
1
1−
3n + 1
and
(2.4)
an ,
Contents
an ,
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∞
X
1
n=1 a1
+
1
a2
n
+ ··· +
1
an
<2
∞
X
n=1
Page 6 of 14
an .
J. Ineq. Pure and Appl. Math. 4(1) Art. 2, 2003
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Indeed, we note that the inequalities (2.2), (2.3), (2.4) are implied from (2.1),
because
(2.5)
∞
X
1
1
2
1−
− 4n
2
3n + 1
k(k + 1) (3k + 1)(3k + 4)
k=n
1
4n
−
2
3n + 1 (n + 1) (3n + 1)(3n + 4)
1
<1−
< 1.
3n + 1
<1−
Note on the Carleman’s
Inequality for a Negative Power
Number
Hence, we obtain from (2.1), (2.5) that
(2.6)
∞
X
1
n=1 a1
+
≤2
1
a2
n
+ ··· +
∞
X
n=1
∞ X
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
1
an
∞
X
1
1
1−
− 4n2
2
3n + 1
k(k + 1) (3k + 1)(3k + 4)
k=n
1
4n
=2
1−
−
2
3n + 1 (n + 1) (3n + 1)(3n + 4)
n=1
!
∞
X
1
− 4n2
an
2 (3k + 1)(3k + 4)
k(k
+
1)
k=n+1
∞ X
4n
1
=2
1−
−
an
2 (3n + 1)(3n + 4)
3n
+
1
(n
+
1)
n=1
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!
an
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−8
∞
X
n=1
∞
X
∞
X
1
2
k(k + 1) (3k + 1)(3k + 4)
k=n+1
!
n 2 an
1
4n
<2
1−
−
2
3n + 1 (n + 1) (3n + 1)(3n + 4)
n=1
∞
X
1
an
< 2
1−
3n
+
1
n=1
<2
∞
X
an
Note on the Carleman’s
Inequality for a Negative Power
Number
an .
n=1
To prove Theorem 2.1, we first prove the following lemma.
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
Lemma 2.2. We have
(2.7)
1
a1
+
1
a2
1
+ ··· +
1
an
a1 b21 + a2 b22 + · · · + an b2n
,
≤
(b1 + b2 + · · · + bn )2
where ak > 0, bk > 0, ∀k = 1, 2, . . . , n.
Proof. This is a simple application of the Cauchy-Schwartz inequality
(2.8)
2
(b1 + b2 + · · · + bn )
2
1 √
1 √
1 √
an b n
a2 b 2 + · · · + √
a1 b 1 + √
≤ √
an
a2
a1
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≤
1
1
1
+
+ ··· +
a1 a2
an
a1 b21 + a2 b22 + · · · + an b2n .
Proof of Theorem 2.1. We prove the theorem by the method of indeterminate
coefficients.
Consider b1 , b2 , . . . to be the positive indeterminate coefficients. Let N =
1, 2, . . . Put
(2.9)
Note on the Carleman’s
Inequality for a Negative Power
Number
N
X
nb2k
Ck =
2 , 1 ≤ k ≤ N.
(b
+
b
+
·
·
·
+
b
)
1
2
n
n=k
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
Applying Lemma 2.2, we obtain
(2.10)
N
X
1
n=1 a1
+
1
a2
n
+ ··· +
1
an
≤
=
=
N X
n
X
n=1 k=1
N X
N
X
nak b2k
(b1 + b2 + · · · + bn )2
nak b2k
2
k=1 n=k
N
X
(b1 + b2 + · · · + bn )
Ck ak .
k=1
Choosing bk = k, k = 1, 2, . . . , we have from (2.9) that
(2.11)
Ck =
N
X
n=k
N
X
1
nk 2
2
.
2 = 4k
(1 + 2 + · · · + n)
n (n + 1)2
n=k
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On the other hand, we have the equality
(2.12)
2n2
1
1
1
−
−
2
2
2
+ 3 n 2(n + 1) + 3 (n + 1) n (n + 1)2
2
> 0, for all n = 1, 2, . . .
=
2
n (n + 1) (3n + 1)(3n + 4)
Hence, it follows from (2.11) that
(2.13)
N
X
n=k
1
n (n + 1)2
1
1
= 2 2 −
2
2k + 3 k 2(N + 1) + 23 (N + 1)
−
N
X
2
n (n + 1) (3n + 1)(3n + 4)
2
n=k
N
X
1
2
≤ 2 2 −
, 1 ≤ k ≤ N.
2
2k + 3 k n=k n (n + 1) (3n + 1)(3n + 4)
Hence, we obtain from (2.10), (2.13) that
(2.14)
Ck = 4k 2
N
X
n=k
≤2−
1
n (n + 1)2
Note on the Carleman’s
Inequality for a Negative Power
Number
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
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N
X
2
1
− 8k 2
.
2
3k + 1
n
(n
+
1)
(3n
+
1)(3n
+
4)
n=k
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(2.15)
N
X
1
n=1 a1
≤
+
N
X
1
a2
n
+ ··· +
1
an
Ck ak
k=1
N
X
N
X
2
1
≤
2−
− 8k 2
2
3k + 1
n (n + 1) (3n + 1)(3n + 4)
k=1
n=k
N
X
2
≤
2−
ak
3k + 1
k=1
!
N
N
X
X
1
−8
k 2 ak
2
n
(n
+
1)
(3n
+
1)(3n
+
4)
k=1
n=k
N
X
2
=
2−
ak
3k + 1
k=1
!
N
n
X
X
k 2 ak
−8
n (n + 1)2 (3n + 1)(3n + 4)
n=1
k=1
N N
X
X
2
=
2−
ak − 8
βn .
3k
+
1
n=1
k=1
ak
Note on the Carleman’s
Inequality for a Negative Power
Number
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
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where
Pn
(2.16)
!
βn =
k 2 ak
.
n (n + 1)2 (3n + 1)(3n + 4)
Page 11 of 14
k=1
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We have
0 < βn
(2.17)
Pn
k 2 ak
n (n + 1)2 (3n + 1)(3n + 4)
P
n2 nk=1 ak
≤
9n5
n
∞
1 X
1 X
= 3
ak ∼ 3
ak , as n → +∞.
9n k=1
9n k=1
k=1
=
Hence, the series
(2.18)
P∞
βn converges. Letting N → +∞ in (2.15), we obtain
n=1
∞
X
Note on the Carleman’s
Inequality for a Negative Power
Number
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
n
1
n=1 a1
+ a12 +
∞ X
≤
=
k=1
∞ X
k=1
··· +
2
2−
3k + 1
2
2−
3k + 1
−8
=
k=1
2−
ak − 8
2
3k + 1
∞
X
Contents
βn
JJ
J
n=1
ak
∞
n
X
X
n=1
∞ X
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1
an
k=1
k 2 ak
n (n + 1)2 (3n + 1)(3n + 4)
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ak
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−8
∞
∞
X
X
k=1
=2
∞
X
k=1
1−
n=k
1
2
n (n + 1) (3n + 1)(3n + 4)
!
k 2 ak
1
3k + 1
−4k 2
∞
X
n=k
1
2
n (n + 1) (3n + 1)(3n + 4)
The proof of Theorem 2.1 is complete.
!
ak .
Note on the Carleman’s
Inequality for a Negative Power
Number
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
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References
[1] G.H. HARDY, J.E. LITTLEWOOD
bridge Univ. Press, London,1952.
AND
G. POLYA, Inequalities, Cam-
[2] BICHENG YANG AND L. DEBNATH, Some inequalities involving the
constant e and an application to Carleman’s inequality, J. Math. Anal. Appl.,
223 (1998), 347–353.
[3] YAN PING AND GUOZHENG SUN, A strengthened Carleman’s inequality, J. Math. Anal. Appl., 240 (1999), 290–293.
Note on the Carleman’s
Inequality for a Negative Power
Number
[4] XIAOJING YANG, On Carleman’s inequality, J. Math. Anal. Appl., 253
(2001), 691–694.
Thanh Long Nguyen, Vu Duy Linh
Nguyen and Thi Thu Van Nguyen
[5] THANH LONG NGUYEN AND VU DUY LINH NGUYEN, The Carleman’s inequality for a negative power number, J. Math. Anal. Appl., 259
(2001), 219–225.
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