10_xxx_xi.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 4, No. 1, 2011, 83-88

ISSN 1307-5543 – www.ejpam.com

On a Strengthened of the More Accurate Hilbert’s Inequality

Gaowen Xi

College of Mathematics and Physics, Chongqing University of Science and Technology, Chongqing,

401331, P. R. China

Abstract. By deducing the inequality of weight coefficient:

ω(n) =

∞
∑

m=0

1

m+ n+ 1
(

2n+ 1

2m+ 1
)

1

2 < π− 5

6(
p

2n+ 1+ 3

4

p

(2n+ 1)−1)
,

where n ∈ N . We obtain on a strengthened of the more accurate Hilbert’s inequality.

2000 Mathematics Subject Classifications: 26D15

Key Words and Phrases: Hilbert’s inequality, weight coefficient, Cauchy’s inequality, strengthen

1. Introduction

Let p > 1, 1

p
+ 1

q
= 1, an ≥ 0, bn ≥ 0, and 0<

∞
∑

n=1−λ
a

p
n <∞, 0<

∞
∑

n=1−λ
b

q
n <∞, (λ= 0,1),

then

∞
∑

n=1−λ

∞
∑

m=1−λ

am bn

m+ n+λ
< π{

∞
∑

n=1−λ
a2

n

∞
∑

n=1−λ
b2

n}
1

2 , (1)

∞
∑

n=1−λ

∞
∑

m=1−λ

ambn

m+ n+λ
<

π

sin(π
p
)
{
∞
∑

n=1−λ
ap

n}
1

p {
∞
∑

n=1−λ
bq

n}
1

q , (2)

where, constant π and π

sin(π
p
)

is best possible. (1) is Hilbert’s type inequality . for λ = 1, we

have

∞
∑

n=0

∞
∑

m=0

am bn

m+ n+ 1
< π{

∞
∑

n=0

a2
n

∞
∑

n=0

b2
n}

1

2 . (3)

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G. Xi / Eur. J. Pure Appl. Math, 4 (2011), 83-88 84

Inequality (3) is named of more accurate Hilbert’s inequality. Inequality (2) is Hardy-Hilbert’s.

For λ= 1, inequality (2) is named of more accurate Hardy-Hilbert’s inequality [1].

In [2], Yang obtained a strengthened of inequality (3):

∞
∑

n=0

∞
∑

m=0

ambn

m+ n+ 1
< {
∞
∑

n=0

[π− θ

(n+ 1)
1

2

]a2
n}

1

2 · {
∞
∑

n=0

[π− θ

(n+ 1)
1

2

]b2
n}

1

2 , (4)

where, θ = π−
∞
∑

m=0

1

(m+1)
3
2

= 0.5292496+.

In [3], by the following inequality of weight coefficient:

ω(n, r) =

∞
∑

m=0

1

m+ n+ 1
(

2n+ 1

2m+ 1
)

1

r <
π

sin(π
r
)
− θ

(2n+ 1)2−
1

r

,

where r > 1, n ∈ N , then

∞
∑

n=0

∞
∑

m=0

am bn

m+ n+ 1
< {

∞
∑

n=0

[
π

sin(π
p
)
− ln 2− C

(2n+ 1)
1+ 1

p

]ap
n}

1

p

·{
∞
∑

n=0

[
π

sin(π
p
)
− ln2− C

(2n+ 1)
1+ 1

q

]bq
n}

1

q ,

where C is Euler constant. In particular, for p = q = 2, Yang obtained again a strengthened of

inequality (3):

∞
∑

n=0

∞
∑

m=0

am bn

m+ n+ 1
< {

∞
∑

n=0

[π− ln2− C

(2n+ 1)
3

2

]a2
n}

1

2

·{
∞
∑

n=0

[π− ln2− C

(2n+ 1)
3

2

]b2
n}

1

2 . (5)

In this paper, by establishing the inequality of the weight coefficient, we will obtain a strength-

ened of inequalities (3), (4) and (5).

2. Some Lemmas

First of all, we give several lemmas which are to be used later.

Lemma 1. Let f (2r)(x) > 0, f (2r+1)(x) < 0, x ∈ [0, ∞), f (r)(∞) = 0 (r = 0, 1, 2, 3),
∫∞

0
f (x)d x <∞. Then

∞
∑

m=0

f (m)<

∫ ∞

0

f (x)d x +
1

2
f (0)− 1

12
f ′(0). (6)

Proof. See [4] or [5] .



G. Xi / Eur. J. Pure Appl. Math, 4 (2011), 83-88 85

Lemma 2. We have

ω(n) =

∞
∑

m=0

1

m+ n+ 1
(

2n+ 1

2m+ 1
)

1

2

< π− 1p
2n+ 1

[
5

6
+

1

6(2n+ 1)
− 2

(2n+ 1)2
], (7)

where n ∈ N.

Proof. Let fn(x) =
1

(x+n+1)
( 2n+1

2x+1
)

1

2 , x ∈ [0,∞), then

fn(0) =

p
2n+ 1

n+ 1

.

f ′n(x) =
p

2n+ 1[− 1

(x + n+ 1) · (2x + 1)
3

2

− 1

(x + n+ 1)2 · (2x + 1)
1

2

].

f ′n(0) =
p

2n+ 1[− 1

n+ 1
− 1

(n+ 1)2
].

∫ ∞

0

fn(x)d x =

∫ ∞

1

2n+1

1

(y + 1)y
1

2

d y

=

∫ ∞

0

1

(y + 1)y
1

2

d y −
∫

1

2n+1

0

1

(y + 1)y
1

2

d y

= π− 2

∫
1

2n+1

0

1

y + 1
d y

1

2

= π− 2[

p
2n+ 1

2(n+ 1)
+

2

3

∫
1

2n+1

0

1

(y + 1)2
d y

3

2 ]

= π− 2[

p
2n+ 1

2(n+ 1)
+

p
2n+ 1

6(n+ 1)2
+

4

3

∫
1

2n+1

0

1

(y + 1)3
d y

3

2 ]

< π− [
p

2n+ 1

(n+ 1)
+

p
2n+ 1

3(n+ 1)2
].

If ω(n) =
∞
∑

m=0

1

m+n+1
( 2n+1

2m+1
)

1

2 , so ω(n) =
∞
∑

m=0

fn(m). By lemma 1, we have

ω(n) =

∞
∑

m=0

fn(m)

<

∫ ∞

0

fn(x)d x +
1

2
fn(0)−

1

12
f ′n(0)



G. Xi / Eur. J. Pure Appl. Math, 4 (2011), 83-88 86

< π− [
p

2n+ 1

(n+ 1)
+

p
2n+ 1

3(n+ 1)2
] +

p
2n+ 1

2(n+ 1)
+

p
2n+ 1

12
[

1

n+ 1
+

1

(n+ 1)2
]

< π−
p

2n+ 1[
5

12(n+ 1)
+

1

4(n+ 1)2
]

< π− 1p
2n+ 1

[
5(2n+ 1)

12(n+ 1)
+

2n+ 1

4(n+ 1)2
].

For n ∈ N , we have

5(2n+ 1)

12(n+ 1)
+

2n+ 1

4(n+ 1)2
=

5

6
(1+

1

2n+ 1
)−1+

1

2n+ 1
(1+

1

2n+ 1
)−2

>
5

6
(1− 1

2n+ 1
) +

1

2n+ 1
(1− 2

2n+ 1
)

>
5

6
+

1

6(2n+ 1)
− 2

(2n+ 1)2
.

The proof of the lemma is completed.

Lemma 3. We have

ω(n) =

∞
∑

m=0

1

m+ n+ 1
(

2n+ 1

2m+ 1
)

1

2 < π− 5

6(
p

2n+ 1+ 3

4

p

(2n+ 1)−1)
, (8)

where n ∈ N.

Proof. Since

[
5

6
+

1

6(2n+ 1)
− 2

(2n+ 1)2
](1+

a

2n+ 1
)

=
5

6
+

1

2n+ 1
[

5a+ 1

6
− 12− a

6(2n+ 1)
− 2a

(2n+ 1)2
]

=
5

6
+

1

2n+ 1
· (5a+ 1)(2n+ 1)2− (12− a)(2n+ 1)− 12a

6(2n+ 1)2
.

For n= 1, a ≥ 3

4
, we have

(5a+ 1)(2n+ 1)2− (12− a)(2n+ 1)− 12a

6(2n+ 1)2
=

45a+ 9− 36+ 3a− 12a

54

=
36a− 27

150
≥ 0.

Then for n≥ 1, n ∈ N and a ≥ 3

4
,

[
5

6
+

1

6(2n+ 1)
− 2

(2n+ 1)2
](1+

a

2n+ 1
)>

5

6
.



G. Xi / Eur. J. Pure Appl. Math, 4 (2011), 83-88 87

For a = 3

4
, n= 0, θ = π−

∞
∑

m=0

1

(m+1)
3
2

= 0.5292496+, we have

θ

(n+ 1)
1

2

>
5

6(
p

2n+ 1+ 3

4

p

(2n+ 1)−1)
,

and

π− θ

(n+ 1)
1

2

< π− 5

6(
p

2n+ 1+ 3

4

p

(2n+ 1)−1)
.

The proof of the lemma is completed.

3. Main Results

Theorem 1. Let an ≥ 0, bn ≥ 0, and 0<
∞
∑

n=1

a2
n <∞, 0<

∞
∑

n=1

b2
n <∞, then

∞
∑

n=0

∞
∑

m=0

ambn

m+ n+ 1
< {

∞
∑

n=0

[π− 5

6(
p

2n+ 1+ 3

4

p

(2n+ 1)−1)
]a2

n

·
∞
∑

n=0

[π− 5

6(
p

2n+ 1+ 3

4

p

(2n+ 1)−1)
]b2

n}
1

2 , (9)

and

∞
∑

n=0

(

∞
∑

m=0

am

m+ n+ 1
)2 < π{

∞
∑

n=0

[π− 5

6(
p

2n+ 1+ 3

4

p

(2n+ 1)−1)
]a2

n. (10)

Proof. By Cauchy’s inequality, we have

∞
∑

n=0

∞
∑

m=0

ambn

m+ n+ 1
=

∞
∑

n=0

∞
∑

m=0

[
am

(m+ n+ 1)
1

2

(
2m+ 1

2n+ 1
)

1

4 ] · [ bn

(m+ n+ 1)
1

2

(
2n+ 1

2m+ 1
)

1

4 ]

≤ {
∞
∑

n=0

∞
∑

m=0

[
a2

m

m+ n+ 1
(
2m+ 1

2n+ 1
)

1

2 ] ·
∞
∑

n=0

∞
∑

m=0

[
b2

n

m+ n+ 1
(

2n+ 1

2m+ 1
)

1

2 ]} 1

2

= {
∞
∑

m=0

[

∞
∑

n=0

1

m+ n+ 1
(
2m+ 1

2n+ 1
)

1

2 ]a2
m ·
∞
∑

n=0

[

∞
∑

m=0

1

m+ n+ 1
(

2n+ 1

2m+ 1
)

1

2 ]b2
n}

1

2

= {
∞
∑

m=0

ω(m)a2
m

∞
∑

n=0

ω(n)b2
n}

1

2 .

By lemma 3, we have inequality (9).



REFERENCES 88

Let bn =
∞
∑

m=0

am

m+n+1
, then 0<

∞
∑

n=0

b2
n =

∞
∑

n=0

(
∞
∑

m=0

am

m+n+1
)2 <∞, so

(

∞
∑

n=0

b2
n)

2 = [

∞
∑

n=0

(

∞
∑

m=0

am

m+ n+ 1
)2]2

= (

∞
∑

n=0

∞
∑

m=0

ambn

m+ n+ 1
)2

<

∞
∑

n=0

[π− 5

6(
p

2n+ 1+ 3

4

p

(2n+ 1)−1)
]a2

n ·
∞
∑

n=0

[π− 5

6(
p

2n+ 1+ 3

4

p

(2n+ 1)−1)
]b2

n

< π

∞
∑

n=0

[π− 5

6(
p

2n+ 1+ 3

4

p

(2n+ 1)−1)
]a2

n ·
∞
∑

n=0

b2
n.

We have inequality (10). The proof of the theorem is completed.

Remark 1. Obviously, inequality (9) is a strengthened of inequality (3). Since, for n ∈ N,

5

6(
p

2n+ 1+ 3

4

p

(2n+ 1)−1)
>

θ

(n+ 1)
1

2

,

and
5

6(
p

2n+ 1+ 3

4

p

(2n+ 1)−1)
>

ln 2− C

3
p

(2n+ 1)2
.

Then inequality (9) is also a strengthened of inequality (4) and (5).

Acknowledgements This research is funded by Research Foundation of Chongqing Univer-

sity of Science and Technology, the project No. is CK2010B03.

References

[1] G. H. Hardy, J. E. Littlewood and G. Polya, Inequalities, Cambridge Univ. Press, 1952.

[2] B. Yang, A refinement of Hilbert’s inequality, Huanghuai Journal, 13.2: 47-51. 1997.

[3] B. Yang, On a strengthened version of the more accurate Hardy-Hilbert’s inequality, Acta

Mathematica Sinica, 42.6: 1103-1110. 1999.

[4] B. Yang and L. Debnath, On a New Generalization of Hardy-Hilbert’s Inequality and Its

Applications, Journal of Mathematical Analysis and Applications, 233, 484-497. 1999.

[5] J. C. Kuang and L. Debnath, On a new generalization of Hilbert’s inequality and their

applications, J. Math. Anal. Appl., 245: 248-265. 2000.