On Carleman's Inequality

Journal of Mathematical Analysis and Applications 253, 691᎐694 Ž2001. doi:10.1006rjmaa.2000.7155, available online at http:rrwww.idealibrary.com on

NOTE On Carleman’s Inequality Xiaojing Yang Department of Mathematics, Tsinghua Uni¨ ersity, Beijing 100084, People’s Republic of China Submitted by A. M. Fink Received May 4, 2000

In this paper the result of P. Yan and G. Sun Ž1999, J. Math. Anal. Appl. 240, 290᎐293. is generalized and a conjecture is proposed. 䊚 2001 Academic Press Key Words: Carleman’s inequality.

Recently, Yan and Sun w1x generalized the strengthened Carleman’s inequality obtained in w2x. The following Carleman’s inequality is well known w3x Let a n G 0, n s 1, 2, . . . , and 0 - Ý⬁ns1 a n - ⬁. Then

THEOREM A.

⬁

Ý Ž a1 a2

⭈⭈⭈ a n .

1rn

-e

ns1

⬁

Ý an .

Ž 1.

ns1

The main result of w1x is the following theorem THEOREM B w2x.

Let a n G 0, n s 1, 2, . . . , and 0 - Ý⬁ns1 a n - ⬁. Then

⬁

Ý Ž a1 a2

⭈⭈⭈ a n .

1rn

-e

ns1

⬁

Ý ns1

ž

1q

y1 r2

1 n q 1r5

/

an .

Ž 2.

The main result of this paper is the following theorem Let a n G 0, n s 1, 2, . . . , and 0 - Ý⬁ns1 a n - ⬁. Then

THEOREM 1. ⬁

Ý Ž a1 a2

⭈⭈⭈ a n .

1rn

ns1

-e

⬁

Ý ns1

ž

1y

1 2 Ž n q 1.

y

1 24 Ž n q 1 .

2

y

1 48 Ž n q 1 .

3

/

an . Ž 3.

691 0022-247Xr01 $35.00 Copyright 䊚 2001 by Academic Press All rights of reproduction in any form reserved.

692

NOTE

To prove Theorem 1, we need the following lemma. For x ) 0, we ha¨ e

LEMMA 1.

ž

1q

1 x

x

/

1

-e 1y

y

2Ž 1 q x .

1

y

2

24 Ž 1 q x .

1 48 Ž 1 q x .

3

.

Ž 4.

Proof. The right side of Ž4. can be written as e

48 y 3 y 24 y 2 y 2 y y 1

where y s x q 1 ) 1.

,

48 y 3

Define pŽ x . s

48 y 3 48 y 3 y 24 y 2 y 2 y y 1

ž

1q

x

1

/

x

y s x q 1.

,

We need to show pŽ x . - e for x ) 0. Let q Ž x . s Ž lnpŽ x .., r Ž y . s 48 y 3 y 24 y 2 y 2 y y 1. Then q Ž x . s Ž ln 1 q x . y lnx q 2ry y r ⬘Ž y .rr Ž y .. By Taylor’s formula ⬁

ln Ž 1 q x . y lnx s

1

Ý

kŽ1 q x.

ks1

k

s

⬁

Ý ks1

1 ky k

we have QŽ y . s q Ž x . s 2ry q Ý⬁ks 1Ž1rky k . y r ⬘rr; therefore Q⬘ Ž y . s y

s

2 y2

y

⬁

1

Ý

y kq1

ks1

y Ž 3 y y 2. y 2 Ž y y 1.

y

y

rr ⬙ y Ž r ⬘ .

rr ⬙ y Ž r ⬘ .

2

r2 2

r2

2 Ž 3 y y 2 . r 2 q y 2 Ž y y 1 . Ž rr ⬙ y Ž r ⬘ . . sy y 2 Ž y y 1. r 2

s

y584 y 3 q 152 y 2 y 5 y y 2 y 2 Ž y y 1. r 2

and Q⬘Ž y . - 0 for y ) 1. Hence Q Ž y . is decreasing on Ž1, ⬁.. As lim Q Ž y . s 0

yª⬁

we have QŽ y . ) 0 which implies p⬘Ž x . ) 0. It is easy to see lim p Ž x . s e.

xª⬁

Hence pŽ x . - e for x g Ž0, ⬁x.

693

NOTE

Proof of Theorem 1. Let c m s Ž1 q m1 . m m, m s 1, 2, . . . . Then by the arithmetic-geometric average inequality, we have ⬁

⭈⭈⭈ a n .

Ý Ž a1 a2

1rn

s

ns1

⬁

Ý ns1

s

c1 a1 c 2 a2 ⭈⭈⭈ c n a n

ž

c1 c 2 ⭈⭈⭈ c n

⬁

Ý Ž c1 c 2

⭈⭈⭈ c n .

y1 rn

⭈⭈⭈ c n .

y1 rn

1rn

/

1rn Ž c1 a1 ⭈⭈⭈ c n an .

ns1

F

⬁

Ý Ž c1 c 2

n

ns1

s

⬁

⬁

1

Ý ck ak Ý ks1

nsk

⬁

⬁

n

1

1

Ý ck ak Ý n Ž n q 1. ks1 nsk

s

Ý

1

ks1

s

k

⬁

Ý ks1

ž

ks1

y1 rn Ž c1 c2 ⭈⭈⭈ c n .

s

⬁

n

Ý ck ak

ck ak

1q

1 k

k

/

ak .

By Lemma 1, we have ⬁

Ý Ž a1 a2

⭈⭈⭈ a n .

1rn

ns1

-e

⬁

Ý ns1

ž

1y

1

y

2 Ž n q 1.

1 24 Ž n q 1 .

2

y

1 48 Ž n q 1 .

Remark 1. It is easy to verify that 0-1y -1y

1 2 Ž n q 1. 1 2 Ž n q 1.

- 1q

ž

1 n q 1r5

y y

1 24 Ž n q 1 .

2

y

1 48 Ž n q 1 .

1 24 Ž n q 1 .

2

y1 r2

/

,

for n s 1, 2, . . . .

Thus, Theorem 1 improves Theorem B and Theorem A.

3

/

an .

694

NOTE

Remark 2. Lemma 1 can be further refined as

ž

1q

1 x

x

/

-e 1y

1 2Ž 1 q x .

y

1 24 Ž 1 q x .

2

y

y

1 48 Ž 1 q x . 11

1280 Ž 1 q x .

y

3

y

5

73 5670 Ž 1 q x .

4

y1945 580608 Ž 1 q x .

6

.

Therefore, Theorem 1 can also be further refined. The proof of the above inequality is similar to the proof of Lemma 1, but it is considerably longer and complicated. Conjecture. Let x ) 0. Then the following equality holds

ž where a1 s 12 , a2 s

1q 1 24

1 x

x

/

, a3 s

se 1y

⬁

Ý ks1

1 48

, a4 s

73 5670

ak

Ž1 q x.

, a5 s

k

11 1280

,

, a6 s

1945 580608

.

Conjecture. Is a k ) 0 for k G 7? ACKNOWLEDGMENT The author thanks the referee for his valuable comments and suggestions.

REFERENCES 1. P. Yan and G. Sun, A strengthened Carleman’s inequality, J. Math. Anal. Appl. 240 Ž1999., 290᎐293. 2. B. Yang and L. Debnath, Some inequalities involving the constant e, and an application to Carleman’s inequality, J. Math. Anal. Appl. 223 Ž1998., 347᎐353. 3. G. H. Hardy, J. E. Littlewood, and G. Polya, ‘‘Inequalities,’’ Cambridge Univ. Press, London, 1952.