IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 2 Ver. V (Mar-Apr. 2014), PP 17-22 www.iosrjournals.org

A Further Refinement of Van Der Corput’s Inequality Amusa I. S1. Mogbademu A2. A. Baiyeri J. Funso3 and Mohammed. M. A4 Ekakitie E.T5 Department Of Mathematics, University Of Lagos1 & 2, Department Of Mathematics, Yaba College Of Technology 3 & 4

Abstract: In this paper, we obtain a further refinement of Van der Corput’s inequality using an analytical technique.

Key words and phrases: Refinement, Van der Corput’s inequality, Harmonic number. I. Let Sn 

n

1

k

Introduction 

be the harmonic number and

an  0 for n  such that 0    n  1 an   . n 1

k 1

The Van der Corput’s inequality states that 1

 n 1/ k  Sn 1  1.1    ak   e   n  1 an , n 1  k 1 n 1  1 where   0.57721566... denotes the Euler-Mascheroni’s constant. The constant e is the best possible. 

In 2003, Hu [5], gave the following version of (1.1): 1

 n 1/ k  Sn 1   ln n     ak   e   n  4  an  n 1  k 1 n 1   

1.2

This inequality is a refinement of (1.1). In 2005, YANG [1] obtained a better result than Hu’s inequality (1.2) as 1

 n 1/ k  Sn 1   ln n     ak   e   n  3  an .  n 1  k 1 n 1   

1.3

This inequality is also a refinement of (1.1). He further extended the original Van der Corput’s inequality (1.1) as follows. Let

 n l 1 an  0 for n  such that 0    n     an   and Tn      .Then 2  n 1  k 1 k  

1

 n k 1 Tn    11      1  e   ak     n     an , 2  n 1  k 1 n 1   n n 1  1  ln  n     for    1,   and Tn  0   Sn   . where  1     lim  n  k 1 k  k 1 k    Setting   0 in (1.4), it becomes 

1

 n 1/ k  Sn 1   1     ak   e   n  2  an .  n 1  k 1 n 1   

1.4

1.5

Clearly, inequality (1.5) improves inequality (1.1) In 2006, Cao et al [3] established another version of (1.1) as follows.

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A Further Refinement of Van Der Corput’s Inequality 

Let

an  0 for n  N such that 0   an   . Then n 1

 n 1/    ak n 1  k 1 

where   0,   ,

1

k  k  

  ln  n  1  Un   11 /3     e n  1 a,   3 1      n n 1   4  n  1  2   n

Un     k 1

1.6

 n  n  and      lim U n     2ln . n  1  n    

1 k k   

Consequently, they established a sharper inequality that further refines (1.1), (1.2), (1.3) and (1.5), which is given by 1

 n ln n   n 1/ k  sn  1 a  e    k n  an .   3n  1/ 4  n 1  k 1 n 1   

1.7 

The aim of this paper is to further refine inequality (1.7) to obtain sharper inequality than that of (1.7). Our main results are the following.

II.

Main Results

n

1 Theorem 2.1: Let an  0 and S n   such that for n  k 1 k  n ln n   0   n   an  . Then 2n  ln n  11/ 6  n 1 

and

1

    6n  112n  11 n ln n  n 1/ k  sn  1 a  e    n     k  an ,    2n  ln n  11/ 6  n 1  k 1 n 1   6n  112n  11  6  6n  1   9    

(2.1) where   0.57721566... is the Euler-Mascheroni’s constant and e Remark 2.2. Let

1

is the best possible.

   6n  112n  11 n ln n  U n  e1    n   2n  ln n  11/ 6    6n  112n  11  6  6n  1   9   For n 

, the numerical computations of n 1 2 3 4 5

U n and Vn

Un 4.4230 8.0200 11.9770 16.1260 20.3990

n ln n   and Vn  e1  n  . 3n  1/ 4  

gives the following table of values:

Vn 4.8415 8.5157 12.7008 17.0810 21.5659

 Vn , for n  1. Also, we consider 2n  ln n  11/ 6 and 3n  1/ 4. Clearly U n

For n  4 : 2n  ln n  116  3n  14

 1

ln n ln n  1 . 11 2n  ln n  6 3n  14

Clearly, inequality (2. 1) is an improvement and the refinement of the (1.7). n

Theorem 2.3: Let

1 such that for n  k 1 k

an  0 and Sn  

and

 n ln n   0   n   an  . Then 2n  ln n  11/ 6  n 1 

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A Further Refinement of Van Der Corput’s Inequality 1

 n ln n  n 1/ k  sn   1 a  e    k n  an ,   2n  ln n  11/ 6  n 1  k 1 n 1   

(2.2)

  0.57721566... is the Euler-Mascheroni’s constant and e1 is the best possible. n ln n n ln n   1  1  Remark 2.4: Let Tn  e  n   and Vn  e  n  . 2n  ln n  11/ 6  3n  1/ 4    where

For n  4 : Numerical computations of Tn and Vn give the following table of values

n 1 2 3 4 5 6 7 8 9 10

Tn

Vn

4.8415 8.6545 12.7379 16.9730 21.3091 25.7177 30.1810 34.6870 39.2273 43.7958

4.8415 8.5157 12.7008 17.0810 21.5659 26.1164 30.7120 35.3405 39.9941 44.6674

Clearly, inequality (2. 2) is a refinement of (1.7) since

Tn  Vn .

In order to prove our main results, we consider the following lemmas. Lemma 2.5 [3]. For n  where the constant

,

1 2n 

1 1

2

 Sn  ln n   

1 , 2n  13

1 1  2 and are the best possible. 1  3

 1   1 Lemma 2.6 [3]. If x  0, then 1    e 1  11   x  2x  6  x

Lemma 2.7.

For n  N ,

  n  1 Sn  1 let An    nSn  

nSn

, then

   6n  112n  11 n ln n  An  e1    n  . 2n  ln n  11/ 6    6n  112n  11  6  6n  1   9   Proof of Lemma 2.7:

We have that

  n  1 Sn  1 An    nSn  

nSn

nSn   Sn 1 n  1 S  1     n        nS n     

Sn 1

.

(2.3) nSn

nSn

nSn

  n  1 Sn  1 Sn 1  n  1 1  Sn 1  Sn  1 Sn 1 Suppose we set Bn      1     nS n nS nSn  n n     Then applying Lemma 2.6 in (1.10), we get      Sn  1 Sn 1  Bn  e 1   e 1   e 1  .     11    2n  11  S  11    2n  11  S  2 n  n n 6 6 6 6       www.iosrjournals.org

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A Further Refinement of Van Der Corput’s Inequality

using Lemma 2.5, in view of (2.3), we obtain   1  An  e 1  11     2n  6   and Sn  ln n   

Sn 1

e

Sn 1

1 2n  13

 1  1  11   2n  6 

 Sn 

Sn 1

 2.4

.

1  ln n   . 2n  13

Hence, Equation  2.4  yields 1

1  ln n   1 2 n  13

 ln n   1

1  1  2n 3 An  e . 1  11   2n  6  To proceed from here, we consider the following inequalities:

m

 1 If m  1, then 1    e.  m

 2.5

 2.6

1 . 1 m m 1  Observe from  2.6  that 1    e1.  m

 2.7 

If m  1, then e m 

ln n

Thus,

 1  1  11   2n  6 

Using  2.7  in  2.8 ,  1  1  11   2n  6 

ln n

e

ln n  2 n  11  2 n  116    6  1    2 n  116   1  e    2n  11   6       we have

ln n

ln n 2 n  11 6

 1  Therefore, 1  11   2n  6  Similarly, 1 

 1  1  11   2n  6 

1 2 n  13

e

And, from  2.7  , 1 

 1  1  11   2n  6 

1 2 n  13

1

1

ln n

1 2 n  13

ln n 2 n 11/6

 1

e

6 6 n 1 9  6 n 112 n 11

Setting m 

e

1 2 n  13

ln n 2n  ln n  116

  n ln n eln n   n  . 11  2n  ln n  6  

 2.8

 2.9 

 2.10 

.

6  6n  1   9 , inequality  2.10  becomes  6n  112n  11

   6n  112n  11      6n  112n  11  6  6n  1   9 

 2.11

Combining inequalities (2.5), (2.9) and (2.11), we obtain

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A Further Refinement of Van Der Corput’s Inequality

An  e

1  ln n   1 2 n  13

1

1  1  2n 3 1   11   2n  6 

 ln n   1

1   1 ln n  1    2 n  13   1 1   2 n  13   1  ln n   e 1  e 1  e   .  11  2n  116    2n  6         6n  112n  11 n ln n  Thus, An  e1   .   n  11    6n  112n  11  6  6n  1   9   2n  ln n  6 

 2.12

Hence Lemma 2.3 is proved. Inequality (2.12) can further be written as:

  6n  112n  11    1  n ln n An  e1   e   n  11  6 n  1 12 n  11 2 n  ln n     6   

  n ln n . n 11  2 n  ln n  6  

I. III.

Proof Of Theorems 2.1 And 2.3

We now give proofsII. of our main results. Proof of Theorem 2.1:

  k  1 Sk  1 Let ck    kSk 1    n 1/ k    ck   k 1 

1 Sn

kSk

, where So is assumed zero i.e. So = 0. Then

1 .  n  1 Sn1

1

1

1

 n  sn   n  sn  n  Sn  n      a1/k k      a1/k k    c1/k k    c1/k k  n 1  k 1 n 1  k 1    k 1   k 1  

1

1 sn  n   n       ak ck  k    c1/k k  n 1  k 1   k 1  

1 Sn

1 Sn

 3.1

By using arithmetic mean – geometric mean inequality and interchanging the order of the inequality (3.1), we have 1

1

 1 sn  n 1/ k  sn  n   n 1/ k  k a  a c    k     k k     ck  n 1  k 1 n 1  k 1    k 1  

1

  n  sn  n      a1/k k      c1/k k  n 1  k 1 n 1  k 1    1 1 Letting   . Sk n k  n  1 Sn1Sn 

1 Sn

1

1 Sn

1 Sn

 n      c1/k k  n 1  k 1  

1 Sn

1 Sn

n

ak ck k 1 k

 ak ck ak ck  1  .    k 1 k k 1 k n  k  n  1 S n 1S n n

 n 1/ k  sn  ak ck    k  1 Sk  1   ak .    ak    kS    kSk n 1  k 1 k 1 k 1   k  

kSk

 3.2 

Applying inequality (2.12) in (3.2), we get

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A Further Refinement of Van Der Corput’s Inequality 1

   k  1 Sk  1   n 1/ k  sn  ak    ak     kSk n 1  k 1 k 1    

kSk

     6k  112k  11 k ln k   e1   a.   k  11  k 2k  ln k  6  k 1   6k  112k  11  6  6k  1   9   (3.3)

Replacing

k by n in the right hand side of (3.3), this becomes 1

 n 1/ k  sn  1     ak    e n 1  k 1 n 1  

   6n  112n  11 n ln n  a    n  11  n 6 n  1 12 n  11  6 6 n  1   9 2 n  ln n       6    

 3.4

This completes the proof of theorem 2.1. Proof of Theorem 2.3: Substituting (2.13) into (3.2), we obtain 1

n ln n   n 1/ k  sn 1       ak   e   n  2n  ln n  11  an . n 1  k 1 n 1   6  

3.4

This proves Theorem 2.3.

References [1]. [2]. [3]. [4]. [5].

B. Ch. Yang, 1 (2005), On an extension and a refinement of Van der Corput’s inquality, Chinese Quart. J. Math, 22, 5 pages Da-Wei Niu, J. Cao and Feng Qi, A refinement of van der Corput's inequality, J. Inequal. Pure Appl. Math. 7 (2006), no. 4, Article 127; Available online at http://www.emis.de/journals/JIPAM/article744.html. J Cao, Da-Wei Niu and Feng Qi, An extension and a refinement of van der Corput's inequality, Internat. J. Math. Math. Sci. 2006 (2006), Article ID 70786, 10 pages; Available online at http://dx.doi.org/10.1155/IJMMS/2006/70786 F. Qi, J. Cao, and D.-W. Niu, A generalization of van der Corput's inequality, RGMIA Res. Rep. Coll. 10 (2007), Suppl., Article 5; Available online at http://rgmia.org/v10(E).php. K. Hu, On Van der Corput’s inequality, Shuxue Zazhi, (J. Maths. (Wuhan)) , 23 (2003) No. 1, 126-128 (Chinese).

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