Cmjv01i02p0103 by Journal of Computer and Mathematical Sciences

J. Comp. & Math. Sci. Vol. 1(2), 103-110 (2010).

A theorem on N , p n k summability of infinite series U.K. MISRA1, S.P. PANDA2 and S.P. PANDA3 1

Department of Mathematics Berhampur University, Berhampur-760007, Orissa (India) 2 Department of Mathematics Khemundi College Digapahandi, Ganjam, Orissa (India) 3 Department of Mathematics Rayagada (Auto) College Rayagada, Orissa (India) ABSTRACT In this paper a theorem on N , p n

summability of an

infinite series has been established.

N , p n k summability..

Key words :

sequence {sn}.

1. INTRODUCTION



Let {sn} denotes the nth partial

The series

a

and let

n 0

{pn} be a sequences of positive real constants such that

Pn  p 0  p1    p n , n  N , Pi  p i  0, i  1.

1 Pn

summable N , p n k , k  1 ([2]), if 

 Pn   n 1  p n

  

k 1

Tn  Tn 1

 .

 pv s v , Pn  0 n

summa-

bility reduces to C ,1 k summability

(1.1)

v 0

N , p 

(1.2)

method.

defines the

is said to be

Taking pn=1 for all n, N , p n

Then the sequence to sequence transformation

Tn 

n 0



sum of an infinite series

a

mean of the

2. Known Result Concerning with the

Journal of Computer and Mathematical Sciences Vol. 1 Issue 2, 31 March, 2010 Pages (103-273)

C ,1 k

104

U.K. Misra et al., J.Comp.&Math.Sci. Vol.1(2), 103-110 (2010). 

summability of an infinite series

a

n 0

Flett has established the following result. He proved: Theorem A:

mean of the sequence na n  and Tn  be the sequence as defined in (1.1) where

p n 

be a sequence of positive real constants satisfying the following conditions.

Let  n and  n denotes the (C,1) mean of the sequence s n  and na n  respectively. That is

1 n  n 1

(i). npn  OPn 

(3.1)

(ii). Pn  O np n 

(3.2)

(iii). n  p n  O  p n  .

(3.3)

and



sv ,

(2.1)

v0



Then

a n 0

n 

1 n 1

a

is summable N , p n k , k  1

if and only if v

(2.2)

v 0



n

 .

(3.4)

n 1



Then the series

a

is summable

n 0

C ,1 k , k  1 if and only if 

n

n

 .

3. Require Lemma We require the following Lemma for the proof of our theorem.

(2.3)

Lemma :

n 1

If p n  be a sequence of positive

3. Main Result The aim of this paper is to establish a similar result for N , p n k summability method. Here we prove the following.

p n 1  O(1) . pn

(3.5)

Proof. From (3.1), we have

Theorem : Let

real constants satisfing (3.1) and (3.2). Then

n  1 pn 1  OPn 1  . t n 

denotes the

N , p  n

Then there exists a positive real

Journal of Computer and Mathematical Sciences Vol. 1 Issue 2, 31 March, 2010 Pages (103-273)

U.K. Misra et al., J.Comp.&Math.Sci. Vol.1(2), 1-8 (2010). constant A and a positive integer n1 such that

n  1 p n1

 A Pn 1 ,

for all n  n1 .

 n  1 p n 1  A p n 1  A Pn , for all n  n1

A  Pn , for all n  n1 . n 1  A

 p n1

tn 

105

1 Pn

 p a .

 0

Then

Pn t n  Pn1 t n 1  p n n a n Pn t n  Pn 1 t n 1 n pn

 an 

(4.1)

Now, we have From (3.2), we have

1 Pn

Tn 

Pn  On p n  .

1  Pn

Then there exists a positive real constant B and a positive integer n2 such that

1 Pn



Pn  B n p n , forall n  n2 Let n0  Max n1 , n2  . Then for n  n0

p n1

1  Pn

ABn A  Pn  pn n 1 A n 1  A

AB A 1 . 1 n

a p   v

v 0

0

 a

a  P  

 a 

 

0

a 

1 Pn



of the sequence n a n , we have e

mean

1 Pn

 a 

 P 1 

0

1 Pn

P  

a

1

0

P  

1

a

1

Then Tn  Tn  Tn 1



Sufficient Part:

v

 0

lemma. 4. Proof of the Theorem

 p

 0

p Hence n 1  O(1) , which proves the pn

v 0

 0



Since t n  is the N , p n

p

 p AB n  n 1   pn n 1 A

n 1

P  

1

a 

1

1 Pn 1

 1 1   a n      Pn 1 Pn 

a   0

n 1

P  

1

a

1

n 1

P  

1

a 

1

Journal of Computer and Mathematical Sciences Vol. 1 Issue 2, 31 March, 2010 Pages (103-273)

Pn1 an Pn

106

U.K. Misra et al., J.Comp.&Math.Sci. Vol.2(1), 103-110 (2010).

pn  an  Pn Pn 1

n 1

P   1

1

P a   n 1 a n Pn

pn  Pn Pn 1



 

P 1 a 

(4.2)

1

Using (4.1) we get

 Tn  

pn Pn Pn 1

pn Pn Pn 1 n

P

v 1

v 1

n 1

t pn  n  n Pn Pn1

 v 0

 v 1

P

v 1

n 1

t pn  n  n Pn Pn 1



pn Pn Pn 1

 v 1



 Pn   n 1  p n

Pv21 t v 1  pv

 v 1

n 1 2 v

P

tn pn  n Pn Pn 1

v 1 n 1

 v 1

pn  Pn Pn 1

 v 1

k 1

Tni

  

  for i  1,2,3,4 .

k 1



 Pn   n 1  p n

Tn1 

 P    n n 1  n p n 

 O (1)

 n 1

  

k 1

  

k 1

n k

, using (3.2)

 O(1) .

 P  Pv  t v  v1    pv (v  1) pv 1 

Now m 1

 Pn   n 2  p n

Pv t v v

  

k 1

Tn 2

 1  1  t v    v p v v  1 p v 1 

Pv t v pn  v Pn Pn 1

n 1

  



Pv 1 Pv t v pn  v pv Pn Pn 1

n 1

tn pn  n Pn Pn 1



Pv t v  Pv 1 t v 1 v pv

Pv2 t v (1  v) p v 1

n 1



 Pn   n 1  p n

Now we have

Pv 1 Pv t v pn  v pv Pn Pn 1



part, by using Minkonski's inequality it is sufficient to show that

p v 1

To complete the proof of the sufficient

P    n n 2  p n k

m 1



 Tn1  Tn 2  Tn3  Tn 4 .

m 1

 pn     Pn Pn 1 

Pv2 t v v 1 v (v  1)

Pv t v  1 1     v  pv p v 1 

 n2

pn Pn Pnk1 m 1

 O (1)  n 2

  

k 1

 n 1 Pv t v   v 1 v 

 n 1 Pv p v t v   v 1 p v v 

   

pn  n1    pv tv  k Pn Pn1  v 1 

(using 3.2)

Journal of Computer and Mathematical Sciences Vol. 1 Issue 2, 31 March, 2010 Pages (103-273)

U.K. Misra et al., J.Comp.&Math.Sci. Vol.1(2), 103-110 (2010). k 1

m 1

n 1 pn  n 1  k p pv t v     v k Pn Pn 1  v 1  v 1 (using Holder's inequality)

 O (1)  n2

m1

pv Pn Pnk1

 n 2

m 1

 O (1)  n 2

pn Pn Pn1

p

 O(1)

n 1

m 1



v 1

n  v 1

p

 O(1)

 pv t v

v 1 m



 O(1)

v 1

 1 1      Pn 1 Pn 

Now, m 1

 Pn   n 2  p n

  

 O(1)

 v 1

m 1

, (using 3.1)





k m 1



 n 2

pn Pn Pnk1

pn Pn Pnk1 m 1



 n 2

  

k 1 k

Tn 3 

m 1

 n 2

2 v

 Pn   pn

  

k 1

  P tv    v 1 v(v  1) p  v  1   n 1

m 1



 n 2

pn Pn Pnk1

m 1

P    n n 2  p n

  

k 1

 pn     Pn Pn 1 

1 1  pv p v 1

   

Now,

 Pn   pn

Tn 4

 n1 Pv2 t v   v 1 v 

 O(1) , as m   .

m 1

k 1

1 Pv

v pv tv Pv v

lines of Tn 2 ).

n 2



 O (1) , as m   ,(going through the

 1 1      Pv Pm 1 

v 1

 O(1)

pn  n    pv t v  k Pn Pn 1  v 1 

n 2

v 1

p

 n1 Pv  Pv 1     pv t v   v1 v pv (v  1) pv 1 

m 1

 O(1)

107

p nk   Pnk Pnk1 

 n1 Pv Pv 1 t v     v 1 v(v  1) p  v 1  

m 1

 O(1)

 n 2

   

 n 1 Pv2 v  1 v p n 1   t v  v 1 v v p v p v 1  

p n  n 1  Pn Pnk1  v 1 

 n1 Pv2 p v 1  pv  t v  v 1 v p v p v 1 

 Pv  v p  

v  1  p v 1 p v 1

   

  tv p  

p n  n1    pv t v  k Pn Pn 1  v 1 

(using 3.2 and 3.3)

 O(1) , as m   (going through the

Journal of Computer and Mathematical Sciences Vol. 1 Issue 2, 31 March, 2010 Pages (103-273)

108

U.K. Misra et al., J.Comp.&Math.Sci. Vol.1(2), 103-110 (2010). lines of Tn 2 ). This proves the sufficient part.

v 1

v 1

To complete the necessary part, by using Minkonski's inequality it is sufficient to show that

From (4.2), we have

Pn 1 Pn  Tn  pn

P

v 1

v 1





Then  P P 1  Pn Pn 1  Tn  n 1 n 2  Tn 1   Pn 1  p n p n 1 

t ni

  for i  1,2,3,4, .

n 1

Now 

t n1



Pn P  Tn  n 2  Tn 1 pn p n 1



 

1 Pn



pv v av



 Tv 



v Pv  Tv 

v 1

1 Pn

 n  Tn 

n 1

1 Pn



1 Pn

 v pv Pv  2  Tv 1  pv 1  n 1



v  1 p v1



1 Pn

k 1

 Tn

v 0

1 Pn

  

k 1 k

 Tn (using 3.1)

 O(1) . Now

Pv 1  Tv

n 1

 v Pv  Tv

m 1

t n2



n 2

m 1



 n 2

1 1  n1    v  Tv p v  k n Pn  v 1 

v 1

v  1 p v 1 pv

v 1

 n  Tn 

n

P  O(1)   n n 1  p n

m 1



 n 1



v 1





n 1

  v P v 1

n k  Tn





1 Pn

n 1

Now

1 tn  Pn

p v 1  Tv pv

P

 t n1  t n 2  t n3  t n 4 .

Necessary Part:

an 

n 1

1 Pn



n 1



Pv 1  Tv

v  Tv Pv  Pv 1 



n2

m 1



v 1

 n 2

m 1

n 1

 p  T P 1 1   1   p   1 





 n 2

k 1

1  n 1  n 1   pv   p v v k  Tv k n Pn  v 1  v 1 (using Holders inequality)

1 n Pn

n 1

p

 k  Tv

v 1

Pn p n n Pn p n Pn1

n 1

p

v 1

Journal of Computer and Mathematical Sciences Vol. 1 Issue 2, 31 March, 2010 Pages (103-273)

 Tv

U.K. Misra et al., J.Comp.&Math.Sci. Vol.1(2), 103-110 (2010). m 1

n 1

pn Pn Pn 1

 O(1)  n 2

 pv  k  Tv

m 1



 O(1)

v 1

n 2

1  n 1    p 1  k n Pn  v 1 

109

k 1

(using 3.1) m

 p v  k  Tv

 O(1)

m 1

v 1



n  v 1

n 1

 Pv   v 1  p v

 1 1      Pn 1 Pn 

  

 Tv

p v 1

Using Holders inequality m

 O (1)



1 1      Pv Pm 1 

p v  Tv

v 1

m 1



 O(1)

n 2

 k pv  Tv Pv

m 1

 O(1)

 v 1

m 1

 O(1)

 v 1

 pv   Pv

  

 O(1)

 Pv   pv

 v 1

  

m 1

 O(1)

 n 2

k 1

 Pv     pv 

 Tv

 O(1)



k 1

 Tv

m 1





n 2



(using 3.1)

 O(1)

n 2

1 n Pnk

m 1



 n 2

p

v 1

v 1

1 n Pnk m 1

 O (1)

   V Pv 1  Tv  v 1 

 n2

m 1

v 1

   Tv 

 Pv   pv

  

pv 1

v 1



 Tv

n  v 1

   

 p v 1    Pv 1  Tv  pv   v 1 k

 O(1)

 v 1

 O(1)

 v 1

m 1

 n v 1

pn Pn Pn1

 1 1      Pv Pm1 

   Tv 

 Pv   pv

  

pn Pn2

 Pv   pv

p v 1

   Tv 

pv 1



pv 1

 Pv   pv

pv 1 pv

 Tv

m 1

   Tv 

v 1

 Pv   pv

 O(1)

n 1

 1  n 1 P   O1 T p 1  k  n  2 n Pn   1 p 

p v 1

 n 1 p   Pv 1  Tv V  1 v 1  v 1 pv 

1 n Pnk

 O(1)

n 1

pv 1  pv    pv 



 Pv     pv 

t n3

n 1

pn Pn Pn

Now m 1

v 1

 O(1) , as m   . k



 Pv   pv

v 1

m 1

n 1

1 n Pn

1 Pv

k 1

 Tv

k 1

 Tv

by Lemma

 O(1) , as m   . Now

Journal of Computer and Mathematical Sciences Vol. 1 Issue 2, 31 March, 2010 Pages (103-273)

110

m 1

 n 2

U.K. Misra et al., J.Comp.&Math.Sci. Vol.1(2), 103-110 (2010).

tn4

m 1



n 2

1 n Pnk

p v 1 Tv pv m 1

 n 2

1 n Pnk

  

This completes the proof of necessary part as such the theorem is established.

 n 1   Pv 1  v 1

REFERENCES

 n 1 pv    Tv p v 1   v 1 p v 1 

 O(1) , as m   , (going through the lines of t n 3 ).

1. Flett, T. M., On an Extension of Absolute Summability and Some Theorem of Littlewood and Paley. Proc. Lond. Math. Soc. 7, p.113-141 (1957). 2. Tripyuthi Sanjaya, A Relation Between Two Summability Methods, Bulletin of Pure and Applied Sciences, Vol. 20E (No.1) p.143-152 (2001).

Journal of Computer and Mathematical Sciences Vol. 1 Issue 2, 31 March, 2010 Pages (103-273)