a theorem on n , pn k summability of infinite series

8/10/2019 A theorem on N , pn k summability of infinite series

1/8

A theorem onkn

pN, summability of infinite series

U.K. MISRA1, S.P. PANDA2 andS.P. PANDA3

1Department of Mathematics Berhampur University,

Berhampur-760007, Orissa (India)2Department of Mathematics Khemundi College Digapahandi,

Ganjam, Orissa (India)3Department of Mathematics Rayagada (Auto)

College Rayagada, Orissa (India)

ABSTRACT

In this paper a theorem onkn

pN, summability of an

infinite series has been established.

Key words :kn

pN, summability..

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

1. INTRODUCTION

Let {sn} denotes the nth partial

sum of an infinite series

0nna and let

{pn} be a sequences of positive real

constants such that

,,10 pPNnpppP iinn

1,0 i .Then the sequence to sequence trans-formation

m

v

nvv

n

n PspP

T0

0,1

(1.1)

defines the npN, mean of the

sequence {sn}.

The series

0nna is said to be

summable 1,, kpNkn

([2]), if

1

1

1

n

k

nn

k

n

n TTp

P. (1.2)

Taking pn=1 for all n,kn

pN, summa-

bility reduces tok

C1, summability

method.

2. Known Result

Concerning with thek

C1,

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


2/8

summability of an infinite series

0nna

Flett1 has established the following

result. He proved:

Theorem A:

Letn

andn

denotes the (C,1)

mean of the sequence ns and nnarespectively. That is

n

v

vn sn 01

1 , (2.1)

n

v

vn an 01

1

. (2.2)

Then the series

0nna is summable

kC1, , 1k if and only if

1

1

n

k

nn

. (2.3)

3. Main Result

The aim of this paper is to establish

a similar result forkn

pN, summability

method. Here we prove the following.

Theorem :

Let nt denotes the npN,

mean of the sequence nna and nTbe the sequence as defined in (1.1) where

np be a sequence of positive realconstants satisfying the followingconditions.

(i). nn POnp (3.1)

(ii). nn npOP (3.2)and

(iii). nn pOpn . (3.3)

Then

0nna is summable

knpN, , 1k

if and only if

1

1

n

k

ntn

. (3.4)

3.Require Lemma

We require the following Lemmafor the proof of our theorem.

Lemma :

If np be a sequence of positivereal constants satisfing (3.1) and (3.2).Then

)1(1 Op

p

n

n . (3.5)

Proof.From (3.1), we have

111 nn POpn .

Then there exists a posit ive real

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



3/8

constant A and a positive integer n1such

that

111 ,1 nnallforPApn nn .

111

,1 nnallforPApApn nnn

11 ,1

nnallforPAn

Ap nn

.

From (3.2), we have

nn pnOP .

Then there exists a posit ive real

constant B and a positive integer n2such

that

2, nnforallpnBP nn

Let 210 , nnMaxn . Then for 0nn

nnn pAn

nBAP

An

Ap

111

n

A

AB

An

nAB

p

p

n

n

11

1

1

.

Hence )1(1

Op

p

n

n

, which proves the

lemma.

4. Proof of the Theorem

Suff icient Part:

Since n

t is thekn

pN, mean

of the sequence nan , we havee

ap

Pt

n

n

n

0

1.

Then

nnnnnn anptPtP 11

n

nnnn

npn

tPtPa 11 (4.1)

Now, we have

n

v

vv

n

n spP

T0

1

n

v

v

v

n

apP 0 0

1

n v

v vn

paP 0

1

n

n

n

PPaP 0

1

1

n n

n

aPP

a0 0

1

1

n n

n

aPP

a0 1

1

1

Then 1 nnn TTT

1

00 1

1

1 nn n

n

aaPP

a

1

1

1

1

1 n

n

aPP

1

1

1

1

1

11 n

n

n

n

nn

n a

P

PaP

PPa

U.K. Misra et al., J .Comp.&Math.Sci. Vol.1(2), 1-8 (2010). 105



4/8

nn

nn

nn

n

n aP

PaP

PP

pa 1

1

1

1

1

n

nn

n aPPP

p

1

1

1

(4.2)

Using (4.1) we get

v

vvvvn

v

v

nn

n

npv

tPtPP

PP

pT 11

1

1

1

n

v v

vv

nn

nn

v v

vvv

nn

n

p

tP

PP

p

pv

tPP

PP

p

1

1

2

1

11

1

1

1

1

1

1

1 nn

nn

v v

vvv

nn

nn

PP

p

pv

tPP

PP

p

n

t

1

0 1

2

)1(

n

v v

vv

pv

tP

1

1 1

1

1 )1(

n

v v

v

v

vvv

nn

nn

pv

P

p

PtP

PP

p

n

t

1

11

n

v

vv

nn

nn

v

tP

PP

p

n

t

1

1 1

2

1 1

11n

v vv

vv

nn

n

pvpvtPPP

p

n

v v

vv

nn

nn

v

vv

nn

nn

vvp

tP

PP

p

v

tP

PP

p

n

t

1 1

2

1

1

11 )1(

1

1

11

11

vv

n

v

vv

nn

n

ppv

tP

PP

p

4321 nnnn TTTT .

To complete the proof of the sufficient

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

1

1

4,3,2,1n

k

ni

k

n

n iforTp

P.

Now we have

k

n

k

n n

n Tp

P1

1

1

k

k

n

k

n n

n

n

t

p

P1

1

n

t

pn

Pk

n

k

n n

n

1

1

1

)1(n

k

n

ntO , using (3.2)

)1(O .

Now

1

2

2

1m

n

k

n

k

n

n Tp

P

m

n

k

n

n

p

P

1

2

1

k

n

v

vv

k

nn

n

v

tP

PP

p

1

11

km

n

n

v v

vvv

k

nn

n

vp

tpP

PP

p

1

2

1

11

km

n

n

v

vvk

nn

n tpPP

pO

1

2

1

11

)1(

(using 3.2)




5/8

1

1

11

2

1

11

)1(n

v

k

vv

km

n

n

v

vk

nn

n tppPP

pO

(using Holder's inequality)

1

1

1

2 1)1(

n

v

k

vv

m

n nn

n

tpPP

pO

m

v

m

vn nn

k

vvPP

tpO1

1

1 1

11)1(

m

v mv

k

vvPP

tpO1 1

11)1(

m

v v

k

vvP

tpO1

1)1(

v

t

P

pvO

k

vm

v v

v

1

)1( , (using 3.1)

m

v

k

v

v

tO

1

)1(

masO ,)1( .

Now,

1

2

3

1m

n

k

n

k

n

n Tp

Pk

n

k

n

k

nm

n

k

n

n

PP

p

p

P

1

1

2

1

kn

v v

vv

pvv

tP

1

1 1

2

)1(

kn

v v

vvvm

nk

nn

n

pvv

tPP

PP

p

1

1 1

11

2 1 )1(

km

n

n

v

vv

v

v

v

v

k

nn

v tppv

P

pv

P

PP

p

1

2

1

1 1

1

1 )1(

1

2 11

)1(m

n

kn

v

vvk

nn

n tpPP

pO

masO ,)1( ,(going through the

lines of 2nT ).

Now,

k

n

km

n n

n Tp

P4

11

2

1

2 1

1

m

n

k

nn

n

k

n

n

PP

p

p

P

2

1

1

1

211

vv

n

v

vv

ppv

tP

km

n

n

v vv

vv

v

v

k

nn

n

pp

ppt

v

P

PP

p

1

2

1

1 1

12

1

km

n

n

v vv

n

v

v

k

nn

n

pp

pv

v

vt

v

P

PP

p

1

2

1

1 1

12

1

1

m

n

n

v

v

v

k

nn

n t

pv

P

PP

p

1

2

1

1

2

1

k

v

v

vp

pvp

1

11

1

2

1

11

)1(m

n

kn

v

vvk

nn

n tpPP

pO

(using 3.2 and 3.3)

masO ,)1( (going through the




6/8

lines of 2nT ).

This proves the sufficient part.

Necessary Part :

From (4.2), we have

n

v

vvn

n

nn aPTp

PP

1

1

1

Then

1

1

211

1

1n

n

nnn

n

nn

n

n Tp

PPT

p

PP

Pa

11

2

nn

n

n

n

n Tp

PT

p

P

Now

n

v

vv

n

n avpP

t1

1

n

v

v

v

vvvv

n

Tp

PpvTPv

P 11

1

21

vv

n

v

n

v v

v

n

vv

n

TPp

pv

PTPv

P

11

1

0

1111

1

1

1 n

vvv

nn

TPvP

Tn

vv

n

v v

v

n

TPp

pv

P

11

1

111

1

1

1

1 n

v

vvv

n

n PPTvP

Tn

1

1

11 1

1 n

n p

pPT

P

v

n

v v

v

v

n

Tp

pP

P

1

1

11

1

4321 nnnn tttt .

To complete the necessary part, byusing Minkonski's inequality it issufficient to show that

,4,3,2,11

iforn

t

n

k

ni.

Now

1 1

1

n n

k

n

kk

n

n

Tn

n

t

1

1

n

k

n

k

Tn

1

1

)1(n

k

n

k

n

n Tp

PO (using 3.1)

)1(O .

Now

1

2

2m

n

k

n

n

t

1

2

1

1

11m

n

kn

v

vvk

n

pTvPn

km

n

n

v

v

k

v

kn

v

vk

n

TvppPn

1

2

1

1

11

1

1

(using Holders inequality)

1

1

1

2

1 n

v

k

v

k

v

m

n n

TpPn

1

2

1

11

m

n

n

v

k

v

k

v

nnn

nn TpPpPn

pP




7/8

1

2

1

11

)1(m

n

n

v

k

v

k

v

nn

n TpPP

pO

(using 3.1)

1

1 11

11)1(

m

vn nn

m

v

k

v

k

v

PP

TpO

11

11)1(

mv

m

v

k

vv

k

PPTpO

1

1

)1(m

v

k

v

v

v

k

TP

pO

k

km

v v

v

k

v

v Tvp

P

P

pO

11

1

)1(

k

v

km

v v

v Tp

PO

11

1

)1( (using 3.1)

masO ,)1( .

Now

1

2

3m

n

k

n

n

t m

n

n

v

vvk

n

TPVPn

1

2

1

1

1

1

k

v

vv

p

pp

1

k

m

n

n

v v

v

vvk

n p

pVTP

Pn

1

2

1

1

1

1 11

m

n

n

v v

v

vvk

n p

pTP

PnO

1

2

1

1

11

1)1(

k

nm

nk

n

pTp

P

PnO

1

1

1

1

2

11

11

2

1

1

1

1)1(

km

n

n

vk

n

pPn

O

1

1

1

v

n

v

k

v

k

v

v pTp

P

Using Holders inequality

1

1

2

1

1

1)1(

v

k

v

km

n

n

v v

v

n

pTp

P

PnO

1

1

2

1

1

)1(

v

k

v

km

n

n

v v

v

nn

n pTp

P

PP

pO

1

1

21

1

)1(m

vn n

n

v

k

v

k

v

vm

v P

ppT

p

PO

1

1 11

1)1(m

vn nn

n

km

v

v

k

v

vv

PP

pT

p

PpO

11

1

11)1(

mv

km

v

v

k

v

v

vPP

Tp

PpO

v

m

v

k

v

k

v

v

vP

Tp

PpO

1)1(

1

1

k

v

k

v

vm

v v

v Tp

P

p

pO

1

1

1)1(

k

v

k

v

vm

v

Tp

PO

1

1

)1( by Lemma

masO ,)1( .

Now




8/8

a theorem on n , pn k summability of infinite series

Documents