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86 SOME NOTES ON ABSOLUTE SUMMABILITY FACTORS

*W. T. Sulaiman

Department of Computer Engineering, College of Engineering, University of Mosul, Iraq

*Author for Correspondence ABSTRACT

In this paper we give improvement to the result of [3] concerning absolute summability of infinite series.

2000 (MSC): 40F05, 40D25.

Key Words: Absolute Summability, Summability Factor INTRODUCTION

Let T be a lower triangular matrix,

  s

n a sequence, and : .

0

n

v v nv

n t s

T (1.1) A series

a

n is said to be summable

T , k  1 ,

k if

1 .

1

1  

n k n

k T

n (1.2)

Given any lower triangular matrix T one can associate the matrices

T

and Tˆ, with entries defined by

nv nv n v

n

v i

ni

nv t n i t t t

t , , 0,1,2..., ˆ 1,

respectively. With

,

0

n

i i i

n

a

s

.

0 0

0 0

0

    

n

i

i i ni n

i v

nv n

i i i n

v v

i i i nv n

v v nv

n t s t a a t t a

t

(1.3)

: ˆ , 1, 0.

0 1

0 , 1 0

1    

  

n n

n

i

i i ni n

i

i i i n n

i

i i ni n

n

n t t t a t a t a ast

Y

(1.4)

We call T a triangle if T is lower triangular and

t

nn

 0

for all n.

We assume that

  p

n is a sequence of positive real numbers such that

P

n

p

0

p

1

 ...  p

n

  as n   .

In the special case when

t

nv

p

v

/ P

n

,

summability

T

k reduces to

n k

p

N , summability.

Rhoads and Savas (2005) proved the following result

Theorem 1.1. Let A be a triangle with nonnegative entries satisfying (i)

a

n0

 1 , n  0 , 1 , ...,

(ii) an1,vanv for nv1, (iii)

na

nn

  ( 1 ), 1   ( na

nn

) ,

(2)

87

(iv)

ˆ ( ) .

0 , 1 nn

n

v

a

vv

a

nv

  a

If

X

n

is a positive monotonic nondecreasing sequence such that (v)

n

X

n

  ( 1 ) ,

(vi) 2 (1),

1  

n m

n nXn

(vii)

mn1

a

nn

t

n k

   X

m

,

where

, 1

1

1

n

k k

n

ka

t n

then the series

a

n

n is summable

A , k  1 .

k

The object of this paper is to give two improvements to theorem 1.1 as follows 1. By weakening the condition (v) ,

2. By weakening the condition (vii), Main Result

In what follows we prove the following

Theorem 2.1. Let A be a triangle with nonnegative entries satisfying (i)

a

n0

 1 , n  0 , 1 , ...,

(ii) an1,vanv for nv1, (iii)

na

nn

  ( 1 ), 1   ( na

nn

) ,

(iv)

ˆ ( ) .

0 , nn

n

v

a

vv

a

nv

  a

(v) 1

ˆ

, 1

( 1 ) .

1

1

 

mn v nv k

k

a

n

If

X

n

is a positive monotonic nondecreasing sequence and if

 

n is a sequence of constants satisfying (vi)

 

n

( 1 ) ,

(vii) 2 (1),

1  

n m

n nXn

(viii) mn 1 k 1

m

,

n k n

nn

X

X t

a  

where

,

1 1

1

 

n

k k

n

ka

t n

then the series

a

n

n is summable

A , k  1 .

k

It may be mentioned as well that whenever

X

n

  ,

condition (viii) of theorem 2.1 is weaker than condition (vii) of theorem 1.1. For if (viii) of theorem 1.1 is satisfied, then

X

n

 

implies that

, 0

n

while if

n

 0 ,

then by choosing

, 0 , , (1/2)

2 /

1  

n Xn n

n

we obtain

n

X

n

  ( n

)   ( 1 ) .

(3)

88

Although we add conditions (v), but this condition with (vi) caused for no losing powers via the proof.

As an example in the proof of theorem 1.1,

n k1 has been replaced by

 ( 1 ),

and this is not the case in theorem 2.1.

Lemma 2.2. Condition (viii) of theorem 2.1 is weaker than (vii) of theorem 1.1 Proof. If (viii) holds, then we have

m

k n m

n k nn m

n k n

k n

nn a t X

X X

t

a  





1 1 1 1

1

1 ,

while if (vii) is satisfied then,

1

1 1 1

nk

m

n

k k n n nn m

n

k n

nn

t X

X t a

a

1

1 1 1

1

1

1 1

 

   

mk

m

n k n

k n nn m

n

k n n

v k v

k v

vv

X

X t X a

X t a

 

1

 

1

1

1

1

nk m mk

m

n

n X X X

X

     

mk

m

n

k n k n

m X X X

X   

1

1

1 1

1 1

     

mk

k k

m

m X X X

X   

1 1 11

 

( m),

k

m X

X 

for k 1.

Lemma 2.3. Conditions (vi)-(vii) of theorem 2.1 implies

nX

n

n

  ( 1 ) ,

(2.1) ,

1

n

n

Xn

(2.2) and (2.2) implies

n

X

n

  ( 1 ) .

(2.3) Proof. Since

n

 0 ,

then

n

 0 ,

and hence

 

  (1)

2  (1)

2 (1).

v n

v v n

v

v n

n v

v n

n

n nX nX vX

nX

m

m

n n n

m

n n

v v m

n

n

n

X X

X  

 

 

 

 

   

1

1

1 1

1

(4)

89

(1) (1) (1)

1

1

2   

m m m

n

n

n mX

nX

As

n

 0 ,

 (1)

  (1).

v n

v v n

v v n

n

n X X

X

Lemma 2.4. Under the conditions of theorem 1.2, 1 ˆ

 

,

0

nn n

v

nv

va  a

(2.4)

1

1

) ˆ (

m

v n

vv nv

va a . (2.5)

ˆ ( 1 ) .

1

1 1

,

 

m

v n

v

a

n (2.6) For the proof see [3].

Proof of Theorem 1.2.

Let

Y

n denote the nth term of A-transform of the series

a

n

n , then

 

n

v

v nv v n

v

v v nv

n

v

a a v a

a Y

1 0

ˆ ˆ

:

 

 

 

 

 

 

 

    

n

a a v v

a a

r

nn n

n

v v v

nv v n

v v

r r

1 1

1 1

ˆ

nv v v nv v nv v n nn n

n

v

v

t a

n a n

a v a v

v t v

v 1

1 ˆ ˆ 1

1 ˆ 1

) 1 ( ) 1 1

(

, 1

1

1

 

 

 

 

 

 

 

n nn n

n

v

v v n v v nv v v v nv

v t a

n a n

t a t a

vt ˆ

ˆ

ˆ

1

1 1

1

1 ,

 



 

    

Y

n1

Y

n2

Y

n3

Y

n4

.

In order to prove the Theorem, by Minkowski's inequality, we have to show that

, 1,2,3,4.

1

1  

Y j

n nj k

n k

By Holder's inequality,

n k

v

v nv v m

n k k n m

n

k

t a

n v Y

n  

1

1 2

1 1

2

1

1 ˆ

(5)

90

1 1

1 1

1

1 2

1

ˆ ˆ

 

    

n k

v

nv vv k

v nv k v k vv n

v k m

n

k

v a t a a a

n

 

nv v k

k v k vv n

v k m

n

k

nn v a t a

na ˆ

) 1

( 1

1

1 2

1

 

1 1

) ˆ 1 (

v n

nv k

v k v k vv m

v

vv va t a

a

v k v k

m

v vv t

a

1

) 1 (

 

1

1

)

1

1

(

v v v k

m

v k v

k v

vv

X

X t

a

v

m

v k v

k v vv

X t

a

1

)

1

1 (

 

 

1

1 1

1

( 1 )

) 1 (

m

v

m v

r k r

k r rr

v

X

t

a

m

v k v

k v vv

X t a

1 1

v m m

m

v

v

X X

 

 

1

1

) 1 (

  ( 1 ) .

in view of (iv), (2.2) ,(2.3) and (2.6) .

n k

v

v nv v v n

k k n n

k

Y n t a

n  

 

1

1 2

1 2

2

1

ˆ 

1 1

1 1

1 2

1

ˆ ˆ

 

 

   

n k

v

nv v k

v nv v n

v k v n

k

t a a

n

 

v nv v k

n

v k v n

k

nn t a

na ˆ

) 1 (

1

1 2

1

 

v nv

v n k v v

k

v a

t ˆ

) 1 (

1 1

 

v k

v

k v vvt

a

1

) 1 (

  ( 1 ) ,

as in the case of

Y

n1

.

n k

v

v v n v m

n k k n m

n

k

Y n t a

n  

 

1

1 1 , 2

1 3

2

1

ˆ

(6)

91

1 1

1 1

1

1 , 1

2

1

ˆ

 

 

   

k

v n

v v v

n

v

k v n k v k v m

n

k

t X a X

n

v

n

v

k v n k v k v m

n

k t X a

n

 

1

1

1 , 1

2

1 ˆ

) 1 (

 

m

v n

k v n k v

m

v

k v k

v X n t

t

1

1 , 1

1

1 ˆ

) 1

(

v

m

v k v

k v

vv v

X t

a

1

) 1

1 (

    

m

v k v

k v vv m v

r k r

k r rr m

v

v

X

t m a

X t v a

1 1 1

1 1

1

) 1 ( )

1

(

m m

m

v

v v m

v

v

v

v X m X

X      

  

) 1 ( )

1 ( )

1 (

1

1

2 1

1

  ( 1 ) .

in view of (2.1), (2.2), (v), and (viii) .

k

n nn n n

k k n n

k t a

n n n Y

n 1

2 1 4

2

1

n k nnk nk

n

k t a

n

1

) 1

1 (

(1) (1),

1

k n k n n

nnt

a

as in the case of

Y

n2

This completes the proof of the Theorem .

Corollary 2.5. Let

(i)

np

n

  ( P

n

) , P

n

  ( np

n

) ,

(ii)

1/

.

1 1

1 k

v k

n n

n v

n

k P

P P

n p  



If

X

n

is a positive nondecreasing sequence and the sequence

 

n is satisfying conditions (vi)-(vii) of theorem 2.1, and

m

m

n k

n n

k n

n

X

X P

t

p  

1 1 , (2.7)

(7)

92

then the series

a

n

n is summable N,p , k1.

n k

REFERENCES

[1] H. Bor (1993). On absolute summability factors. Proceedings of the American Mathematical Society. 118. 71-75.

[2] T. M. Flett (1957). On an extension of absolute summability and some theorems of Littlewood and Paley. Proceedings of the London Mathematical Society. Third Series 7.

113-141.

[3] B. E. Rhoads and E. Savas (2005). A note on absolute summability factors. Periodica Mathematica Hungarica, 51 (1). 53-60.

References

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