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1. Introduction. Let P a n be a given infinite series with the sequence of partial sums (s n ). By σ δn we denote the nth Ces` aro mean of order δ > −1 of the sequence (s n ),

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VOL. 80 1999 NO. 2

A GENERAL THEOREM COVERING MANY ABSOLUTE SUMMABILITY METHODS

BY

W. T. S U L A I M A N (QATAR)

Abstract. A general theorem concerning many absolute summability methods is proved.

1. Introduction. Let P a n be a given infinite series with the sequence of partial sums (s n ). By σ δ n we denote the nth Ces` aro mean of order δ > −1 of the sequence (s n ),

σ δ n = 1 A δ n

n

X

v=1

A δ−1 n−v s v . Here A δ k = k+δ k 

= (δ + 1) . . . (δ + k)/k!. It may be easily verified that A δ k ∼ k δ . The series P a n is said to be |C, δ| k summable, k ≥ 1, if

X

n=1

n k−1n δ − σ n−1 δ | k < ∞.

Let (p n ) be a sequence of positive numbers such that P n =

n

X

v=0

p v → ∞ as n → ∞.

The transformation

t n = 1 P n

X

v=0

p v s v

defines the sequence (t n ) of the Riesz means of the sequence (s n ) generated by the sequence of coefficients (p n ) (see [4]). The series P a n is said to be

|R, p n | k summable, k ≥ 1, if

X

n=1

n k−1 |t n − t n−1 | k < ∞.

1991 Mathematics Subject Classification: 40D15, 40F15, 40F05.

Key words and phrases: summability, series, sequence.

[245]

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The series P a n is said to be |N , p n | k summable, k ≥ 1 (Bor [1]), if

X

n=1

 P n

p n

 k−1

|t n − t n−1 | k < ∞.

In the special case when p n = 1 for all values of n, both |R, p n | k and

|N , p n | k summability are the same as |C, 1| k summability.

The series P a n is said to be |N, p n | summable if (1)

X

n=1

|T n − T n−1 | < ∞, where

T n = 1 P n

n

X

v=0

p n−v s v . The sequence class M is defined by

M =



p = {p n } : p n > 0 & p n+1

p n

≤ p n+2

p n+1

≤ 1, n = 0, 1, . . . , P n → ∞

 . It is known (Das [3]) that for p ∈ M , (1) holds iff

X

n=1

1 nP n

n

X

v=1

p n−v va v

< ∞.

For p ∈ M , the series P a n is said to be |N, p n | k summable, k ≥ 1 (Sulaiman [5]), if

X

n=1

1 nP n k

n

X

v=1

p n−v va v

k

< ∞.

In the special case in which p n = A r−1 n , r > −1, |N, p n | k summability is equivalent to |C, r| k summability. The series P a n is said to be |R, log n, 1| k

summable if it is |N , p n | k summable with p n = 1/(n+1) and P n ∼ log(n+1).

For any sequence {f n }, we define ∆f n = f n − f n+1 . 2. Main result. We prove the following:

Theorem 1. Let {f n }, {g n }, {G n }, and {H n } be sequences of positive constants such that {f n } ∈ M and F n = P n

v=1 f v → ∞ as n → ∞. Let {ε n } be a sequence of constants. Given a sequence {x n } define

X n = 1 G n

n

X

v=1

g v x v , Y n = 1 F n−1 H n

n

X

v=1

vf n−v x v ε n

and assume

g n+1 = O(g n ), (2)

H n+1 F n+1

G n+1

= O  H n F n

G n



,

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∆g n = O  g n+1

n

 , (4)

∆  g n H n F n

nG n



= O  g n H n

nG n

 , (5)

 nG n

g n H n F n

ε n



= O

 1 F n

 , (6)

X

n=v+1

f n−v

F n−1 H n k = O

 1 H v k

 . (7)

Let k ≥ 1. Then a necessary and sufficient condition for the implication:

if X

|X n | k < ∞ then X

|Y n | k < ∞ to hold (for any sequence {x n }) is

(i) ε n = O(g n H n F n /(nG n )), and (ii) ∆ε n = O(g n+1 H n /(nG n )).

3. Lemmas

Lemma 1 (Bor [2]). Let k ≥ 1, and let A = (a nv ) be an infinite matrix that maps ` k into ` k . Then a nv = O(1) for all n and v.

P r o o f. By the Closed Graph Theorem, A defines a bounded linear map- ping in ` k . Then the bound |a nv | ≤ C follows, where C is the norm of A.

Lemma 2 (Sulaiman [6]). Let p ∈ M. Then for 0 < r ≤ 1,

X

n=v+1

p n−v−1

n r P n−1

= O(v −r ).

Lemma 3. Suppose that ε n = O(α n β n ), α n , β n > 0, α n+1 β n+1 = O(α n β n ), ∆(α n β n ) = O(α n ) and ∆(ε n /(α n β n )) = O(1/β n ). Then ∆ε n = O(α n ).

P r o o f. We have ε n = k n α n β n where k n = ε n /(α n β n ) = O(1). Therefore

∆ε n = k n ∆(α n β n ) + ∆k n (α n+1 β n+1 )

= O(1)O(α n ) + O(1/β n )O(α n β n ) = O(α n ).

4. Proof of Theorem 1. Sufficiency. We have via Abel’s transforma- tion:

Y n = 1 F n−1 H n

n

X

v=1

g v x v

 v f n−v

g v

ε v



= 1

F n−1 H n

 n−1 X

v=1

 X v

r=1

g r x r



∆ v

 v f n−v

g v

ε v

 +

 X n

r=1

g r x r

 n f 0

g n

ε n



(4)

= 1 F n−1 H n

n−1

X

v=1

G v X v



− f n−v

g v

ε v + (v + 1)∆g −1 v f n−v ε v

+ (v + 1)g v+1 −1 ∆ v f n−v ε v + (v + 1)g v+1 −1 f n−v−1 ∆ε v



+ nG n X n f 0

F n−1 H n g n

ε n

= Y n,1 + Y n,2 + Y n,3 + Y n,4 + Y n,5 , say.

By Minkowski’s inequality,

m

X

n=1

|Y n,1 | k = O(1)

m

X

n=1 5

X

r=1

|Y n,r | k . Applying H¨ older’s inequality gives

m+1

X

n=2

|Y n,1 | k =

m+1

X

n=2

1 F n−1 k H n k

n−1

X

v=1

f n−v

G v

g v

X v ε v

k

m+1

X

n=2

1 F n−1 H n k

×

 n−1 X

v=1

f n−v

 G v

g v

 k

|X v | kv | k

 n−1 X

v=1

f n−v

F n−1

 k−1

≤ O(1)

m

X

v=1

 G v

g v

 k

|X v | kv | k

m+1

X

n=v+1

f n−v

F n−1 H n k

≤ O(1)

m

X

v=1

1 H v k

 v F v

 k

 G v

g v

 k

|X v | kv | k ,

m+1

X

n=2

|Y n,2 | k =

m+1

X

n=2

1 F n−1 k H n k

n−1

X

v=1

(v + 1)G v ∆g v −1 f n−v X v ε v

k

≤ O(1)

m+1

X

n=2

1 F n−1 H n k

× n n−1 X

v=1

v k G k v |∆g −1 v | k f n−v |X v | k |ε v | k o  n−1

X

v=1

f n−v

F n−1

 k−1

≤ O(1)

m

X

v=1

v k G k v |∆g −1 v ||X v | kv | k

m+1

X

n=v+1

f n−v

F n−1 H n k

= O(1)

m

X

v=1

 v H v

 k

G k v |∆g v | k

g v k g k v+1 |X v | k |ε v | k

(5)

≤ O(1)

m

X

v=1

1 H v k

 v F v

 k

 G v

g v

 k

|X v | kv | k ,

m+1

X

n=2

|Y n,3 | k =

m+1

X

n=2

1 F n−1 k H n k

n−1

X

v=1

(v + 1)g v+1 −1 G v ∆ v f n−v X v ε v

k

m+1

X

n=2

1 F n−1 k H n k

×

 n−1 X

v=1

v k

 G v

g v+1

 k

|∆ v f n−v ||X v | kv | k

 n n−1 X

v=1

|∆f n−v | o k−1

≤ O(1)

m

X

v=1

v k  G v

g v

 k

|X v | kv | k

m+1

X

n=v+1

|∆ v f n−v | F n−1 k H n k

≤ O(1)

m

X

v=1

1 H v k

 v F v

 k

 G v

g v

 k

|X v | kv | k ,

m+1

X

n=2

|Y n,4 | k =

m+1

X

n=2

1 F n−1 k H n k

n−1

X

v=1

vg v+1 −1 f n−v−1 G v X v ∆ε v

k

m+1

X

n=2

1 F n−1 H n k

×

 n−1

X

v=1

v k

 G v

g v+1

 k

f n−v−1 |X v | k |∆ε v | k

 n−1

X

v=1

f n−v−1

F n−1

 k−1

≤ O(1)

m

X

v=1

v k

 G v

g v+1

 k

|X v | k |∆ε v | k

m+1

X

n=v+1

f n−v−1

F n−1 H n k

≤ O(1)

m

X

v=1

 v H v

 k  G v

g v+1

 k

|X v | k |∆ε v | k ,

m

X

n=1

|Y n,5 | k =

m

X

n=1

nG n X n f 0 ε n

F n−1 H n g n

k

≤ O(1)

m

X

n=1

 n F n

 k

 G n

g n

 k

1

H n k |X n | kn | k .

Necessity of (i). By the result of Bor [1], the transformation from (X n )

into (Y n ) maps ` k into ` k and hence the diagonal elements of this transfor-

mation are bounded (by Lemma 1), so (i) is necessary.

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Necessity of (ii). This follows from Lemma 3 and necessity of (i) by taking α n ≡ g n H n /(nG n ) and β n ≡ F n , using (2).

This completes the proof of the theorem.

Remark. (1) If we put x n = a n , f n = p n and H n = n 1/k in the formula defining Y n , p ∈ M , then the condition P |Y n | k < ∞ is equivalent to |N, p n | k summability of P a n ε n (note that P n /P n−1 is a bounded sequence).

(2) If we put x n = a n , Q n = q 0 + . . . + q n , g n = Q n−1 and G n = Q n−1 (Q n /q n ) 1/k in the formula defining X n , then the condition P |X n | k <

∞ simply means |N , q n | k summability of P a n .

(3) If we put x n = a n , Q n = q 0 + . . . + q n , g n = Q n−1 and G n = n 1/k−1 Q n Q n−1 /q n in the formula defining X n , then the condition P |X n | k

< ∞ means |R, q n | k summability of P a n .

5. Applications. Throughout the rest of the paper we assume that P n → ∞ and Q n → ∞ as n → ∞.

Theorem 2. Let p ∈ M and let nq n = O(Q n ), Q n = O(Q n−1 ), and P n+1

P n

= O

 Q n

Q n−1

 q n Q n+1

q n+1 Q n

 1/k  ,

∆  P n

n

 nq n

Q n

 1/k 

= O  1 n

 nq n

Q n

 1/k  ,

∆  n P n

 Q n

nq n

 1/k

ε n



= O

 1 P n

 .

Then a necessary and sufficient condition that P a n ε n be |N, p n | k summable whenever P a n is |N , q n | k summable, k ≥ 1, is

ε n = O  P n

n

 nq n

Q n

 1/k 

, ∆ε n = O  1 n

 nq n

Q n

 1/k  . Theorem 3. Let p ∈ M and let nq n = O(Q n ), Q n = O(Q n−1 ), and

P n+1

P n

= O  q n Q n+1

q n+1 Q n



, ∆  P n q n

Q n



= O  q n

Q n

 ,

 Q n

P n q n

ε n



= O

 1 P n

 .

Then a necessary and sufficient condition that P a n ε n be |N, p n | k summable whenever P a n is |R, q n | k summable, k ≥ 1, is

ε n = O(P n q n /Q n ), ∆ε n = O(q n /Q n ).

The following results are consequences of Theorem 2.

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Corollary 4. A necessary and sufficient condition that P a n ε n be

|C, α| k summable, 0 < α < 1, whenever P a n is |C, 1| k summable, k ≥ 1, is ε n = O(n α−1 ), ∆ε n = O(n −1 ),

provided that ∆(n 1−α ε n ) = O(n −α ).

Corollary 5. A necessary and sufficient condition that P a n ε n be

|N, 1/(n + 1)| k summable whenever P a n is |C, 1| k summable, k ≥ 1, is ε n = O(log n/n), ∆ε n = O(n −1 ),

provided that

 n log n ε n



= O

 1 log n

 .

Corollary 6. A necessary and sufficient condition that P a n ε n be

|N, 1/(n + 1)| k summable whenever P a n is |R, log n, 1| k summable, k ≥ 1, is

ε n = O{(log n) 1−1/k /n}, ∆ε n = O{1/n(log n) 1/k }, provided that

 n

(log n) 1−1/k ε n



= O

 1 log n

 .

Corollary 7. A necessary and sufficient condition that P a n ε n be

|C, α| k summable, 0 < α ≤ 1, whenever P a n is |R, log n, 1| k summable, k ≥ 1, is

ε n = O{n α−1 /(log n) 1/k }, ∆ε n = O{1/(n(log n) 1/k )}, provided that ∆(n 1−α (log n) 1/k ε n ) = O(n −α ).

Acknowledgments. The author is grateful to the referee for his kind advice and valuable suggestions during the preparation of this paper.

REFERENCES

[1] H. B o r, On |N , p

n

|

k

summability factors, Proc. Amer. Math. Soc. 94 (1985), 419–

422.

[2] —, On the relative strength of two absolute summability methods, ibid. 113 (1991), 1009–1012.

[3] G. D a s, Tauberian theorems for absolute N¨ orlund summability , Proc. London Math.

Soc. 19 (1969), 357–384.

[4] G. H. H a r d y, Divergent Series, Oxford Univ. Press, Oxford, 1949.

[5] W. T. S u l a i m a n, Notes on two summability methods, Pure Appl. Math. Sci. 31

(1990), 59–68.

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[6] W. T. S u l a i m a n, Relations on some summability methods, Proc. Amer. Math. Soc.

118 (1993), 1139–1145.

Department of Mathematics College of Science

University of Qatar Qatar

E-mail: waad@qu.edu.qa

Received 23 January 1998;

revised 2 December 1998

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