In this paper we have proved two theorems concerning an inclusion between two absolute summability methods by using any absolute summability factor

http://jipam.vu.edu.au/

Volume 4, Issue 4, Article 66, 2003

INCLUSION THEOREMS FOR ABSOLUTE SUMMABILITY METHODS

H. S. ÖZARSLAN DEPARTMENT OFMATHEMATICS,

ERCIYESUNIVERSITY, 38039 KAYSERI,

TURKEY.

seyhan@erciyes.edu.tr

Received 07 April, 2003; accepted 30 April, 2003 Communicated by L. Leindler

ABSTRACT. In this paper we have proved two theorems concerning an inclusion between two absolute summability methods by using any absolute summability factor.

Key words and phrases: Absolute summability, Summability factors, Infinite series.

2000 Mathematics Subject Classification. 40D15, 40F05, 40G99.

1. INTRODUCTION

LetP

a_n be a given infinite series with partial sums(s_n), andr_n = na_n. By u_nandt_nwe denote then-th(C,1)means of the sequences (sn)and(rn), respectively. The series P

anis said to be summable|C,1|_k, k ≥1, if (see [4])

(1.1)

∞

X

n=1

n^k−1|u_n−u_n−1|^k<∞.

But sincet_n =n(u_n−un−1)(see [7]), the condition (1.1) can also be written as (1.2)

∞

X

n=1

1

n |t_n|^k <∞.

The seriesP

a_nis said to be summable|C,1;δ|_k k≥1andδ≥0, if (see [5]) (1.3)

∞

X

n=1

n^δk−1|t_n|^k<∞.

If we takeδ= 0, then|C,1;δ|_ksummability is the same as|C,1|_k summability.

ISSN (electronic): 1443-5756

c 2003 Victoria University. All rights reserved.

046-03

Let(p_n)be a sequence of positive numbers such that

(1.4) P_n =

n

X

v=0

p_v → ∞ as n → ∞, (P−i =p−i = 0, i≥1).

The sequence-to-sequence transformation

(1.5) T_n= 1

P_n

n

X

v=0

p_vs_v

defines the sequence(T_n)of the( ¯N , p_n)mean of the sequence(s_n), generated by the sequence of coefficients(p_n)(see [6]).

The seriesP

a_nis said to be summable N , p¯ _n

k, k ≥1,if (see [1]) (1.6)

∞

X

n=1

P_n pn

^k−1

|∆Tn−1|^k <∞ and it is said to be summable

N , p¯ n;δ

k,k≥1andδ≥0,if (see [3]) (1.7)

∞

X

n=1

P_n p_n

δk+k−1

|∆Tn−1|^k <∞,

where

(1.8) ∆T_n−1 =− p_n

PnPn−1 n

X

v=1

P_v−1a_v, n≥1.

In the special case whenδ= 0(resp. p_n = 1for all values ofn)

N , p¯ _n;δ

ksummability is the same as

N , p¯ _n

k(resp. |C,1;δ|_k) summability.

Concerning inclusion relations between |C,1|_k and N , p¯ _n

k summabilities, the following theorems are known.

Theorem 1.1. ([1]).Let k ≥ 1 and let (p_n) be a sequence of positive numbers such that as n→ ∞

(1.9) (i)P_n =O(np_n), (ii)np_n=O(P_n).

If the seriesP

a_nis summable|C,1|_k,then it is also summable N , p¯ _n

k.

Theorem 1.2. ([2]). Let k ≥ 1 and let (p_n) be a sequence of positive numbers such that condition (1.9) of Theorem 1.1 is satisfied. If the seriesP

a_n is summable N , p¯ _n

k,then it is also summable|C,1|_k.

2. THEMAINRESULT

The aim of this paper is to generalize the above theorems for|C,1;δ|_kand

N , p¯ _n;δ

ksumma- bilities, by using a summability factors. Now, we shall prove the following theorems.

Theorem 2.1. Letk≥1and0≤δk <1. Let(pn)be a sequence of positive numbers such that Pn =O(npn)and

(2.1)

∞

X

n=v+1

P_n p_n

^δk−1 1

P_n−1 =O (

P_v p_v

δk

1 P_v

) .

LetP

an be summable|C,1;δ|_k. Then P

anλn is summable

N , p¯ n;δ

k, if(λn)satisfies the following conditions:

(2.2) n∆λ_n=O

Pn

np_n ^1−δk_k

,

(2.3) λ_n=O

P_n npn

_k¹ .

Theorem 2.2. Letk ≥ 1and 0 ≤ δk < 1. Let(p_n) be a sequence of positive numbers such thatnp_n = O(P_n)and satisfies the condition (2.1). LetP

a_n be summable

N , p¯ _n;δ

k.Then Pa_nλ_nis summable|C,1;δ|_k,if(λ_n)satisfies the following conditions:

(2.4) n∆λ_n=O

np_n P_n

^1−δk_k

(2.5) λ_n=O

np_n Pn

_k¹ .

Remark 2.3. It may be noted that, if we takeλ_n = 1andδ = 0in Theorem 2.1 and Theorem 2.2, then we get Theorem 1.1 and Theorem 1.2, respectively. In this case condition (2.1) reduces

to m+1

X

n=v+1

p_n P_nPn−1

=

m+1

X

n=v+1

1 Pn−1

− 1 P_n

=O 1

P_v

as m→ ∞, which always holds.

Proof of Theorem 2.1. Since

t_n = 1 n+ 1

n

X

v=1

va_v we have that

an = n+ 1

n tn−tn−1. Let(T_n)denote the( ¯N , p_n)mean of the seriesP

a_nλ_n. Then, by definition and changing the order of summation, we have

T_n = 1 P_n

n

X

v=0

p_v

v

X

i=0

a_iλ_i = 1 P_n

n

X

v=0

(P_n−Pv−1)a_vλ_v.

Then, forn ≥1, we have

T_n−Tn−1 = pn

P_nPn−1 n

X

v=1

Pv−1a_vλ_v.

By Abel’s transformation, we have T_n−Tn−1 = p_n

P_nP_n−1

n−1

X

v=1

P_v∆λ_vv+ 1

v t_v− p_n P_nP_n−1

n−1

X

v=1

p_vλ_vv+ 1 v t_v

+ p_n PnPn−1

n−1

X

v=1

P_v

v λ_v+1t_v+ p_n Pn

λ_nn+ 1 n t_n

=T_n,1+T_n,2+T_n,3 +T_n,4, say.

Since

|T_n,1+T_n,2+T_n,3+T_n,4|^k≤4^k(|T_n,1|^k+|T_n,2|^k+|T_n,3|^k+|T_n,4|^k),

to complete the proof of the theorem, it is enough to show that

(2.6)

∞

X

n=1

Pn

p_n

δk+k−1

|T_n,r|^k<∞ for r= 1,2,3,4.

Now, whenk > 1, applying Hölder’s inequality with indicesk andk⁰, where ¹_k + _k¹0 = 1, we get

m+1

X

n=2

P_n p_n

δk+k−1

|T_n,1|^k

≤

m+1

X

n=2

P_n p_n

δk−1

1 P_n−1^k

n−1

X

v=1

P_vv+ 1

v |t_v| |∆λ_v|

!k

=O(1)

m+1

X

n=2

P_n p_n

δk−1

1 P_n−1^k

n−1

X

v=1

|tv| |v∆λv| P_v vp_vpv

!k

=O(1)

m+1

X

n=2

P_n pn

^δk−1 1 P_n−1^k

n−1

X

v=1

|t_v| |v∆λ_v|p_v

!k

=O(1)

m+1

X

n=2

P_n p_n

δk−1

1 Pn−1

n−1

X

v=1

|t_v|^k|v∆λ_v|^kp_v 1 Pn−1

n−1

X

v=1

p_v

!^k−1

=O(1)

m

X

v=1

|t_v|^k|v∆λ_v|^kp_v

m+1

X

n=v+1

P_n p_n

δk−1

1 Pn−1

=O(1)

m

X

v=1

|t_v|^k|v∆λ_v|^k P_v

p_v δk−1

=O(1)

m

X

v=1

v^δk−1|t_v|^k =O(1) as m→ ∞.

by virtue of the hypotheses of Theorem 2.1.

Again using Hölder’s inequality,

m+1

X

n=2

P_n p_n

δk+k−1

|T_n,2|^k

=O(1)

m+1

X

n=2

P_n p_n

δk−1

1 P_n−1

n−1

X

v=1

p_v|t_v|^k|λ_v|^k 1 P_n−1

n−1

X

v=1

p_v

!^k−1

=O(1)

m

X

v=1

p_v|t_v|^k|λ_v|^k

m+1

X

n=v+1

P_n p_n

^δk−1 1 P_n−1

=O(1)

m

X

v=1

P_v p_v

δk−1

|t_v|^k|λ_v|^k

=O(1)

m

X

v=1

v^δk−1|t_v|^k|λ_v|^k

=O(1)

m

X

v=1

v^δk−1|t_v|^k =O(1) as m→ ∞, by virtue of the hypotheses of Theorem 2.1.

Also

m+1

X

n=2

P_n pn

^δk+k−1

|T_n,3|^k

≤

m+1

X

n=2

P_n p_n

δk−1

1 P_n−1^k

n−1

X

v=1

P_v|λ_v+1| v |t_v|

!k

=O(1)

m+1

X

n=2

P_n p_n

δk−1

1 P_n−1^k

n−1

X

v=1

vpv

|λ_v+1| v |tv|

!k

=O(1)

m+1

X

n=2

P_n p_n

δk−1

1 Pn−1

n−1

X

v=1

pv|λv+1|^k|tv|^k 1 Pn−1

n

X

v=1

pv

!^k−1

=O(1)

m

X

v=1

p_v|λ_v+1|^k|t_v|^k

m+1

X

n=v+1

Pn

p_n δk−1

1 Pn−1

=O(1)

m

X

v=1

|tv|^k Pv

p_v δk−1

|λv+1|^k

=O(1)

m

X

v=1

v^δk−1|t_v|^k =O(1) as m → ∞, by virtue of the hypotheses of Theorem 2.1.

Lastly

m

X

n=1

P_n p_n

δk+k−1

|T_n,4|^k=O(1)

m

X

n=1

P_n p_n

δk−1

|λ_n|^k|t_n|^k

=O(1)

m

X

n=1

P_n np_n

δk−1

|λn|^k|tn|^kn^δk−1

=O(1)

m

X

n=1

n^δk−1|t_n|^k =O(1) as m→ ∞, by virtue of the hypotheses of Theorem 2.1.

This completes the proof of Theorem 2.1.

Proof of Theorem 2.2. Let(T_n)denotes the( ¯N , p_n)mean of the seriesP

a_n. We have T_n= 1

Pn n

X

v=0

p_vs_v = 1 Pn

n

X

v=0

(P_n−P_v−1)a_v.

Since

T_n−T_n−1 = p_n PnPn−1

n

X

v=0

P_v−1a_v,

we have that

(2.7) an = −P_n

p_n ∆Tn−1+ P_n−2 pn−1

∆Tn−2.

Let

t_n= 1 n+ 1

n

X

v=1

va_vλ_v.

By using (2.7) we get t_n = 1

n+ 1

n

X

v=1

v

−P_v

p_v ∆T_v−1 +Pv−2

pv−1

∆T_v−2

λ_v

= 1 n+ 1

n−1

X

v=1

(−v)P_v

p_v∆T_v−1λ_v − nP_nλ_n

(n+ 1)p_n∆T_n−1 + 1 n+ 1

n

X

v=1

vPv−2

pv−1

∆T_v−2λ_v

= 1 n+ 1

n−1

X

v=1

(−v)P_v

p_v∆T_v−1λ_v + 1 n+ 1

n−1

X

v=1

(v + 1)P_v−1

p_v ∆T_v−1λ_v+1− nP_nλ_n

(n+ 1)p_n∆T_n−1

= 1 n+ 1

n−1

X

v=1

∆T_v−1

p_v {−vλvPv+ (v+ 1)λv+1Pv−1} − nP_nλ_n

(n+ 1)p_n∆Tn−1

= 1 n+ 1

n−1

X

v=1

∆T_v−1

p_v {−vλvPv+ (v+ 1)λv+1(Pv−pv)} − nP_nλ_n

(n+ 1)p_n∆Tn−1

= 1 n+ 1

n−1

X

v=1

∆Tv−1

p_v {−vλvPv+ (v+ 1)λv+1Pv−(v+ 1)λv+1pv} − nPnλn

(n+ 1)p_n∆Tn−1

= 1 n+ 1

n−1

X

v=1

∆Tv−1

p_v {−(∆vλ_v)P_v−(v+ 1)λ_v+1p_v} − nPnλn

(n+ 1)p_n∆Tn−1

= 1 n+ 1

n−1

X

v=1

−Pv

p_v ∆Tv−1{v∆λ_v −λ_v+1} − 1 n+ 1

n−1

X

v=1

∆Tv−1(v+ 1)λ_v+1

− nP_nλ_n

(n+ 1)p_n∆T_n−1

=− 1 n+ 1

n−1

X

v=1

vP_v

p_v ∆λ_v∆Tv−1 + 1 n+ 1

n−1

X

v=1

P_v

p_v∆Tv−1λ_v+1

− 1

n+ 1

n−1

X

v=1

(v+ 1)λ_v+1∆Tv−1− nP_nλ_n

(n+ 1)p_n∆Tn−1

=t_n,1+t_n,2+t_n,3+t_n,4, say.

Since

|t_n,1+t_n,2+t_n,3+t_n,4|^k ≤4^k(|t_n,1|^k+|t_n,2|^k+| |t_n,3|^k+|t4|^k),

to complete the proof of Theorem 2.2, it is enough to show that

∞

X

n=1

n^δk−1|t_n,r|^k <∞ f or r= 1,2,3,4.

Now, when k > 1applying Hölder’s inequality with indicesk and k⁰, where ¹_k + _k¹0 = 1, we have that

m+1

X

n=2

n^δk−1|t_n,1|^k ≤

m+1

X

n=2

n^δk−1 1 n^k

n−1

X

v=1

P_v

p_vv|∆λ_v| |∆T_v−1|

!^k

≤

m+1

X

n=2

1 n^2−δk

n−1

X

v=1

P_v p_v

k

|∆λ_v|^kv^k|∆Tv−1|^k 1 n

n−1

X

v=1

1

!^k−1

≤

m+1

X

n=2

1 n^2−δk

n−1

X

v=1

P_v p_v

k

|∆λ_v|^kv^k|∆Tv−1|^k

=O(1)

m

X

v=1

P_v pv

k

|∆λ_v|^kv^k|∆Tv−1|^k

m+1

X

n=v+1

1 n^2−δk

=O(1)

m

X

v=1

P_v p_v

k

|∆λ_v|^kv^k|∆Tv−1|^k 1 v^1−δk

=O(1)

m

X

v=1

P_v p_v

δk+k−1

|∆Tv−1|^k

=O(1) as m→ ∞, by virtue of the hypotheses of Theorem 2.2.

Again using Hölder’s inequality,

m+1

X

n=2

n^δk−1|t_n,2|^k ≤

m+1

X

n=2

n^δk−1−k

n−1

X

v=1

|λ_v+1|P_v

p_v |∆Tv−1|

!k

≤

m+1

X

n=2

1 n^2−δk

n−1

X

v=1

Pv

p_v k

|λv+1|^k|∆Tv−1|^k 1 n

n−1

X

v=1

1

!^k−1

=O(1)

m

X

v=1

Pv

p_v k

|λ_v+1|^k|∆Tv−1|^k

m+1

X

n=v+1

1 n^2−δk

=O(1)

m

X

v=1

Pv

p_v k

|λ_v+1|^k|∆Tv−1|^k 1 v^1−δk

=O(1)

m

X

v=1

P_v p_v

)^δk+k−1|∆Tv−1|^k

=O(1) as m→ ∞,

Also, we have that

m+1

X

n=2

n^δk−1|t_n,3|^k≤

m+1

X

n=2

n^δk−1−k

n−1

X

v=1

(v+ 1)|λ_v+1| |∆T_v−1|

!^k

=O(1)

m+1

X

n=2

n^δk−1−k

n−1

X

v=1

v|λ_v+1| |∆Tv−1|

!k

=O(1)

m+1

X

n=2

1 n^2−δk

n−1

X

v=1

v^k|λ_v+1|^k|∆Tv−1|^k 1 n

n−1

X

v=1

1

!^k−1

=O(1)

m+1

X

n=2

1 n^2−δk

n−1

X

v=1

v^k|λ_v+1|^k|∆Tv−1|^k

=O(1)

m

X

v=1

v^k|λ_v+1|^k|∆Tv−1|^k

m+1

X

n=v+1

1 n^2−δk

=O(1)

m

X

v=1

v^δk+k−1|∆Tv−1|^k

=O(1)

m

X

v=1

P_v pv

^δk+k−1

|∆Tv−1|^k

=O(1) as m→ ∞, by virtue of the hypotheses of Theorem 2.2.

Finally, we have that

m

X

n=1

n^δk−1|t_n,4|^k=O(1)

m

X

n=1

P_n p_n

k

n^δk−1|λ_n|^k|∆Tn−1|^k

=O(1)

m

X

n=1

Pn

p_n

δk+k−1

|∆Tn−1|^k

=O(1) as m→ ∞, by virtue of the hypotheses of Theorem 2.2.

Therefore, we get that

m

X

n=1

n^δk−1|t_n,r|^k=O(1) as m→ ∞, for r= 1,2,3,4.

This completes the proof of Theorem 2.2.

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