Quasiβ-Power Increasing Sequences Santosh Kr. Saxena vol. 10, iss. 2, art. 56, 2009
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ON QUASI β -POWER INCREASING SEQUENCES
SANTOSH Kr. SAXENA
H. N. 419, Jawaharpuri, Badaun Department of Mathematics Teerthanker Mahaveer University Moradabad, U.P., India
EMail:ssumath@yahoo.co.in
Received: 31 January, 2008
Accepted: 15 May, 2009
Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: 40D05, 40F05.
Key words: Absolute Summability, Summability Factors, Infinite Series.
Abstract: In this paper we prove a general theorem on
N , p¯ αn;δ
k summability, which generalizes a theorem of Özarslan [6] on
N , p¯ n;δ
ksummability, under weaker conditions and by using quasiβ-power increasing sequences instead of almost increasing sequences.
Acknowledgements: The author wishes to express his sincerest thanks to Dr. Rajiv Sinha and the referees for their valuable suggestions for the improvement of this paper.
Quasiβ-Power Increasing Sequences Santosh Kr. Saxena vol. 10, iss. 2, art. 56, 2009
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Contents
1 Introduction 3
2 Main Result 6
Quasiβ-Power Increasing Sequences Santosh Kr. Saxena vol. 10, iss. 2, art. 56, 2009
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1. Introduction
A positive sequence(γn)is said to be a quasiβ-power increasing sequence if there exists a constantK =K(β, γ)≥1such that
(1.1) Knβγn≥mβγm
holds for alln≥m ≥1. It should be noted that every almost increasing sequence is a quasiβ-power increasing sequence for any non-negative β, but the converse need not be true as can be seen by taking the example, sayγn = n−β forβ > 0. So we are weakening the hypotheses of the theorem of Özarslan [6], replacing an almost increasing sequence by a quasiβ-power increasing sequence.
Let P
an be a given infinite series with partial sums (sn) and let(pn) be a se- quence withp0 >0,pn≥0forn >0andPn=Pn
ν=0pν. We define (1.2) pαn =
n
X
ν=0
Aα−1n−νpν, Pnα =
n
X
ν=0
pαν, P−iα =pα−i = 0, i≥1 ,
where
Aα0 = 1, Aαn = (α+ 1) (α+ 2)· · ·(α+n)
n! , (α >−1, n= 1,2,3, ...) (1.3)
The sequence-to-sequence transformation
(1.4) Unα = 1
Pnα
n
X
ν=0
pανsν
defines the sequence(Unα)of the N , p¯ αn
mean of the sequence(sn), generated by the sequence of coefficients(pαn)(see [7]).
Quasiβ-Power Increasing Sequences Santosh Kr. Saxena vol. 10, iss. 2, art. 56, 2009
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The seriesP
anis said to be summable N , p¯ αn
k,k ≥1, if (see [2]) (1.5)
∞
X
n=1
Pnα pαn
k−1
Unα−Un−1α
k <∞,
and it is said to be summable
N , p¯ αn;δ
k,k ≥1andδ ≥0,if (see [7]) (1.6)
∞
X
n=1
Pnα pαn
δk+k−1
Unα−Un−1α
k <∞.
In the special case when δ = 0, α = 0 (respectively, pn = 1 for all values of n)
N , p¯ αn;δ
ksummability is the same as N , p¯ n
k(respectively|C,1;δ|k) summability.
Mishra and Srivastava [4] proved the following theorem for|C,1|ksummability.
Later on Bor [3] generalized the theorem of Mishra and Srivastava [4] for N , p¯ n
k
summability.
Quite recently Özarslan [6] has generalized the theorem of Bor [3] under weaker conditions. For this, Özarslan [6] used the concept of almost increasing sequences.
A positive sequence (bn) is said to be almost increasing if there exists a positive increasing sequence(cn)and two positive constantsAandBsuch thatAcn ≤bn ≤ Bcn(see [1]). Obviously every increasing sequence is an almost increasing sequence but the converse needs not be true as can be seen from the examplebn=ne(−1)n. Theorem 1.1. Let(Xn)be an almost increasing sequence and the sequences (ρn) and(λn)such that the conditions
|∆λn| ≤ρn, (1.7)
ρn→0 as n→ ∞, (1.8)
|λn|Xn=O(1), as n→ ∞, (1.9)
Quasiβ-Power Increasing Sequences Santosh Kr. Saxena vol. 10, iss. 2, art. 56, 2009
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∞
X
n=1
n|∆ρn|Xn<∞.
(1.10)
are satisfied. If(pn)is a sequence such that the condition
(1.11) Pn=O(npn), as n→ ∞,
is satisfied and
(1.12)
m
X
n=1
Pn pn
δk−1
|sn|k=O(Xm), as m→ ∞,
(1.13)
∞
X
n=ν+1
Pn pn
δk−1 1
Pn−1 =O (
Pν pν
δk
1 Pν
) ,
then the seriesP
anλnis summable
N , p¯ n;δ
kfork ≥1and0≤δ < 1k.
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2. Main Result
The aim of this paper is to generalize Theorem1.1for
N , p¯ αn;δ
ksummability under weaker conditions by using quasi β-power increasing sequences instead of almost increasing sequences. Now, we will prove the following theorem.
Theorem 2.1. Let(Xn)be a quasiβ-power increasing sequence for some0< β <1 and the sequences(ρn)and(λn)such that the conditions (1.7) – (1.10) of Theorem 1.1are satisfied. If pαn
is a sequence such that
(2.1) Pnα=O(npαn), as n→ ∞, is satisfied and
(2.2)
m
X
n=1
Pnα pαn
δk−1
|sn|k=O(Xm), as m→ ∞,
(2.3)
∞
X
n=ν+1
Pnα pαn
δk−1
1
Pn−1α =O (
Pνα pαν
δk
1 Pνα
) , then the seriesP
anλnis summable
N , p¯ αn;δ
kfork≥1and0≤δ < k1.
Remark 1. It may be noted that, if we take(Xn)as an almost increasing sequence and α = 0 in Theorem2.1, then we get Theorem1.1. In this case, conditions (2.1) and (2.2) reduce to conditions (1.11) and (1.12) respectively and condition (2.3) reduces to (1.13). If additionallyδ = 0, relation (2.3) reduces to
(2.4)
∞
X
n=ν+1
pn PnPn−1
=O 1
Pν
, which always holds.
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We need the following lemma for the proof of our theorem.
Lemma 2.2 ([5]). Under the conditions on (Xn), (βn) and (λn) as taken in the statement of the theorem, the following conditions hold
(2.5) nρnXn=O(1), as n→ ∞,
(2.6)
∞
X
n=1
ρnXn <∞.
Proof of Theorem2.1. Let(Tnα)be the N , p¯ αn
mean of the seriesP
anλn.Then by definition, we have
Tnα= 1 Pnα
n
X
ν=0
pαν
ν
X
w=0
awλw = 1 Pnα
n
X
ν=0
Pnα−Pν−1α aνλν.
Then, forn≥1, we get
Tnα−Tn−1α = pαn PnαPn−1α
n
X
ν=1
Pν−1α aνλν.
Applying Abel’s transformation, we have Tnα−Tn−1α = pαn
PnαPn−1α
n−1
X
ν=1
∆ Pν−1α λν
sν + pαn Pnαsnλn
=− pαn PnαPn−1α
n−1
X
ν=1
pανsνλν + pαn PnαPn−1α
n−1
X
ν=1
Pναsν∆λν + pαn Pnαsnλn
=Tn,1α +Tn,2α +Tn,3α , say.
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Since
Tn,1α +Tn,2α +Tn,3α
k≤3k Tn,1α
k+ Tn,2α
k+ Tn,3α
k , to complete the proof of Theorem2.1, it is sufficient to show that
∞
X
n=1
Pnα pαn
δk+k−1
Tn,wα
k <∞, for w= 1,2,3.
Now, whenk >1,applying Hölder’s inequality with indiceskandk0, wherek1+k10 = 1, and using|λn|=O
1 Xn
=O(1), by (1.9), we have
m+1
X
n=2
Pnα pαn
δk+k−1
Tn,1α
k =
m+1
X
n=2
Pnα pαn
δk+k−1
pαn PnαPn−1α
n−1
X
ν=1
pανsνλν
k
≤
m+1
X
n=2
Pnα pαn
δk−1
1 Pn−1α
n−1
X
ν=1
pαν |sν|k|λν|k 1 Pn−1α
n−1
X
ν=1
pαν
!k−1
=O(1)
m
X
ν=1
pαν|sν|k|λν|k
m+1
X
n=ν+1
Pnα pαn
δk−1 1 Pn−1α
=O(1)
m
X
ν=1
pαν|sν|k|λν|k Pνα
pαν δk
1 Pνα
=O(1)
m
X
ν=1
Pνα pαν
δk−1
|sν|k|λν|k
=O(1)
m
X
ν=1
Pνα pαν
δk−1
|sν|k|λν| |λν|k−1
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=O(1)
m
X
ν=1
Pνα pαν
δk−1
|sν|k|λν|
=O(1)
m−1
X
ν=1
∆|λν|
ν
X
u=1
Puα pαu
δk−1
|su|k+ O(1)|λm|
m
X
ν=1
Pνα pαν
δk−1
|sν|k
=O(1)
m−1
X
ν=1
|∆λν|Xν +O(1)|λm|Xm
=O(1)
m−1
X
ν=1
ρνXν +O(1)|λm|Xm
=O(1), as m→ ∞,
by virtue of the hypotheses of Theorem2.1and Lemma2.2. Sinceνρν =O
1 Xν
= O(1),by (2.5), using the fact that|∆λn| ≤ ρnby (1.7) andPnα =O(npαn)by (2.1) and after applying the Hölder’s inequality again, we obtain
m+1
X
n=2
Pnα pαn
δk+k−1
Tn,2α
k
≤
m+1
X
n=2
Pnα pαn
δk−1 1 Pn−1α
k(n−1 X
ν=1
Pνα|∆λν| |sν| )k
≤
m+1
X
n=2
Pnα pαn
δk−1 1 Pn−1α
(n−1 X
ν=1
pαν (νρν)k|sν|k
) ( 1 Pn−1α
n−1
X
ν=1
pαν )k−1
=O(1)
m
X
ν=1
pαν (νρν)k|sν|k
m+1
X
n=ν+1
Pnα pαn
δk−1
1 Pn−1α
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=O(1)
m
X
ν=1
Pνα pαν
δk−1
(νρν)k|sν|k
=O(1)
m−1
X
ν=1
∆ (νρν)
ν
X
w=1
Pwα pαw
δk−1
|sw|k+O(1)mρm
m
X
ν=1
Pνα pαν
δk−1
|sν|k
=O(1)
m−1
X
ν=1
|∆ (νρν)|Xν +O(1)mρmXm
=O(1)
m−1
X
ν=1
ν|∆ρν|Xν+O(1)
m−1
X
ν=1
ρν+1Xν+1+O(1)mρmXm
=O(1), as m → ∞,
by the virtue of the hypotheses of Theorem 2.1and Lemma 2.2. Finally, using the fact thatPnα =O(npαn),by (2.1) as inTn,1α , we have
m
X
n=1
Pnα pαn
δk+k−1
Tn,3α
k =O(1)
m
X
n=1
Pnα pαn
δk−1
|sn|k|λn|
=O(1), asm → ∞.
Therefore, we get
∞
X
n=1
Pnα pαn
δk+k−1
Tn,wα
k=O(1), as m → ∞, for w= 1,2,3.
This completes the proof of Theorem2.1.
If we takepn = 1andα= 0for all values ofnin Theorem2.1, then we obtain a result concerning the|C,1, δ|k summability.
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[7] S.K. SAXENAANDS.K. SAXENA, A note on
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