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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2007, Article ID 86095, 8 pages
doi:10.1155/2007/86095
Research Article
ANoteon
|A|
k
Summability Factors for Infinite Series
Ekrem Savas¸ and B. E. Rhoades
Received 9 November 2006; Accepted 29 March 2007
Recommended by Martin J. Bohner
We obtain sufficient conditions on a nonnegative lower triangular matrix A and a se-
quence λ
n
for the series
a
n
λ
n
/na
nn
to be absolutely summable of order k ≥ 1byA.
Copyright © 2007 E. Savas¸ and B. E. Rhoades. This is an open access article distributed
under the Creative Commons Attribution License, which permits unrestricted use, dis-
tribution, and reproduction in any medium, provided the original work is properly cited.
A weighted mean matrix, denoted by (
N, p
n
), is a lower triangular matrix with entries
p
k
/P
n
,where{p
k
} is a nonnegative sequence with p
0
> 0, and P
n
:=
n
k
=0
p
k
.
Mishra and Srivastava [1]obtainedsufficient conditions on a sequence
{p
k
} and a
sequence
{λ
n
} for the series
a
n
P
n
λ
n
/np
n
to be absolutely summable by the weighted
mean matrix (
N, p
n
). Bor [2] extended this result to absolute summability of order k ≥ 1.
Unfortunately, an incorrect definition of absolute summability was used.
In this note, we establish the corresponding result for a nonnegative triangle, using
the correct definition of absolute summability of order k
≥ 1, (see [3]). As a corollary, we
obtain the corrected version of Bor’s result.
Let A be an infinite lower triangular matr ix. We may associate with A two lower trian-
gular matrices
A and
A, whose entries are defined by
a
nk
=
n
i=k
a
ni
, a
nk
= a
nk
− a
n−1,k
,(1)
respectively. The motivation for these definitions will become clear as we proceed.
Let A be an infinite matrix. The series
a
k
is said to be absolutely summable by A,of
order k
≥ 1, written as |A|
k
,if
∞
k=0
n
k−1
Δt
n−1
k
< ∞,(2)
2 Journal of Inequalities and Applications
where Δ is the forward differ ence operator and t
n
denotes the nth term of the matrix
transform of the sequence
{s
n
},wheres
n
:=
n
k
=0
a
k
.
Thus
t
n
=
n
k=0
a
nk
s
k
=
n
k=0
a
nk
k
ν=0
a
ν
=
n
ν=0
a
ν
n
k=ν
a
nk
=
n
ν=0
a
nν
a
ν
,
t
n
− t
n−1
=
n
ν=0
a
nν
a
ν
−
n−1
ν=0
a
n−1,ν
a
ν
=
n
ν=0
a
nν
a
ν
,
(3)
since
a
n−1,n
= 0.
Theresulttobeprovedisthefollowing.
Theorem 1. Let A be a triangle with nonnegative entries satisfy ing
(i)
a
n0
= 1, n = 0,1, ,
(ii) a
n−1,ν
≥ a
nν
for n ≥ ν +1,
(iii) na
nn
O(1),
(iv) Δ(1/a
nn
) = O(1),
(v)
n
ν
=0
a
νν
|a
n,ν+1
|=O(a
nn
).
If
{X
n
} is a positive nondecreasing sequence and the sequences {λ
n
} and {β
n
} satisfy
(vi)
|Δλ
n
|≤β
n
,
(vii) limβ
n
= 0,
(viii)
|λ
n
|X
n
= O(1),
(ix)
∞
n=1
nX
n
|Δβ
n
| < ∞,
(x) T
n
:=
n
ν
=1
(|s
ν
|
k
/ν) = O(X
n
),
then the series
∞
ν=1
a
n
λ
n
/na
nn
is summable |A|
k
, k ≥ 1.
The proof of the theorem requires the following lemma.
Lemma 2 (see Mishra and Srivastava [1]). Let
{X
n
} be a positive nondecreasing sequence
and the sequences
{β
n
}, {λ
n
} satisfy conditions (vi)–(ix) of Theorem 1. Then
nX
n
β
n
= O(1), (4)
∞
n=1
β
n
X
n
< ∞. (5)
Since {X
n
} is nondecreasing, X
n
≥ X
0
, which is a positive constant. Hence condition
(viii) implies that λ
n
is bounded. It also follows from (4)thatβ
n
= O(1/n), and hence that
Δλ
n
= O(1/n) by condition (iv).
Proof. Let T
n
denote the nth term of the A-transform of the series
(a
n
λ
n
)/(na
nn
). Then
we may write
T
n
=
n
ν=0
a
nν
ν
i=0
a
i
λ
i
a
ii
i
=
m
i=0
a
i
λ
i
a
ii
i
n
ν=i
a
nν
=
n
i=0
a
ni
a
i
λ
i
a
ii
i
. (6)
E. Savas¸ and B. E. Rhoades 3
Thus,
T
n
− T
n−1
=
n
i=0
a
ni
a
i
λ
i
a
ii
i
−
n−1
i=0
a
n−1,i
a
i
λ
i
a
ii
i
=
n
i=0
a
ni
− a
n−1,i
a
i
λ
i
a
ii
i
=
n
i=0
a
ni
a
i
λ
i
a
ii
i
=
n
i=0
a
ni
λ
i
a
ii
i
s
i
− s
i−1
=
n−1
i=0
a
ni
λ
i
a
ii
i
s
i
+ a
nn
λ
n
a
nn
n
s
n
−
n
i=0
a
ni
λ
i
s
i−1
a
ii
i
=
n−1
i=0
a
ni
λ
i
a
ii
i
s
i
+ a
nn
λ
n
a
nn
n
s
n
−
n−1
i=0
a
n,i+1
λ
i+1
s
i
(i +1)a
i+1,i+1
=
n
i=0
a
ni
λ
i
a
ii
i
− a
n,i+1
λ
i+1
(i +1)a
i+1,i+1
s
i
+ a
nn
λ
n
na
nn
.
(7)
We may wr i te
a
ni
λ
i
ia
ii
−
a
n,i+1
λ
i+1
(i +1)a
i+1,i+1
=
a
ni
λ
i
ia
ii
−
a
n,i+1
λ
i+1
(i +1)a
i+1,i+1
+
a
n,i+1
λ
i
(i +1)a
i+1,i+1
−
a
n,i+1
λ
i
(i +1)a
i+1,i+1
= Δ
i
a
ni
ia
ii
λ
i
+
a
n,i+1
(i +1)a
i+1,i+1
Δ
λ
i
.
(8)
Also we may write
Δ
i
a
ni
ia
ii
λ
i
=
a
ni
ia
ii
λ
i
−
a
n,i+1
(i +1)a
i+1,i+1
λ
i
−
a
n,i+1
ia
ii
λ
i
+
a
n,i+1
ia
ii
λ
i
=
Δ
i
a
ni
λ
i
ia
ii
+ a
n,i+1
λ
i
1
ia
ii
−
1
(i +1)a
i+1,i+1
.
(9)
Hence,
T
n
− T
n−1
=
n−1
i=0
Δ
i
a
ni
ia
ii
λ
i
s
i
+
n−1
i=0
a
n,i+1
λ
i
1
ia
ii
−
1
(i +1)a
i+1,i+1
s
i
+
n−1
i=0
a
n,i+1
(i +1)a
i+1,i+1
Δ
i
(λ
i
)s
i
+
λ
n
n
s
n
= T
n1
+ T
n2
+ T
n3
+ T
n4
,say.
(10)
To finish the proof of the theorem, it will be sufficient to show that
∞
n=1
n
k−1
T
nr
k
< ∞,forr = 1,2,3,4. (11)
4 Journal of Inequalities and Applications
Using H
¨
older’s inequality and (iii),
I
1
=
m+1
n=1
n
k−1
T
n1
k
≤
m+1
n=1
n
k−1
n−1
i=0
Δ
i
a
ni
ia
ii
λ
i
s
i
k
= O(1)
m+1
n=1
n
k−1
n−1
i=0
Δ
i
a
ni
λ
i
s
i
k
= O(1)
m+1
n=1
n
k−1
n−1
i=0
Δ
i
a
ni
λ
i
k
s
i
k
n−1
i=0
Δ
i
a
ni
k−1
.
(12)
But using (ii),
Δ
i
a
ni
=
a
ni
− a
n,i+1
= a
ni
− a
n−1,i
− a
n,i+1
+ a
n−1,i+1
= a
ni
− a
n−1,i
≤ 0.
(13)
Thus using (i),
n−1
i=0
Δ
i
a
ni
=
n−1
i=0
a
n−1,i
− a
ni
=
1 − 1+a
nn
= a
nn
. (14)
From (viii), it follows that λ
n
= O(1). Using (iii), (vi), (x), and property (5)of
Lemma 2,
I
1
= O(1)
m+1
n=1
na
nn
k−1
n
−1
i=0
λ
i
k
s
i
k
Δ
i
a
ni
=
O(1)
m+1
n=1
na
nn
k−1
n−1
i=0
λ
i
k−1
λ
i
Δ
i
a
ni
s
i
k
=
O(1)
m
i=0
λ
i
s
i
k
m+1
n=i+1
na
nn
k−1
Δ
i
a
ni
=
O(1)
m
i=0
λ
i
s
i
k
a
ii
=
λ
0
s
0
k
a
00
+ O(1)
m
i=1
λ
i
s
i
k
i
= O(1) + O(1)
m
i=1
λ
i
i
r=1
s
r
k
r
−
i−1
r=1
s
r
k
r
=
O(1)
m
i=1
λ
i
i
r=1
s
r
k
r
−
m−1
j=0
λ
j+1
j
r=1
s
r
k
r
=
O(1)
m−1
i=1
Δ
λ
i
i
r=1
1
r
s
r
k
+ O(1)
λ
m
m
i=1
s
i
k
i
E. Savas¸ and B. E. Rhoades 5
= O(1)
m−1
i=1
Δ
λ
i
X
i
+ O(1)
λ
m
X
m
= O(1)
m
i=1
β
i
X
i
+ O(1)
λ
m
X
m
= O(1),
I
2
=
m+1
n=1
n
k−1
T
n2
k
=
m+1
n=1
n
k−1
n−1
i=0
a
n,i+1
λ
i
Δ
1
ia
ii
s
i
k
= O(1)
m+1
n=1
n
k−1
n−1
i=0
a
n,i+1
λ
i
Δ
1
ia
ii
s
i
k
.
(15)
Now
Δ
1
ia
ii
=
1
ia
ii
−
1
(i +1)a
i+1,i+1
=
1
ia
ii
−
1
(i +1)a
i+1,i+1
+
1
(i +1)a
ii
−
1
(i +1)a
ii
=
1
(i +1)
1
a
ii
−
1
a
i+1,i+1
+
1
a
ii
1
i
−
1
i +1
=
1
(i +1)
Δ
1
a
ii
+
1
ia
ii
.
(16)
Thus using (iv) and (ii),
Δ
1
ia
ii
=
1
i +1
Δ
1
a
ii
+
1
ia
ii
≤
1
i +1
a
i+1,i+1
− a
ii
a
ii
a
i+1,i+1
+
1
ia
ii
=
1
i +1
O(1) + O(1)
.
(17)
Hence, using H
¨
older’s inequality, (v) and (iii),
I
2
= O(1)
m+1
n=1
n
k−1
n−1
i=0
a
n,i+1
λ
i
1
i +1
s
i
k
= O(1)
m+1
n=1
n
k−1
n−1
i=0
a
n,i+1
a
ii
λ
i
s
i
k
= O(1)
m+1
n=1
n
k−1
n−1
i=0
a
n,i+1
a
ii
λ
i
k
s
i
k
n−1
i=0
a
ii
a
n,i+1
k−1
= O(1)
m+1
n=1
na
nn
k−1
n
−1
i=0
a
n,i+1
a
ii
λ
i
k
s
i
k
6 Journal of Inequalities and Applications
= O(1)
m
i=0
λ
i
k
s
i
k
a
ii
m+1
n=i+1
na
nn
k−1
a
n,i+1
=
O(1)
m
i=0
λ
i
k
s
i
k
a
ii
m+1
n=i+1
a
n,i+1
.
(18)
From [4],
m+1
n=i+1
a
n,i+1
≤
1. (19)
Hence,
I
2
= O(1)
m
i=1
λ
i
k
s
i
k
a
ii
= O(1)
m
i=1
λ
i
λ
i
k−1
s
i
k
1
i
=
m
i=1
λ
i
s
i
k
i
= O(1), (20)
as in the proof of I
1
.
Using (iii), H
¨
older’s inequalit y, and (v),
I
3
=
m+1
n=1
n
k−1
T
n3
k
=
m+1
n=1
n
k−1
n−1
i=0
a
n,i+1
Δλ
i
s
i
(i +1)a
i+1,i+1
k
= O(1)
m+1
n=1
n
k−1
n−1
i=0
a
n,i+1
Δλ
i
s
i
k
= O(1)
m+1
n=1
n
k−1
n−1
i=0
a
ii
a
ii
a
n,i+1
Δλ
i
s
i
k
= O(1)
m+1
n=1
n
k−1
n−1
i=0
a
ii
a
n,i+1
a
k
ii
Δλ
i
k
s
i
k
n−1
i=0
a
ii
a
n,i+1
k−1
= O(1)
m+1
n=1
na
nn
k−1
n
−1
i=0
a
ii
a
n,i+1
a
k
ii
Δλ
i
k
s
i
k
= O(1)
m+1
n=1
n
−1
i=0
a
n,i+1
Δλ
i
k
s
i
k
1
a
k
ii
a
ii
= O(1)
m
i=0
a
ii
a
k
ii
Δλ
i
k
s
i
k
m+1
n=i+1
a
n,i+1
=
O(1)
m
i=0
Δλ
i
a
ii
k−1
Δλ
i
s
i
k
= O(1)
m
i=0
Δλ
i
s
i
k
= O(1)
m
i=0
s
i
k
β
i
.
(21)
E. Savas¸ and B. E. Rhoades 7
Since
|s
i
|
k
= i(T
i
− T
i−1
)by(x),wehave
I
3
= O(1)
m
i=1
i
T
i
− T
i−1
β
i
. (22)
Using Abel’s transformation, (vi), and (5),
I
3
= O(1)
m−1
i=1
T
i
Δ
iβ
i
+ O(1)mT
n
β
n
= O(1)
m−1
i=1
i
Δβ
i
X
i
+ O(1)
m−1
i=1
X
i
β
i
+ O(1)mX
n
β
n
= O(1).
(23)
Using (viii) and (x),
I
4
=
m+1
n=1
n
k−1
T
n4
k
=
m+1
n=1
n
k−1
s
n
λ
n
n
k
=
m+1
n=1
s
n
k
λ
n
k
1
n
=
m+1
n=1
s
n
k
n
λ
n
λ
n
k−1
= O(1),
(24)
as in the proof of I
1
.
Corollary 3. Let {p
n
} be a positive sequence such that P
n
=
n
k
=0
p
k
→∞and satisfies
(i) np
n
O(P
n
);
(ii) Δ(P
n
/p
n
) = O(1).
If
{X
n
} is a positive nondecreasing sequence and the sequences {λ
n
} and {β
n
} are such that
(iii)
|Δλ
n
|≤β
n
,
(iv) β
n
→ 0 as n →∞,
(v)
|λ
n
|X
n
= O(1) as n →∞,
(vi)
∞
n=1
nX
n
|Δβ
n
| < ∞,
(vii) T
n
=
n
ν
=1
|s
ν
|
k
/ν = O(X
n
),
then the series
(a
n
P
n
λ
n
)/(np
n
) is summable |N, p
n
|
k
, k ≥ 1.
Proof. Conditions (iii)–(vii) of Corollary 3 are, respectively, conditions (vi)–(x) of Theo-
rem 1 .
Conditions (i), (ii), and (v) of Theorem 1 are automatically satisfied for any weighted
mean method. Conditions (iii) and (iv) of Theorem 1 become, respectively, conditions
(i) and (ii) of Corollary 3.
Acknowledgment
The first author received support from the Scientific and Technical Research Council of
Turkey.
8 Journal of Inequalities and Applications
References
[1] K. N. Mishra and R. S. L. Srivastava, “On |
––
N
, p
n
| summability factors of infinite series,” Indian
Journal of Pure and Applied Mathematics, vol. 15, no. 6, pp. 651–656, 1984.
[2] H.Bor,“Anoteon
|
––
N
, p
n
|
k
summability factors of infinite series,” Indian Journal of Pure and
Applied Mathematics, vol. 18, no. 4, pp. 330–336, 1987.
[3] T. M. Flett, “On an extension of absolute summability and some theorems of Littlewood and
Paley,” Proceedings of the London Mathematical Society. Third Series, vol. 7, pp. 113–141, 1957.
[4] B. E. Rhoades and E. Savas¸, “A note on absolute summability factors,” Periodica Mathematica
Hungarica, vol. 51, no. 1, pp. 53–60, 2005.
Ekrem Savas¸: Department of Mathematics, Faculty of Sciences and Arts, Istanbul Ticaret University,
Uskudar, 34672 Istanbul, Turkey
Email addresses: ;
B. E. Rhoades: Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA
Email address:
