Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2007, Article ID 86095, 8 pages
doi:10.1155/2007/86095
Research Article
A Note on |A|k Summability Factors for Infinite Series
Ekrem Savaş 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 λn for the series an λn /nann to be absolutely summable of order k ≥ 1 by A.
Copyright © 2007 E. Savaş and B. E. Rhoades. This is an open access article distributed
under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
A weighted mean matrix, denoted by (N, pn ), is a lower triangularmatrix with entries
pk /Pn , where { pk } is a nonnegative sequence with p0 > 0, and Pn := nk=0 pk .
Mishra and Srivastava [1]obtained sufficient conditions on a sequence { pk } and a
sequence {λn } for the series an Pn λn /npn to be absolutely summable by the weighted
mean matrix (N, pn ). 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 matrix. We may associate with A two lower trian whose entries are defined by
gular matrices A and A,
ank =
n
ani ,
ank = ank − an−1,k ,
(1)
i =k
respectively. The motivation for these definitions
will become clear as we proceed.
Let A be an infinite matrix. The series ak is said to be absolutely summable by A, of
order k ≥ 1, written as |A|k , if
∞
k =0
k
nk−1 Δtn−1 < ∞,
(2)
2
Journal of Inequalities and Applications
where Δ is the forward difference operator
and tn denotes the nth term of the matrix
transform of the sequence {sn }, where sn := nk=0 ak .
Thus
tn =
n
ank sk =
k =0
n
ank
k =0
tn − tn−1 =
n
ν =0
k
ν =0
aν =
anν aν −
n
−1
ν =0
n
ν =0
aν
n
ank =
k =ν
an−1,ν aν =
n
ν =0
n
ν =0
anν aν ,
(3)
anν aν ,
since an−1,n = 0.
The result to be proved is the following.
Theorem 1. Let A be a triangle with nonnegative entries satisfying
(i) an0 = 1, n = 0,1,...,
(ii) an−1,ν ≥ anν for n ≥ ν + 1,
(iii) nann O(1),
(iv) Δ(1/ann ) = O(1),
(v) nν=0 aνν |an,ν+1 | = O(ann ).
If {Xn } is a positive nondecreasing sequence and the sequences {λn } and {βn } satisfy
(vi) |Δλn | ≤ βn ,
(vii) limβn = 0,
|λn |Xn = O(1),
(viii) |Δβn | < ∞,
(ix) ∞
n=1 nX
n
(x) Tn := nν=1 (|sν |k /ν) = O(Xn ),
then the series ∞
ν=1 an λn /nann is summable |A|k , k ≥ 1.
The proof of the theorem requires the following lemma.
Lemma 2 (see Mishra and Srivastava [1]). Let {Xn } be a positive nondecreasing sequence
and the sequences {βn }, {λn } satisfy conditions (vi)–(ix) of Theorem 1. Then
nXn βn = O(1),
∞
n =1
(4)
βn Xn < ∞.
(5)
Since {Xn } is nondecreasing, Xn ≥ X0 , which is a positive constant. Hence condition
(viii) implies that λn is bounded. It also follows from (4) that βn = O(1/n), and hence that
Δλn = O(1/n) by condition (iv).
Proof. Let Tn denote the nth term of the A-transform of the series
we may write
Tn =
n
ν =0
anν
ν
ai λ i
i =0
aii i
=
m
n
ai λ i i =0
aii i
ν =i
anν =
n
i=0
ani
(an λn )/(nann ). Then
ai λ i
.
aii i
(6)
E. Savaş and B. E. Rhoades 3
Thus,
Tn − Tn−1 =
=
n
i=0
n
i=0
=
=
n
n
n
aλ
a λ
ai λ i a λ −
an−1,i i i =
ani − an−1,i i i = ani i i
aii i i=0
aii i i=0
aii i i=0 aii i
ani
λs
λi λ
λ
si − si−1 =
ani i si + ann n sn − ani i i−1
aii i
aii i
ann n
aii i
i =0
i=0
n
−1
i =0
n −1
ani
ani
n −1
n
−1
(7)
λi
λ
λi+1 si
si + ann n sn − an,i+1
aii i
ann n
(i
+
1)ai+1,i+1
i =0
i=0
n
ani
λi
λi+1
λ
− an,i+1
si + ann n .
aii i
(i + 1)ai+1,i+1
nann
We may write
an,i+1 λi+1
an,i+1 λi+1
an,i+1 λi
an,i+1 λi
ani λi
ani λi
−
=
−
+
−
iaii
(i + 1)ai+1,i+1
iaii
(i + 1)ai+1,i+1 (i + 1)ai+1,i+1 (i + 1)ai+1,i+1
= Δi
an,i+1
ani
λi +
Δ λi .
iaii
(i + 1)ai+1,i+1
(8)
Also we may write
Δi
an,i+1
a
a
ani
a
λi = ni λi −
λi − n,i+1 λi + n,i+1 λi
iaii
iaii
(i + 1)ai+1,i+1
iaii
iaii
Δi ani λi
1
1
=
+ an,i+1 λi
−
.
iaii
iaii (i + 1)ai+1,i+1
(9)
Hence,
Tn − Tn−1 =
n
−1
i=0
n
−1
Δi ani
1
1
λi si + an,i+1 λi
−
si
iaii
ia
(i
+
1)a
ii
i+1,i+1
i=0
+
n
−1
i =0
an,i+1
λ
Δi (λi )si + n sn
(i + 1)ai+1,i+1
n
= Tn1 + Tn2 + Tn3 + Tn4 ,
(10)
say.
To finish the proof of the theorem, it will be sufficient to show that
∞
n =1
k
nk−1 Tnr < ∞,
for r = 1,2,3,4.
(11)
4
Journal of Inequalities and Applications
Using Hölder’s inequality and (iii),
I1 =
m+1
n =1
n
k −1 n −1 k m+1
Δi ani
k −1
Tn1 ≤
n
λ
s
i
i
iaii
n =1
i =0
n −1
m+1
Δi ani λi si = O(1)
n k −1
n =1
= O(1)
m+1
n =1
k
k
(12)
i =0
n k −1
n −1
n −1
k k
Δi ani λi si Δi ani i =0
k −1
.
i=0
But using (ii),
Δi ani = ani − an,i+1 = ani − an−1,i − an,i+1 + an−1,i+1 = ani − an−1,i ≤ 0.
(13)
Thus using (i),
n
−1
i=0
−1
n
Δi ani =
an−1,i − ani = 1 − 1 + ann = ann .
(14)
i =0
From (viii), it follows that λn = O(1). Using (iii), (vi), (x), and property (5) of
Lemma 2,
I1 = O(1)
= O(1)
m+1
nann
n =1
m+1
−1
k k k−1 n
λi si Δi ani i =0
−1
k−1 k k−1 n
λi λi Δi ani si nann
n =1
i=0
m
k m+1
k−1 Δi ani nann
= O(1) λi si i=0
= O(1)
m
i=0
n=i+1
m λi si k
k
λi si aii = λ0 s0 k a00 + O(1)
= O(1) + O(1)
m
i =1
i
λ i r =1
k
sr r
−
i −1
r =1
i =1
k
sr r
j m
i −1
sr k m
sr k
λi
−
λ j+1
= O(1)
i=1
= O(1)
m
−1
i =1
r =1
Δ λ i r
j =0
r =1
r
i
m si k
1
sr k + O(1)λm r =1
r
i=1
i
i
E. Savaş and B. E. Rhoades 5
= O(1)
m
−1
i =1
= O(1)
m
i=1
I2 =
m+1
n =1
n
= O(1)
Δ λi Xi + O(1)λm Xm
βi Xi + O(1)λm Xm = O(1),
k −1 n −1
k
k m+1
1
k −1 Tn2 =
n a
n,i+1 λi Δ
s
i
iaii n =1
m+1
n =1
n
k −1
n
−1
i =0
i =0
an,i+1 λi Δ 1
iaii
si k
.
(15)
Now
Δ
1
iaii
=
1
1
−
iaii (i + 1)ai+1,i+1
=
1
1
1
1
−
+
−
iaii (i + 1)ai+1,i+1 (i + 1)aii (i + 1)aii
(16)
1 1
1
1
1
1
=
−
+
−
(i + 1) aii ai+1,i+1
aii i i + 1
1
1
1
Δ
+
.
=
(i + 1)
aii
iaii
Thus using (iv) and (ii),
1
Δ
iaii
1
1
=
Δ
i+1
aii
+
1
≤
i+1
1
iaii
ai+1,i+1 − aii 1
+
aii ai+1,i+1 iaii
1 O(1) + O(1) .
=
i+1
(17)
Hence, using Hölder’s inequality, (v) and (iii),
I2 = O(1)
= O(1)
= O(1)
= O(1)
m+1
n =1
m+1
n =1
m+1
n =1
m+1
n =1
n
k −1
n
−1
i =0
n
k −1
n
−1
i =0
an,i+1 λi 1 si k
i+1
an,i+1 aii λi si k
n −1
n −1
k k
k −1
n
an,i+1 aii λi si
aii an,i+1 i =0
nann
k−1
i =0
−1
n
i =0
an,i+1 aii λi k si k
k −1
6
Journal of Inequalities and Applications
= O(1)
m+1
m
k k
k−1 λi si aii
an,i+1 nann
i =0
= O(1)
n=i+1
m
m+1
k k
λi si aii
an,i+1 .
i =0
n=i+1
(18)
From [4],
m+1
n=i+1
an,i+1 ≤ 1.
(19)
Hence,
I2 = O(1)
m
m
m
k k
k
λi si aii = O(1) λi λi k−1 si k 1 = λi si = O(1),
i=1
i
i =1
i=1
i
(20)
as in the proof of I1 .
Using (iii), Hölder’s inequality, and (v),
I3 =
m+1
n =1
n
k −1 n −1
k
a
k m+1
Δλ
si n,i+1
i
k
−
1
Tn3 =
n (i + 1)ai+1,i+1 n =1
i =0
n −1
m+1
an,i+1 Δλi si = O(1)
n k −1
n =1
= O(1)
= O(1)
m+1
n =1
m+1
n =1
k
i=0
n
k −1
n
−1
i =0
n
k −1
n
−1
i =0
aii an,i+1 Δλi si aii
k
n −1
k−1
an,i+1 k k
aii
Δλi si
aii an,i+1
k
aii
m+1
−1
an,i+1 k−1 n
Δλi k si k
= O(1)
nann
aii
n =1
= O(1)
= O(1)
m+1
−1
n
n=1 i=0
an,i+1 Δλi k si k 1 aii
k
aii
aii
m Δλi i =0
= O(1)
akii
m
m+1
aii Δλi k si k
an,i+1 k
i =0
= O(1)
i=0
m
i =0
aii
n=i+1
k −1
Δλi si k
m
Δλi si k = O(1) si k βi .
i =0
i=0
(21)
E. Savaş and B. E. Rhoades 7
Since |si |k = i(Ti − Ti−1 ) by (x), we have
I3 = O(1)
m
i=1
i Ti − Ti−1 βi .
(22)
Using Abel’s transformation, (vi), and (5),
I3 = O(1)
m
−1
i=1
= O(1)
m
−1
i=1
Ti Δ iβi + O(1)mTn βn
m
−1
iΔβi Xi + O(1)
Xi βi + O(1)mXn βn = O(1).
(23)
i =1
Using (viii) and (x),
I4 =
=
m+1
n =1
m+1
n =1
as in the proof of I1 .
n
k −1 s λ k m+1
k k 1
k m+1
n n
k −1 sn λn =
Tn4 =
n n n
n =1
n =1
(24)
k
sn k−1
λn λn = O(1),
n
n
Corollary 3. Let { pn } be a positive sequence such that Pn = k=0 pk → ∞ and satisfies
(i) npn O(Pn );
(ii) Δ(Pn / pn ) = O(1).
If {Xn } is a positive nondecreasing sequence and the sequences {λn } and {βn } are such that
(iii) |Δλn | ≤ βn ,
(iv) βn → 0 as n → ∞,
|λn |Xn = O(1) as n → ∞,
(v) nXn |Δβn | < ∞,
(vi) ∞
n =1 k
(vii) Tn = nν
=1 |sν | /ν = O(Xn ),
then the series (an Pn λn )/(npn ) is summable |N, pn |k , k ≥ 1.
Proof. Conditions (iii)–(vii) of Corollary 3 are, respectively, conditions (vi)–(x) of Theorem 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, pn | summability factors of infinite series,” Indian
Journal of Pure and Applied Mathematics, vol. 15, no. 6, pp. 651–656, 1984.
––
[2] H. Bor, “A note on | N, pn |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. Savaş, “A note on absolute summability factors,” Periodica Mathematica
Hungarica, vol. 51, no. 1, pp. 53–60, 2005.
Ekrem Savaş: Department of Mathematics, Faculty of Sciences and Arts, Istanbul Ticaret University,
Uskudar, 34672 Istanbul, Turkey
Email addresses: ekremsavas@yahoo.com; esavas@iticu.edu.tr
B. E. Rhoades: Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA
Email address: rhoades@indiana.edu