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 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 λn for the series an λn /nann to be absolutely summable of order k ≥ 1 by A. 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, 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)
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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. Savas¸ 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
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
λ λi+1 λ ani i − an,i+1 si + ann n . aii i (i + 1)ai+1,i+1 nann
i=0
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
(8)
an,i+1 ani λi + Δ λi . iaii (i + 1)ai+1,i+1
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
(9)
Δi ani λi 1 1 = + an,i+1 λi − . iaii iaii (i + 1)ai+1,i+1 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 H¨older’s inequality and (iii),
I1 =
m+1 n =1
n
k −1
k n −1 k m+1 Δi ani k −1 Tn1 ≤ n λ s i i iaii n =1
i =0
n −1 m+1 k Δi ani λi si = O(1) n k −1 n =1
= O(1)
m+1 n =1
(12)
i =0
n k −1
n −1 n−1 k −1 k k Δi ani λi si Δi ani . i =0
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
λ i
i r =1
k sr
r
i =1
−
i −1 r =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. Savas¸ and B. E. Rhoades 5 = O(1)
m −1 i =1
= O(1)
m i=1
I2 =
m+1 n =1
n
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
i =0
k
n −1 m+1 1 k −1 = O(1) n an,i+1 λi Δ si . n =1
iaii
i =0
(15) Now
1 Δ 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
1 1 1 1 1 1 = − + − (i + 1) aii ai+1,i+1 aii i i + 1
(16)
1 1 1 Δ + . = (i + 1) aii iaii Thus using (iv) and (ii),
ai+1,i+1 − aii 1 1 1 1 1 1 + Δ = ≤ Δ + iaii i + 1 aii iaii i + 1 aii ai+1,i+1 iaii 1 O(1) + O(1) . = i+1
(17)
Hence, using H¨older’s inequality, (v) and (iii), I2 = O(1)
m+1 n =1
n
k −1
k
n −1 an,i+1 λi 1 si
i+1
i =0
k
n −1 m+1 k −1 = O(1) n an,i+1 aii λi si n =1
= O(1) = O(1)
m+1 n =1 m+1 n =1
i =0
n −1 n−1 k −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
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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), H¨older’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
k n −1 m+1 an,i+1 Δλi si = O(1) n k −1 n =1
i=0
n =1
i =0
n −1 k m+1 aii k −1 = O(1) n an,i+1 Δλi si
= O(1)
m+1 n =1
n
k −1
n −1 i =0
aii
n−1 k−1 an,i+1 k k aii Δλi si aii an,i+1
akii
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
i=0
akii
an,i+1 Δλi k si k 1 aii k
aii
m m+1 aii Δλi k si k an,i+1 k i =0
aii
n=i+1
m Δλi k−1 Δλi si k = O(1) i =0
= O(1)
m i =0
aii
m Δλi si k = O(1) si k βi . i =0
i=0
(21)
E. Savas¸ 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.
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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. 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:
[email protected];
[email protected] B. E. Rhoades: Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA Email address:
[email protected]
