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: 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 . 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. 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. 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