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Annals of the Alexandru Ioan Cuza University - Mathematics
, Volume 61 (1) – Jan 1, 2015

/lp/de-gruyter/a-new-application-of-almost-increasing-sequences-wzyxqYinw1

- Publisher
- de Gruyter
- Copyright
- Copyright © 2015 by the
- ISSN
- 1221-8421
- eISSN
- 1221-8421
- DOI
- 10.2478/aicu-2013-0049
- Publisher site
- See Article on Publisher Site

Bor h proved a main theorem dealing with |N , pn |k summability factors of infinite series. In this paper, we have generalized this theorem to the - |A, pn |k summability factors, under weaker conditions by using an almost increing sequence instead of a positive monotonic non-decreing sequence. Mathematics Subject Clsification 2010: 40D25, 40F05, 40G99. Key words: absolute matrix summability, almost increing sequences, infinite series. 1. Introduction A positive sequence (bn ) is said to be almost increing if there exists a positive increing sequence (cn ) and two positive constants A and B such that Acn bn Bcn (see [1]). Let an be a given infinite series with the partial sums (sn ). Let (pn ) be a sequence of positive numbers such that (1) Pn = pv n , (P-i = p-i = 0, i 1) . The sequence-to-sequence transformation (2) 1 n = Pn pv s v ¯ defines the sequence (n ) of the N , pn mean of the sequence (sn ), generated by the sequence of coefficients (pn ) (see [5]). The series an is said ¯ to be summable N , pn k , k 1, if (see [2]) (3) Pn pn k-1 |n - |k < . Let A = (anv ) be a normal matrix, i.e., a lower triangular matrix of nonzero diagonal entries. Then A defines the sequence-to-sequence transformation, mapping the sequence s = (sn ) to = (An (s)), where (4) The series (5) An (s) = anv sv , n = 0, 1, ... an is said to be summable |A, pn |k , k 1, if (see [7]) Pn pn k-1 ¯ |An (s)|k < , ¯ where An (s) = An (s) - A (s). Let (n ) be any sequence of positive real numbers. We say that the series an is summable - |A, pn |k , k 1, if (6) k-1 ¯ n An (s) < . n If we take n = Pn , then - |A, pn |k summability reduces to |A, pn |k p p n ¯ summability. Also, if we take n = Pn and anv = Pv , then we get N , pn k p n pv summability. Furthermore, if we take n = n and anv = Pn and pn = 1 for all values of n, then - |A, pn |k summability reduces to |C, 1|k summability (see [4]). Before stating the main theorem we must first introduce some further notations. Given a normal matrix A = (anv ), we sociate two lower semi matrices ¯ ^ A = (¯nv ) and A = (^nv ) follows: a a (7) (8) anv = ¯ i=v ani , n, v = 0, 1, ... anv = anv - a,v , ^ ¯ ¯ n = 1, 2, ... a00 = a00 = a00 , ^ ¯ ¯ ^ It may be noted that A and A are the well-known matrices of series-tosequence and series-to-series transformations, respectively. Then, we have n n (9) (10) An (s) = n anv sv = anv av , ¯ ¯ An (s) = anv av . ^ 2. Known result ¯ Bor [3] h proved the following theorem for N , pn summability method. Theorem A. Let (pn ) be a sequence of positive numbers such that (11) Pn = O(npn ) n . If (Xn ) is a positive monotonic non-decreing sequence such that (12) (13) m m Xm = O(1) nXn |2 n | = O(1), pn |tn |k = O(Xm ) Pn (14) 1 where tn = n+1 n v=1 vav then the series ¯ an n is summable |N , pn |k , k1. 3. The main result The aim of this paper is to generalize Theorem A to - |A, pn |k summability. Now we shall prove the following theorem. Theorem. Let A = (anv ) be a positive normal matrix such that (15) (16) (17) (18) an0 = 1, ¯ n = 0, 1, ..., f or n v + 1, a,v anv , ann = O pn , Pn |^n,v+1 | = O (v |v anv |) . a ^ np Let (Xn ) be an almost increing sequence and ( Pnn ) be a non-increing sequence. If conditions (11)-(13) of Theorem A and (19) k-1 n pn Pn |tn |k = O(Xm ) are satisfied, then the series an n is summable - |A, pn |k , k 1. It should be noted that if we take (Xn ) a positive monotonic nonp n decreing sequence, anv = Pv and n = Pn in this theorem, then we get p n Theorem A. In this ce, condition (19) reduces to condition (14) and the np condition "( Pnn ) is a non-increing sequence" is automatically satisfied. We require the following lemma for the proof of the theorem. Lemma ([6]). Under the conditions on (Xn ) and (n ) which are taken in the statement of our theorem, then we have the following: (20) (21) nXn |n | = O(1) n , Xn |n | < . 4. Proof of the theorem Let (Tn ) denotes A-transform of the series (10), we have n n an n . Then, by (9) and ¯ Tn = v=1 anv av v = ^ v=1 anv v ^ vav . v Applying Abel's transformation to this sum, we get ¯ Tn v=1 anv v ^ v rar + r=1 ann n ^ n rar r=1 v+1 n+1 ann n tn + v (^nv ) v tv a n v v=1 v=1 v+1 1 an,v+1 v tv + ^ an,v+1 v+1 tv ^ v v v=1 say. Tn,1 + Tn,2 + Tn,3 + Tn,4 , To prove the theorem, by Minkowski's inequality, it is sufficient to show k-1 that n |Tn,r |k < , for r = 1, 2, 3, 4. Firstly, we have that m k-1 n |Tn,1 |k = m m k-1 n n+1 ann n tn k k k-1 n |n |k-1 |n ||tn |k m-1 n = O(1) k k-1 n |n ||tn |k m pn Pn |n | m-1 r=1 k-1 r pr Pr |tr | + O(1)|m | v=1 k-1 v |tv |k = O(1) m-1 |n |Xn + O(1)|m |Xm |n |Xn + O(1)|m |Xm = O(1) = O(1) by virtue of the hypotheses of the theorem and lemma. Applying H¨lder's inequality with indices k and k , where k > 1 and o 1 1 + k = 1, in Tn,1 , we have that k m+1 k-1 n |Tn,2 |k = +1 n=2 k-1 n +1 +1 k-1 n v=1 v+1 v (^nv )v tv a v k k k-1 |v (^nv )||v | |tv | a k-1 v=1 |v (^nv )| a k-1 |v (^nv )||v |k |tv |k a +1 k n=v+1 = O(1) |v | |v ||tv | k-1 |v (^nv )| a = O(1) |v | |tv |k v k-1 m+1 |v (^nv )| a n=v+1 k = O(1) v=1 k-1 |v | |tv |k v = O(1) by virtue of the hypotheses of the theorem and lemma. 1 Now, since v|v | = O Xv = O(1), by (20), we have that m+1 k-1 n |Tn,3 |k +1 m+1 k-1 n n=2 k-1 n +1 v=1 k-1 n +1 v=1 v=1 v+1 an,v+1 v tv ^ v k k-1 = O(1) |^n,v+1 ||v ||tv | a v=1 |^n,v+1 ||v | a k-1 v|v anv ||v ||tv | ^ k-1 v=1 v|v anv ||v | ^ v|v anv ||v ||tv |k ^ +1 n=v+1 = O(1) v|v ||tv |k v|v ||tv |k k-1 |v anv | ^ |v anv | ^ = O(1) v=1 v k-1 m+1 n=v+1 k-1 v|v ||tv |k v -1 v k-1 r | (v|v |)| r=1 m k-1 v -1 pr Pr |tr |k + O(1)m|m | = O(1) v=1 |tv |k m-1 vXv | v | + O(1) v=1 Xv |v+1 | + O(1)m|m |Xm = O(1) by virtue of the hypotheses of the theorem and lemma. Finally, using the fact Pn = O(npn ), by (11), in Tn,1 , we have that m+1 k-1 n |Tn,4 |k +1 k-1 n n=2 v=1 1 an,v+1 v+1 tv ^ v m+1 k-1 n +1 v=1 k-1 n +1 v=1 k-1 = O(1) 1 |^n,v+1 ||v+1 |k |tv |k a v 1 |^n,v+1 ||v+1 |k |tv |k a v v=1 v=1 1 |^n,v+1 | a v |v anv | ^ k-1 v=1 k-1 1 |^n,v+1 ||v+1 |k |tv |k a v m+1 n=v+1 = O(1) 1 |v+1 |k-1 |v+1 ||tv |k v 1 |v+1 ||tv |k v 1 |v+1 ||tv |k v v v Pv v Pv k-1 |^n,v+1 | a k-1 m+1 = O(1) |^n,v+1 | a n=v+1 k-1 = O(1) = O(1) k-1 |v+1 ||tv |k v = O(1) v=1 k-1 |v+1 ||tv |k v = O(1) by virtue of the hypotheses of the theorem and lemma. This completes the proof of the theorem. Corollary 1. If we take n = |A, pn |k summability. Corollary 2. If we take anv = ¯ with |N , pn , n |k summability. Pn pn , then we get a result concerning the pv Pn , then we have another a result dealing p Corollary 3. If we take anv = Pv and pn = 1 for all values of n, then n we get a result dealing with |C, 1, n |k summability. p Corollary 4. If we take n = n, anv = Pv and pn = 1 for all values of n n, then we get a result for |C, 1|k summability. p Corollary 5. If we take k = 1 and anv = Pv , then we get a result n np ¯ for |N , pn | summability and in this ce the condition "( Pnn ) is a nonincreing sequence" is not needed.

Annals of the Alexandru Ioan Cuza University - Mathematics – de Gruyter

**Published: ** Jan 1, 2015

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