Access the full text.
Sign up today, get DeepDyve free for 14 days.
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
You can share this free article with as many people as you like with the url below! We hope you enjoy this feature!
Read and print from thousands of top scholarly journals.
Already have an account? Log in
Bookmark this article. You can see your Bookmarks on your DeepDyve Library.
To save an article, log in first, or sign up for a DeepDyve account if you don’t already have one.
Copy and paste the desired citation format or use the link below to download a file formatted for EndNote
Access the full text.
Sign up today, get DeepDyve free for 14 days.
All DeepDyve websites use cookies to improve your online experience. They were placed on your computer when you launched this website. You can change your cookie settings through your browser.