International Journal of Analysis and Applications Volume 17, Number 2 (2019), 226-233 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-17-2019-226 A NOTE ON GENERALIZED INDEXED NORLUND SUMMABILITY FACTOR OF AN INFINITE SERIES B. P. PADHY∗, P. TRIPATHY AND B. B. MISHRA Department of Mathematics, School of Applied Sciences, KIIT, Deemed to be University, Bhubaneswar-24, Odisha, India ∗Corresponding author: birupakhya.padhyfma@kiit.ac.in Abstract. In the present article, we have established a result on generalized indexed absolute Norlund summability factor by generalizing results of Mishra and Srivastava on indexed absolute Cesaro summabilty factors and Padhy et.al. on the absolute indexed Norlund summability. 1. Introduction In 1930, J.M.Whittaker [18] was the 1st to establish a result on the absolute summability of Fourier series and in 1932, M. Fekete [6] established a result on generalized indexed summability. Later on the researchers like Daniel [4] in 1964, Das [5] in 1966, Siya Ram [15] in 1969, Mazhar [11] in 1971, Mishra and Srivastava [13] in 1984, Sulaiman [16] in 2011 etc. have established results on indexed summability factors of an infinite series. Let ∑ an be a given infinite series with sequence of partial sums {sn} . Let tnα be the nth (C,α) mean (with order α > −1) of the sequence {sn} and is given by tαn = 1 Aαn n∑ k=0 Aα−1n−ksk, n ∈ N,where A α n = Γ(n + α + 1) Γ(α + 1)Γ(n + 1) , Received 2018-11-21; accepted 2018-12-18; published 2019-03-01. 2010 Mathematics Subject Classification. 40D15, 40F05, 40G99. Key words and phrases. absolute summability; summability factors; infinte series. c©2019 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 226 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-17-2019-226 Int. J. Anal. Appl. 17 (2) (2019) 227 then the series ∑ anis said to be summable |C,α|k,k ≥ 1, [7] if ∞∑ n=1 (n)k−1|tαn − t α n−1| k < ∞. Let tn be the nth (C, 1)- mean of the sequence {sn} and is given by tn = 1 n + 1 n∑ k=0 sk, then the series ∑ anis said to be summable |C, 1|k,k ≥ 1, [3] if ∞∑ n=1 (n)k−1|tn − tn−1|k < ∞. (1.1) Suppose {qn} be a sequence of real numbers with qn > 0, such that Qn = n∑ ν=0 qν →∞, as n →∞(Q−i = q−i = 0, i ≥ 1) (1.2) The sequence to sequence transformation Tn = 1 Qn n∑ ν=0 qn−νsν (1.3) defines the sequence {Tn} of the (N,qn)- means of the sequence {sn} generated by the sequence of coefficients {qn} . The series ∑ an is said to be summable |N,qn| if the sequence {Tn} is of bonded variation i.e; ∑ |Tn −Tn−1| is convergent. The series ∑ an is said to be summable |N,qn|k,k ≥ 1 ,if (see [8]) ∞∑ n=1 ( Qn qn )k−1 |Tn −Tn−1| k < ∞ (1.4) Clearly,|N,qn|k-summabiity is same as |C, 1|-summabiity,when qn = 1, for all values of n. Further any sequence {αn} of positive numbers the series ∑ an is said to be summable |N,qn,αn|k,k ≥ 1 if ∞∑ n=1 (αn) k−1|Tn −Tn−1| k < ∞ (1.5) and is said to be summable |N,qn,αn; δ|k,k ≥ 1,δ ≥ 0 if ∞∑ n=1 (αn) δk+k−1|Tn −Tn−1| k < ∞ (1.6) For any sequence {µn}, ∑∞ n=1 anµn is an infinite series. We define ∆µn = µn −µn−1, |∆µn| = |µn −µn−1| Also, for any sequence {µn}, by µn = O(n), we mean that the sequence {µnn } is bounded. Int. J. Anal. Appl. 17 (2) (2019) 228 2. Known Theorems Concerning with |C, 1| and |N,qn| summability Kishore [10] has proved the following theorem: Theorem 2.1. Let q0 > 0,qn ≥ 0 and (qn) be a non-decreasing sequence. If ∑ an is summable |C, 1| then the series ∑ anQn(n + 1) −1 is summable |N,qn|. Later on Ram [15] has proved the following theorem related to absolute Norlund factors of infinite series. Theorem 2.2. Let (qn) be a non-increasing sequence with q0 > 0,qn ≥ 0. If n∑ k=1 1 k |sk| = O(Yn) as n →∞; where (Yn) is a positive non-decreasing sequence and (µn) is a sequence such that ∞∑ n=1 n|∆2µn|Yn < ∞; |µn|Yn = O(1) as n →∞, then the series ∑ anQn(n + 1) −1 is summable |N,qn|. Also verma [17] has proved the following summability factor theorem: Theorem 2.3. Let (qn) be a non-increasing sequence with q0 > 0,qn ≥ 0. If ∑ an is summable |C, 1|k then the series ∑ anQn(n + 1) −1 is summable |N,qn|k,k ≥ 1. In 1984, Mishra and Srivatava [13] proved the following theorem for |C, 1|k summability. Theorem 2.4. Let (Yn) be a positive non-decreasing sequence and let there be sequnces {βn} and {µn} such that |∆µn| ≤ βn; (2.1) βn → 0 as n →∞; (2.2) |µn|Yn = O(1) as n →∞; (2.3) ∞∑ n=1 n|∆βn|Yn < ∞; (2.4) ∞∑ n=1 1 n |sn| k = O(Ym) as m →∞, (2.5) then the series ∑∞ n=1 anµn is summable |C, 1|k,k ≥ 1. Int. J. Anal. Appl. 17 (2) (2019) 229 Very recently, Padhy et al. [14] have proved a theorem on |N,qn|k-summability by extending theorem 2.4, in the following form: Theorem 2.5. Let for a positive non-decreasing sequence (Yn),there be sequences {βn} and {µn} satisfying the conditions 2.1 to 2.5 and {qn} be a sequence with {qn}∈ R+ such that Qn = O(nqn); (2.6) ∞∑ n=1 qn Qn |sn| k = O(Ym) as m →∞; (2.7) Qn−r−1 Qn = O ( qn−r−1 Qn Qr qr ) ; (2.8) m+1∑ n=r+1 ( Qn qn )k−1 qn−r Qn = O ( qr Qr ) , (2.9) then the series ∑∞ n=1 anµn is summable |N,qn|k,k ≥ 1,where 0 ≤ r ≤ n. It should be noted that if we take qn = 1∀n then condition 2.7 will be reduced to 2.5. In what follows, we have generalized known theorems 2.4 and 2.5 to |N,qn,αn; δ|k - summability in the form of the following theorem after studying [1] and [2] : 3. Main Theorem Theorem 3.1. Let (Yn) be a positive non-decreasing sequence and there be sequences {βn} and {µn} such that the conditions 2.1 to 2.5 are satisfied.Further let {qn} be a sequence of real numbers with qn > 0, such that Qn = O(nqn); (3.1) ∞∑ n=1 qn Qn |sn| k = O(Ym) as m →∞; (3.2) Qn−r−1 Qn = O ( qn−r−1 Qn Qr qr ) ; (3.3) m+1∑ n=r+1 (αn) δk+k−1 qn−r Qn = O ( qr Qr ) , (3.4) then the series ∑∞ n=1 anµn is summable |N,qn,αn; δ|k,k ≥ 1,δ ≥ 0. We require the below mentioned lemma to prove our main theorem: Int. J. Anal. Appl. 17 (2) (2019) 230 4. Lemma [5] Let (Yn) be a positive non decreasing sequence and there be sequences {βn} and {µn} such that the conditions 2.1 to 2.5 are satisfied.Then βnYn = O(1) as n →∞, (4.1) ∞∑ n=1 βnYn < ∞. (4.2) 5. Proof of the Main theorem Suppose (τn) refers to the (N,qn)- mean of the series ∑∞ n=1 anµn. Then by definition, we have τn = 1 Qn n∑ r=0 qn−r r∑ s=0 asµs = 1 Qn n∑ s=0 asµs n∑ r=s qn−r = 1 Qn n∑ s=0 asµsQn−s = 1 Qn n∑ r=0 arµrQn−r Thus τn − τn−1 = 1 Qn n∑ r=1 Qn−rarµr − 1 Qn−1 n−1∑ r=1 Qn−r−1arµr = n∑ r=1 ( Qn−r Qn − Qn−r−1 Qn−1 ) arµr = 1 QnQn−1 n∑ r=1 (Qn−rQn−1 −Qn−r−1Qn)arµr = 1 QnQn−1 [ n−1∑ r=1 ∆{(Qn−rQn−1 −Qn−r−1Qn) µr} ] n∑ ν=1 aν, with p0 = 0 = 1 QnQn−1 [ n−1∑ r=1 (qn−rQn−1 −qn−r−1Qn) µrsr + n−1∑ r=1 (Qn−r−1Qn−1 −Qn−r−2Qn) ∆µrsr ] (By Abel’s transformation) = Tn,1 + Tn,2 + Tn,3 + Tn,4 (say) Now, to show ∑∞ n=1 anµn is summable |N,qn,αn; δ|k,k ≥ 1,δ ≥ 0, by 1.6,we need to show that ∞∑ n=1 (αn) δk+k−1|τn − τn−1| k < ∞. Int. J. Anal. Appl. 17 (2) (2019) 231 i.e; to show that ∞∑ n=1 (αn) δk+k−1|Tn,1 + Tn,2 + Tn,3 + Tn,4| k < ∞. It will be enough to show that ∞∑ n=1 (αn) δk+k−1|Tn,j| k < ∞ for j = 1, 2, 3, 4. to establish the main theorem by using the inequality given by Minkowski. Now we have m+1∑ n=2 (αn) δk+k−1|Tn,1| k m+1∑ n=2 (αn) δk+k−1 | 1 QnQn−1 n−1∑ r=1 qn−rQn−1µrsr| k ≤ m+1∑ n=2 (αn) δk+k−1 1 Qn ( n−1∑ r=1 qn−r|µr| k|sr| k )( 1 Qn n−1∑ r=1 qn−r )k−1 (Using Holder′s inequality) = O(1) m∑ r=1 |µr| k|sr| k m+1∑ n=r+1 (αn) δk+k−1 ( qn−r Qn ) = O(1) m∑ r=1 |µr| k|sr| k qr Qr , by 3.4 = O(1) m∑ r=1 qr Qr |sr| k|µr||µr| k−1 = O(1) m−1∑ r=1 ∆|µr| r∑ w=1 qw Qw |sw| k + O(1)|µm| m∑ r=1 qr Qr |sr| k = O(1) m−1∑ r=1 |∆µr|Yr + O(1)|µm|Ym , by 3.2 = O(1), as m →∞ (By the lemma and 2.3) Next, m+1∑ n=2 (αn) δk+k−1 |Tn,2| k = m+1∑ n=1 (αn) δk+k−1 | 1 QnQn−1 n−1∑ r=1 qn−r−1Qnµrsr| k ≤ m+1∑ n=2 (αn) δk+k−1 1 Qn−1 ( n−1∑ r=1 qn−r−1|µr| k|sr| k )( 1 Qn−1 n−1∑ r=1 qn−r−1 )k−1 (5.1) Int. J. Anal. Appl. 17 (2) (2019) 232 = O(1) m∑ r=1 |µr| k|sr| k m+1∑ n=r+1 (αn) δk+k−1 ( qn−r−1 Qn−1 ) = O(1) m∑ r=1 |µr| k|sr| k qr Qr = O(1), as m →∞, As in proof of the 1st part. Further, m+1∑ n=2 (αn) δk+k−1 |Tn,3| k = m+1∑ n=1 (αn) δk+k−1 | 1 QnQn−1 n−1∑ r=1 Qn−r−1Qn−1∆µrsr| k ≤ m+1∑ n=2 (αn) δk+k−1 1 Qn ( n−1∑ r=1 Qn−r−1|∆µr||sr| k )( 1 Qn n−1∑ r=1 Qn−r−1|∆µr| )k−1 Since, ( 1 Qn n−1∑ r=1 Qn−r−1|∆µr| ) ≤ n−1∑ r=1 |∆µn| ≤ n|∆µr| ≤ nβn Therefore, m+1∑ n=2 (αn) δk+k−1 |Tn,3| k ≤ O(1) m∑ r=1 (rβr) k−1|∆µr||sr| k m+1∑ n=r+1 (αn) δk+k−1 Qn−r−1 Qn = O(1) m∑ r=1 |∆µr||sr| k qr Qr ≤ O(1) m∑ r=1 βr|sr| k qr Qr = O(1) m−1∑ r=1 ∆ (βr) r∑ w=1 qw Qw |sw| k + O(1)(βm) m∑ r=1 qr Qr |sr| k = O(1) m−1∑ r=1 |∆βr|Yr + O(1)(βm)Ym = O(1) as m →∞ Now, m+1∑ n=2 (αn) δk+k−1 |Tn,4| k = m+1∑ n=2 (αn) δk+k−1| 1 QnQn−1 n−1∑ r=1 Qn−r−2Qn∆µrsr| k (5.2) Int. J. Anal. Appl. 17 (2) (2019) 233 ≤ m+1∑ n=2 (αn) δk+k−1 1 Qn−1 ( n−1∑ r=1 Qn−r−2|∆µr||sr| k ) 1 Qn−1 n−1∑ r=1 Qn−r−2|∆µr| k−1 = O(1) m∑ r=1 (rβr) k−1|∆µr||sr| k m+1∑ n=r+1 (αn) δk+k−1 ( Qn−r−1 Qn ) , (as above) = O(1) m∑ r=1 |∆µr||sr| k qr Qr = O(1) as m →∞. (as above) This completes the proof of the theorem. 6. 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