In this paper, a general theorem on $|A,delta|_k$- summability factors of infinite series has been proved under weaker conditions

Ekrem Savaş

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MATHEMATICAL COMMUNICATIONS 323Math. Commun., Vol. 15, No. 2, pp. 323-329 (2010)A note on generalized absolute summability factorsEkrem Savaş1,∗1 Department of Mathematics, Istambul Ticaret University, Üküdar 36 472, Istambul,TurkeyReceived April 22, 2009; accepted November 15, 2009Abstract. In this paper, a general theorem on |A, δ|k- summability factors of infinite serieshas been proved under weaker conditions.AMS subject classifications: 40F05, 40D25Key words: absolute summability, summability factors, almost increasing sequence1. IntroductionRhoades and Savas [4] recently have obtained sufficient conditions for the series∑anλn to be absolutely summable of order k by a triangular matrix.In this paper we generalize the result of Rhoades and Savas under weaker condi-tions for |A, δ|k, k ≥ 1, 0 ≤ δ < 1/k.A positive sequence {bn} is said to be almost increasing if there exists an in-creasing sequence {cn} and positive constants A and B such that Acn ≤ bn ≤ Bcn,(see, [1]). Obviously every increasing sequence is almost increasing. However, theconverse need not be true as can be seen by taking the example, say bn = e(−1)nn.Let A be a lower triangular matrix, {sn} a sequence. ThenAn :=n∑ν=0anνsν .A series∑an is said to be summable |A|k, k ≥ 1 if∞∑n=1nk−1|An −An−1|k < ∞. (1)and it is said to be summable |A, δ|k, k ≥ 1 and δ ≥ 0 if (see,[2])∞∑n=1nδk+k−1|An −An−1|k < ∞. (2)∗Corresponding author. Email address: ekremsavas@yahoo.com (E. Savaş)http://www.mathos.hr/mc c©2010 Department of Mathematics, University of Osijek324 E. SavaşWe may associate with A two lower triangular matrices A and Â defined asfollows:ānν =n∑r=νanr, n, ν = 0, 1, 2, . . . ,andânν = ānν − ān−1,ν , n = 1, 2, 3, . . . .With sn :=∑ni=0 λiai.yn :=n∑i=0anisi =n∑i=0anii∑ν=0λνaν=n∑ν=0λνaνn∑i=νani =n∑ν=0ānνλνaνandYn := yn − yn−1 =n∑ν=0(ānν − ān−1,ν)λνaν =n∑ν=0ânνλνaν . (3)Theorem 1. Let A be a lower triangular matrix satisfying(i) ān0 = 1, n = 0, 1, . . . ,,(ii) an−1,ν ≥ anν for n ≥ ν + 1, and(iii) nann ³ O(1)(iv)n−1∑ν=1aνν |ânν+1| = O(ann),(v)m+1∑n=ν+1nδk|∆ν ânν | = O(νδkaνν)and(vi)m+1∑n=ν+1nδk|ânν+1| = O(νδk).If {Xn} is an almost increasing sequence such that(vii) λmXm = O(1),(viii)m∑n=1(nXn)|∆2λn| = O(1), and(ix)m∑n=1nδkann|tn|k = O(Xm), where tn := 1n + 1n∑k=1kak,then the series∑anλn is summable |A, δ|k, k ≥ 1, 0 ≤ δ < 1/k.Absolute summability methods 325Lemma 1 (see [4]). If (Xn) is an almost increasing sequence, then under the con-ditions of the theorem we have that(i)∞∑n=1Xn|∆λn| < ∞ and(ii) nXn|∆λn| = O(1).Proof. From (3) we may writeYn =n∑ν=1( ânνλνν)νaν=n∑ν=1( ânνλνν)[ ν∑r=1rar −ν−1∑r=1rar]=n−1∑ν=1∆ν( ânνλνν) ν∑r=1rar +ânnλnnn∑ν=1νaν=n−1∑ν=1(∆ν ânν)λνν + 1νtν +n−1∑ν=1ân,ν+1(∆λν)ν + 1νtν+n−1∑ν=1ân,ν+1λν+11νtν +(n + 1)annλntnn= Tn1 + Tn2 + Tn3 + Tn4, say.To finish the proof it is sufficient, by Minkowski’s inequality, to show that∞∑n=1nδk+k−1|Tnr|k < ∞, for r = 1, 2, 3, 4.Using Hölder’s inequality and (iii),I1 :=m∑n=1nδk+k−1|Tn1|k =m∑n=1nδk+k−1∣∣∣n−1∑ν=1∆ν ânνλνν + 1νtν∣∣∣k= O(1)m+1∑n=1nδk+k−1( n−1∑ν=1|∆ν ânν ||λν ||tν |)k= O(1)m+1∑n=1nδk+k−1( n−1∑ν=1|∆ν ânν ||λν |k|tν |k)( n−1∑ν=1|∆ν ânν |)k−1.Using the fact that, from (vii), {λn} is bounded, and condition (i) of Lemma 1,326 E. Savaşand (v)I1 = O(1)m+1∑n=1nδk(nann)k−1n−1∑ν=1|λν |k|tν |k|∆ν ânν |= O(1)m+1∑n=1nδk(nann)k−1( n−1∑ν=1|λν |k−1|λν ||∆ν ânν ||tν |k)= O(1)m∑ν=1|λν ||tν |km+1∑n=ν+1nδk(nann)k−1|∆ν ânν |= O(1)m∑ν=1|λν ||tν |km+1∑n=ν+1nδk|∆ν ânν |= O(1)m∑ν=1νδk|λν |aνν |tν |k= O(1)m∑ν=1|λν |[ ν∑r=1arr|tr|krδk −ν−1∑r=1arr|tr|krδk]= O(1)[ m−1∑ν=1∆(|λν |)ν∑r=1arr|tr|krδk + |λm|m∑r=1arr|tr|krδk]= O(1)m−1∑ν=1|∆λν |Xν + O(1)|λm|Xm= O(1).Using Hölder’s inequality, (iii), and (iv),I2 :=m+1∑n=2nδk+k−1|Tn2|k =m+1∑n=2nδk+k−1∣∣∣n−1∑ν=1ân,ν+1(∆λν)ν + 1νtν∣∣∣k≤m+1∑n=2nδk+k−1[ n−1∑ν=1|ân,ν+1||∆λν |ν + 1ν|tν |]k= O(1)m+1∑n=2nδk+k−1[ n−1∑ν=1|ân,ν+1||∆λν ||tν |]k= O(1)m+1∑n=2nδk+k−1[ n−1∑ν=1(ν)|∆λν ||tν |aνν |ân,ν+1|]k= O(1)m+1∑n=2nδk+k−1n−1∑ν=1(ν|∆λν |)k|tν |kaνν |ân,ν+1|]×[ n−1∑ν=1aνν |ân,ν+1|]k−1Absolute summability methods 327= O(1)m+1∑n=2nδk(nann)k−1n−1∑ν=1(ν|∆λν |)k|tν |kaνν |ân,ν+1|= O(1)m+1∑n=2nδk(nann)k−1n−1∑ν=1(ν|∆λν |)k−1(ν|∆λν |)aνν |ân,ν+1||tν |kConclusion (ii) of Lemma 1 implies that ν|∆λν | = O(1). Therefore, using (iii),(v) and (vi)I2 := O(1)m∑ν=1ν|∆λν |aνν |tν |km+1∑n=ν+1nδk(nann)k−1|âνν+1|= O(1)m∑ν=1ν|∆λν |aνν |tν |km+1∑n=ν+1nδk|ân,ν+1|.Therefore,I2 := O1)m∑ν=1νδkν|∆λν |aνν |tν |k.Using summation by parts and (ix),I2 = O(1)m∑ν=1ν|∆λν |[ ν∑r=1arr|tr|krδk −ν−1∑r=1arr|tr|krδk]= O(1)m−1∑ν=1|∆(ν∆λν)|Xν + O(1).But∆(ν∆λν) = ν∆λν − (ν + 1)∆λν+1 = ν∆2λν −∆λν+1.Using (viii) and property (i) from Lemma 1, and the fact that {Xn} is almostincreasing,I2 = O(1)m−1∑ν=1ν|∆2λν |Xν + O(1)m−1∑ν=1|∆λν+1|Xν+1 = O(1).Using (iii), Hölder’s inequality, (iv), summation by parts, property (i) of Lemma 1,(vi), (vii) and (ix)m+1∑n=2nδk+k−1|Tn3|k =m+1∑n=2nδk+k−1∣∣∣n−1∑ν=1ân,ν+1λν+11νtν∣∣∣k≤m+1∑n=2nδk+k−1[ n−1∑ν=1|λν+1| ân,ν+1ν|tν |]k= O(1)m+1∑n=2nδk+k−1[ n−1∑ν=1|λν+1||ân,ν+1||tν |aνν]k328 E. Savaş= O(1)m+1∑n=2nδk+k−1[ n−1∑ν=1|λν+1|kaνν |tν |k|ân,ν+1|]×[ n−1∑ν=1aνν |ân,ν+1|]k−1= O(1)m+1∑n=2nδk(nann)k−1n−1∑ν=1|λν+1|k−1|λν+1|aνν |tν |k|ân,ν+1|= O(1)m∑ν=1|λν+1||tν |km+1∑n=ν+1nδk|ân,ν+1|= O(1)m∑ν=1|λν+1|aνν |tν |kνδk= O(1)m∑ν=1|λν+1|[ ν∑r=1arr|tr|krδk −ν−1∑r=1arr|tr|krδk]= O(1)[ m−1∑ν=1|∆λν+1|ν∑r=1arr|tr|krδk + |λm+1|ν∑r=1arr|tr|krδk]= O(1)m−1∑ν=1|∆λν+1|Xν + O(1)|λν+1|Xm= O(1).Finally, using (iii), summation by parts, property (i) of Lemma 1 and (vii),m∑n=1nδk+k−1|Tn4|k =m∑n=1nδk+k−1∣∣∣ (n + 1)annλntnn∣∣∣k= O(1)m∑n=1nδk+k−1|ann|k|λn|k|tn|k= O(1)m∑n=1nδk(nann)k−1ann|λn|k−1|λn||tn|k= O(1)m∑n=1nδkann|λn||tn|k,as in the proof of I1.Setting δ = 0 in the theorem yields the following corollary.Corollary 1. Let A be a triangle satisfying conditions (i)-(iv) of Theorem 1 and let{Xn} be an almost increasing sequence satisfying conditions (vii)-(viii). If(ix)∑mn=1 ann|tn|k = O(Xm),then the series∑anλn is summable |A|k, k ≥ 1.Absolute summability methods 329Corollary 2. Let {pn} be a positive sequence such thatPn :=∑nk=0 pk →∞, and satisfies(i) npn ³ O(Pn),(ii)m+1∑n=ν+1nδk| pnPnPn−1| = O(νδkPν).If {Xn} is an almost increasing sequence such that(iii) λmXm = O(1),(iv)m∑n=1nXn|∆2λn| = O(1), and(v)∞∑n=1nδk−1|tn|k = O(Xm),then the series∑anλn is summable |N̄ , p, δ|k, k ≥ 1 for 0 ≤ δ < 1/k.Proof. Conditions (iii) and (iv) of Corollary 2 are conditions (vii) and (viii) ofTheorem 1, respectively.Conditions (i), (ii) and (iv) of Theorem 1 are automatically satisfied for anyweighted mean method. Condition (iii) and (ix) of Theorem 1 become conditions (i)and (v) of Corollary 2 and conditions (v) and (vi) of Theorem 1 become condition(ii) of Corollary 2.AcknowledgementThe author wishes to thank the referees for their careful reading of the manuscriptand their helpful suggestions.References[1] S.Alijančić, D.Arendelović, O−regularly varying functions, Publ. Inst. Math.(Beograd) (N.S) 36(1977), 5–22.[2] T.M.Fleet, On an extension of absolute summability and some theorems of Little-wood and Paley, Proc. London Math. Soc. 3(1957), 113–141.[3] B.E.Rhoades, Inclusion theorems for absolute matrix summability methods, J. Math.Anal. Appl. 118(1993), 71–75.[4] B.E.Rhoades, E. Savaş, A note on absolute summability factors, Periodica Math.Hung. 51(2005), 53–60.[5] B.E.Rhoades, E. Savaş, On |A|k summability factors, Acta Math. Hung. 112(2006),15–23.

A note on generalized absolute summability factors

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