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KYUNGPOOK Math. J. 2019; 59(2): 233-240

Published online June 23, 2019

Copyright © Kyungpook Mathematical Journal.

An Application of Absolute Matrix Summability using Almost Increasing and δ-quasi-monotone Sequences

Hikmet Seyhan Özarslan

Department of Mathematics, Erciyes University, 38039 Kayseri, Turkey
e-mail : seyhan@erciyes.edu.tr and hseyhan38@gmail.com

Received: October 31, 2017; Revised: August 9, 2018; Accepted: August 13, 2018

In the present paper, absolute matrix summability of infinite series is studied. A new theorem concerning absolute matrix summability factors, which generalizes a known theorem dealing with absolute Riesz summability factors of infinite series, is proved using almost increasing and δ-quasi-monotone sequences. Also, a result dealing with absolute Cesàro summability is given.

Keywords: summability factors, almost increasing sequences, absolute matrix summability, quasi-monotone sequences, infinite series, Ho¨lder inequality, Minkowski inequality.

A positive sequence (vn) is said to be almost increasing if there exists a positive increasing sequence (cn) and two positive constants K and L such that KcnvnLcn (see [1]). A sequence (yn) is said to be δ-quasi-monotone, if yn → 0, yn > 0 ultimately and Δyn ≥ −δn, where Δyn=yn – yn+1 and δ = (δn) is a sequence of positive numbers (see [2]). Let ∑an be a given infinite series with partial sums (sn). By (un) and (tn) we denote the n-th (C, 1) means of the sequences (sn) and (nan), respectively. The series ∑an is said to be |C, 1|k summable, k ≥ 1, if (see [6], [8])

n=1nk-1un-un-1k=n=11ntnk<.

Let (pn) be a sequence of positive numbers such that

Pn=v=0npv         as         n,         (P-i=p-i=0,         i1).

The sequence-to-sequence transformation

zn=1Pnv=0npvsv

defines the sequence (zn) of the Riesz mean of the sequence (sn), generated by the sequence of coefficients (pn) (see [7]). The series ∑an is said to be |N̄, pn|k summable, k ≥ 1, if (see [3])

n=1(Pnpn)k-1Δzn-1k<,

where

Δzn-1=-pnPnPn-1v=1nPv-1av,         n1.

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 As = (An(s)), where

An(s)=i=0nanisi,         n=0,1,

The series ∑an is said to be |A, pn|k summable, k ≥ 1, if (see [9])

n=1(Pnpn)k-1Δ¯An(s)k<,

where

Δ¯An(s)=An(s)-An-1(s).

When we take anv=pvPn, then |A, pn|k summability is the same as |N̄, pn|k summability. Also, when we take anv=pvPn and pn = 1 for all values of n, |A, pn|k reduces to |C, 1|k summability.

Let A = (anv) be a normal matrix. Lower semimatrices Ā = (ānv) and  = (ânv) are defined as follows:

a¯nv=i=vnani,         n,v=0,1,

and

a^00=a¯00=a00,         a^nv=a¯nv-a¯n-1,v,         n=1,2,

Ā and  are the well-known matrices of series-to-sequence and series-to-series transformations, respectively. Then, we write

An(s)=i=0nanisi=i=0na¯niai

and

Δ¯An(s)=i=0na^niai.

In [4, 5], the following theorem dealing with |N̄, pn|k summability factors of infinite series has been proved by Bor.

Theorem 2.1

Let (Xn) be an almost increasing sequence such thatXn| = O(Xn/n) and λn → 0 as n → ∞. Suppose that there exists a sequence of numbers (An) such that it is δ-quasi-monotone withnXnδn < ∞, ∑AnXn is convergent andλn| ≤ |An| for all n. If

n=1m1nλn=O(1)         as         m,n=1m1ntnk=O(Xm)         as         m,

and

n=1mpnPntnk=O(Xm)         as         m,

then the seriesanλn is |N̄, pn|k summable, k ≥ 1.

The aim of this paper is to prove following more general theorem dealing with |A, pn|k summability.

Theorem 3.1

Let A = (anv) be a positive normal matrix such that

a¯n0=1,         n=0,1,,an-1,vanv,         for         nv+1,ann=O(pnPn).

If all conditions of Theorem 2.1 are satisfied, then the seriesanλn is |A, pn|k summable, k ≥ 1.

Lemma 3.2.([4])

Under the conditions of Theorem 3.1, we have

λnXn=O(1)         as         n.

Lemma 3.3.([5])

Let (Xn) be an almost increasing sequence such that nXn |= O (Xn). If (An) is a δ-quasi monotone withnXnδn < ∞, andAnXn is convergent, then

nAnXn=O(1)         as         n,n=1nXnΔAn<.

Let (Mn) denotes A-transform of the series ∑anλn. Then, by (1.10) and (1.11), we have

Δ¯Mn=v=0na^nvavλv=v=1na^nvλvvvav.

Applying Abel’s transformation to above sum, we get

Δ¯Mn=v=1n-1Δv(a^nvλvv)r=1vrar+a^nnλnnr=1nrar=n+1nannλntn+v=1n-1v+1vΔv(a^nv)λvtv+v=1n-1v+1va^n,v+1Δλvtv+v=1n-1a^n,v+1λv+1tvv=Mn,1+Mn,2+Mn,3+Mn,4.

To prove Theorem 3.1, we will show that

n=1(Pnpn)k-1Mn,rk<,         for         r=1,2,3,4.

First, by using (2.3), (3.3) and (3.4), we have

n=1m(Pnpn)k-1Mn,1k=n=1m(Pnpn)k-1|n+1nannλntn|k=O(1)n=1m(Pnpn)k-1annkλnktnk=O(1)n=1mpnPnλnk-1λntnk=O(1)n=1mpnPnλntnk=O(1)n=1m-1Δλnr=1nprPrtrk+O(1)λmn=1mpnPntnk=O(1)n=1m-1ΔλnXn+O(1)λmXm=O(1)n=1m-1AnXn+O(1)λmXm=O(1)         as         m.

Now, as in Mn,1, we have

n=2m+1(Pnpn)k-1Mn,2k=O(1)n=2m+1(Pnpn)k-1(v=1n-1Δv(a^nv)λvtv)k=O(1)n=2m+1(Pnpn)k-1(v=1n-1Δv(a^nv)λvktvk)×(v=1n-1Δv(a^nv))k-1.

Since

Δv(a^nv)=a^nv-a^n,v+1=a¯nv-a¯n-1,v-a¯n,v+1+a¯n-1,v+1=anv-an-1,v

by (1.8) and (1.9), we have

v=1n-1Δv(a^nv)=v=1n-1(an-1,v-anv)ann

by using (1.8), (3.1) and (3.2). Hence, we get

n=2m+1(Pnpn)k-1Mn,2k=O(1)v=1mλvk-1λvtvkn=v+1m+1(Pnpn)k-1annk-1Δv(a^nv)=O(1)v=1mλvtvkn=v+1m+1Δv(a^nv).

Now, using (3.2) and (4.1), we obtain

n=v+1m+1Δv(a^nv)=n=v+1m+1(an-1,v-anv)avv,

then

n=2m+1(Pnpn)k-1Mn,2k=O(1)v=1mλvtvkavv=O(1)v=1mλvtvkpvPv=O(1)         as         m,

by virtue of the hypotheses of Theorem 3.1 and Lemma 3.2. Also, we have

n=2m+1(Pnpn)k-1Mn,3k=n=2m+1(Pnpn)k-1|v=1n-1v+1va^n,v+1Δλvtv|k=O(1)n=2m+1(Pnpn)k-1(v=1n-1a^n,v+1Δλvtv)k=O(1)n=2m+1(Pnpn)k-1(v=1n-1a^n,v+1Avtv)k=O(1)n=2m+1(Pnpn)k-1(v=1n-1a^n,v+1Avtvk)×(v=1n-1a^n,v+1Av)k-1=O(1)v=1mAvtvkn=v+1m+1a^n,v+1.

By (1.8), (1.9), (3.1) and (3.2), we obtain

a^n,v+1=i=0v(an-1,i-ani).

Thus, using (1.8) and (3.1), we have

n=v+1m+1a^n,v+1=n=v+1m+1i=0v(an-1,i-ani)1,

then we get

n=2m+1(Pnpn)k-1Mn,3k=O(1)v=1mAvtvk=O(1)v=1mvAv1vtvk=O(1)v=1m-1Δ(vAv)r=1v1rtrk+O(1)mAmv=1m1vtvk=O(1)v=1m-1vΔAvXv+O(1)v=1m-1AvXv+O(1)mAmXm=O(1)         as         m,

by virtue of the hypotheses of Theorem 3.1 and Lemma 3.3. Again, operating Hölder’s inequality, we have

n=2m+1(Pnpn)k-1Mn,4kn=2m+1(Pnpn)k-1(v=1n-1a^n,v+1λv+1tvv)kn=2m+1(Pnpn)k-1(v=1n-1a^n,v+1λv+1tvkv)×(v=1n-1a^n,v+1λv+1v)k-1=O(1)v=1mλv+1vtvkn=v+1m+1a^n,v+1=O(1)v=1mλv+1vtvk=O(1)v=1m-1Δλv+1r=1v1rtrk+O(1)λm+1v=1m1vtvk=O(1)v=1m-1Δλv+1Xv+1+O(1)λm+1Xm+1=O(1)v=1m-1Av+1Xv+1+O(1)λm+1Xm+1=O(1)         as         m,

by (2.1), (2.2), (3.3) and (3.4). This completes the proof of Theorem 3.1.

If we take anv=pvPn in this theorem, then we get Theorem 2.1. If we take anv=pvPn and pn = 1 for all values of n, then we get a result for |C, 1|k summability.

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