# A new note on generalized absolute matrix summability

## Abstract

This paper gives necessary and sufficient conditions in order that a series $âˆ‘{a}_{n}{\mathrm{Î»}}_{n}$ should be summable $|B{|}_{k}$, $kâ‰¥1$, whenever $âˆ‘{a}_{n}$ is summable $|A|$. Some new results have also been obtained.

MSC:40D25, 40F05, 40G99.

## 1 Introduction

Let $âˆ‘{a}_{n}$ be a given infinite series with the partial sums $\left({s}_{n}\right)$. Let $\left({p}_{n}\right)$ be a sequence of positive numbers such that

${P}_{n}=\underset{v=0}{\overset{n}{âˆ‘}}{p}_{v}â†’\mathrm{âˆž}\phantom{\rule{1em}{0ex}}\text{as}\left(nâ†’\mathrm{âˆž}\right),\phantom{\rule{2em}{0ex}}\left({P}_{âˆ’i}={p}_{âˆ’i}=0,iâ‰¥1\right).$
(1)

The sequence-to-sequence transformation

${t}_{n}=\frac{1}{{P}_{n}}\underset{v=0}{\overset{n}{âˆ‘}}{p}_{v}{s}_{v}$
(2)

defines the sequence $\left({t}_{n}\right)$ of the Riesz means of the sequence $\left({s}_{n}\right)$ generated by the sequence of coefficients $\left({p}_{n}\right)$ (see [3]). The series $âˆ‘{a}_{n}$ is said to be summable $|R,{p}_{n}{|}_{k}$$kâ‰¥1$, if (see [1])

$\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}|{t}_{n}âˆ’{t}_{nâˆ’1}{|}^{k}<\mathrm{âˆž}.$
(3)

Let $A=\left({a}_{nv}\right)$ 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 $s=\left({s}_{n}\right)$ to $As=\left({A}_{n}\left(s\right)\right)$, where

${A}_{n}\left(s\right)=\underset{v=0}{\overset{n}{âˆ‘}}{a}_{nv}{s}_{v},\phantom{\rule{1em}{0ex}}n=0,1,â€¦.$
(4)

The series $âˆ‘{a}_{n}$ is said to be summable $|A{|}_{k}$$kâ‰¥1$, if (see [7])

$\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}|\stackrel{Â¯}{\mathrm{Î”}}{A}_{n}\left(s\right){|}^{k}<\mathrm{âˆž},$
(5)

where

$\stackrel{Â¯}{\mathrm{Î”}}{A}_{n}\left(s\right)={A}_{n}\left(s\right)âˆ’{A}_{nâˆ’1}\left(s\right).$

If we take ${a}_{nv}=\frac{{p}_{v}}{{P}_{n}}$, then $|A{|}_{k}$ summability is the same as $|R,{p}_{n}{|}_{k}$ summability.

Before stating the main theorem we must first introduce some further notations.

Given a normal matrix $A=\left({a}_{nv}\right)$, we associate two lover semimatrices $\stackrel{Â¯}{A}=\left({\stackrel{Â¯}{a}}_{nv}\right)$ and $\stackrel{Ë†}{A}=\left({\stackrel{Ë†}{a}}_{nv}\right)$ as follows:

${\stackrel{Â¯}{a}}_{nv}=\underset{i=v}{\overset{n}{âˆ‘}}{a}_{ni},\phantom{\rule{1em}{0ex}}n,v=0,1,â€¦$
(6)

and

${\stackrel{Ë†}{a}}_{00}={\stackrel{Â¯}{a}}_{00}={a}_{00},\phantom{\rule{2em}{0ex}}{\stackrel{Ë†}{a}}_{nv}={\stackrel{Â¯}{a}}_{nv}âˆ’{\stackrel{Â¯}{a}}_{nâˆ’1,v},\phantom{\rule{1em}{0ex}}n=1,2,â€¦.$
(7)

It may be noted that $\stackrel{Â¯}{A}$ and $\stackrel{Ë†}{A}$ are the well-known matrices of series-to-sequence and series-to-series transformations, respectively. Then, we have

${A}_{n}\left(s\right)=\underset{v=0}{\overset{n}{âˆ‘}}{a}_{nv}{s}_{v}=\underset{v=0}{\overset{n}{âˆ‘}}{\stackrel{Â¯}{a}}_{nv}{a}_{v}$
(8)

and

$\stackrel{Â¯}{\mathrm{Î”}}{A}_{n}\left(s\right)=\underset{v=0}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{a}}_{nv}{a}_{v}.$
(9)

If A is a normal matrix, then ${A}^{â€²}=\left({a}_{nv}^{â€²}\right)$ will denote the inverse of A. Clearly if A is normal, then $\stackrel{Ë†}{A}=\left({\stackrel{Ë†}{a}}_{nv}\right)$ is normal and has two-sided inverse ${\stackrel{Ë†}{A}}^{â€²}=\left({\stackrel{Ë†}{a}}_{nv}^{â€²}\right)$, which is also normal (see [2]).

SarÄ±gÃ¶l [6] has proved the following theorem for $|R,{p}_{n}{|}_{k}$ summability method.

Theorem A Suppose that$\left({p}_{n}\right)$and$\left({q}_{n}\right)$are positive sequences with${P}_{n}â†’\mathrm{âˆž}$and${Q}_{n}â†’\mathrm{âˆž}$as$nâ†’\mathrm{âˆž}$. Then$âˆ‘{a}_{n}{\mathrm{Î»}}_{n}$is summable$|R,{q}_{n}{|}_{k}$, $kâ‰¥1$whenever$âˆ‘{a}_{n}$is summable$|R,{p}_{n}|$, if and only if

(10)
(11)
(12)

provided that

${W}_{n}={\left\{\underset{v=n+1}{\overset{\mathrm{âˆž}}{âˆ‘}}{v}^{kâˆ’1}{\left(\frac{{q}_{v}}{{Q}_{v}{Q}_{vâˆ’1}}\right)}^{k}\right\}}^{\frac{1}{k}}<\mathrm{âˆž}.$

Theorem B The$|R,{p}_{n}|$summability implies the$|R,{q}_{n}{|}_{k}$, $kâ‰¥1$, summability if and only if the following conditions hold:

(13)
(14)
(15)

where

${W}_{v}={\left\{\underset{i=v+1}{\overset{\mathrm{âˆž}}{âˆ‘}}{i}^{kâˆ’1}{\left(\frac{{q}_{i}}{{Q}_{i}{Q}_{iâˆ’1}}\right)}^{k}\right\}}^{\frac{1}{k}}<\mathrm{âˆž}$

and we regarded that the above series converges for each v and Î” is the forward difference operator.

It may be remarked that the above theorem has been proved by Orhan and SarÄ±gÃ¶l [5].

Lemma ([4])

$A=\left({a}_{nv}\right)âˆˆ\left({l}_{1},{l}_{k}\right)$ if and only if

$\underset{v}{sup}\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{|{a}_{nv}|}^{k}<\mathrm{âˆž}$
(16)

for the cases$1â‰¤k<\mathrm{âˆž}$, where$\left({l}_{1},{l}_{k}\right)$denotes the set of all matrices A which map${l}_{1}$into${l}_{k}=\left\{x=\left({x}_{n}\right):âˆ‘{|{x}_{n}|}^{k}<\mathrm{âˆž}\right\}$.

## 2 Main theorem

The aim of this paper is to generalize Theorem A for the $|A|$ and ${|B|}_{k}$ summabilities. Therefore we shall prove the following theorem.

Theorem Let$kâ‰¥1$, $A=\left({a}_{nv}\right)$and$B=\left({b}_{nv}\right)$be two positive normal matrices such that

(17)
(18)

Then, in order that$âˆ‘{a}_{n}{\mathrm{Î»}}_{n}$is summable$|B{|}_{k}$whenever$âˆ‘{a}_{n}$is summable$|A|$, it is necessary that

(19)
(20)
(21)

Also (19)-(21) and

(22)
(23)
(24)

are sufficient for the consequent to hold.

It should be noted that if we take ${a}_{nv}=\frac{{p}_{v}}{{P}_{n}}$ and ${b}_{nv}=\frac{{q}_{v}}{{Q}_{n}}$, then we get Theorem A. Also if we take ${\mathrm{Î»}}_{n}=1$, then we get Theorem B.

Proof of the Theorem Necessity. Let $\left({x}_{n}\right)$ and $\left({y}_{n}\right)$ denote A-transform and B-transform of the series $âˆ‘{a}_{n}$ and $âˆ‘{a}_{n}{\mathrm{Î»}}_{n}$, respectively. Then, by (8) and (9), we have

$\stackrel{Â¯}{\mathrm{Î”}}{x}_{n}=\underset{v=0}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{a}}_{nv}{a}_{v}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\stackrel{Â¯}{\mathrm{Î”}}{y}_{n}=\underset{v=0}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{b}}_{nv}{a}_{v}{\mathrm{Î»}}_{v}.$
(25)

For $kâ‰¥1$, we define

$A=\left\{\left({a}_{i}\right):âˆ‘{a}_{i}\text{is summable}|A|\right\},\phantom{\rule{2em}{0ex}}B=\left\{\left({a}_{i}{\mathrm{Î»}}_{i}\right):âˆ‘{a}_{i}{\mathrm{Î»}}_{i}\text{is summable}{|B|}_{k}\right\}.$

Then it is routine to verify that these are BK-spaces, if normed by

$âˆ¥Xâˆ¥=\left\{\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}|\stackrel{Â¯}{\mathrm{Î”}}{x}_{n}|\right\}$
(26)

and

$âˆ¥Yâˆ¥={\left\{\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}{|\stackrel{Â¯}{\mathrm{Î”}}{y}_{n}|}^{k}\right\}}^{\frac{1}{k}}$
(27)

respectively. Since $âˆ‘{a}_{n}$ is summable $|A|$ implies $âˆ‘{a}_{n}{\mathrm{Î»}}_{n}$ is summable ${|B|}_{k}$, by the hypothesis of the theorem,

$âˆ¥Xâˆ¥<\mathrm{âˆž}\phantom{\rule{1em}{0ex}}â‡’\phantom{\rule{1em}{0ex}}âˆ¥Yâˆ¥<\mathrm{âˆž}.$

Now consider the inclusion map $c:Aâ†’B$ defined by $c\left(x\right)=x$. This is continuous, which is immediate as A and B are BK-spaces. Thus there exists a constant M such that

$âˆ¥Yâˆ¥â‰¤Mâˆ¥Xâˆ¥.$
(28)

By applying (25) to ${a}_{v}={e}_{v}âˆ’{e}_{v+1}$ (${e}_{v}$ is the v th coordinate vector), we have

$\stackrel{Â¯}{\mathrm{Î”}}{x}_{n}=\left\{\begin{array}{cc}0,\hfill & \text{if}nv\hfill \end{array}$

and

$\stackrel{Â¯}{\mathrm{Î”}}{y}_{n}=\left\{\begin{array}{cc}0,\hfill & \text{if}nv.\hfill \end{array}$

So (26) and (27) give us

$âˆ¥Xâˆ¥=\left\{{a}_{vv}+\underset{n=v+1}{\overset{\mathrm{âˆž}}{âˆ‘}}|{\mathrm{Î”}}_{v}{\stackrel{Ë†}{a}}_{nv}|\right\}$

and

$âˆ¥Yâˆ¥={\left\{{v}^{kâˆ’1}{b}_{vv}{|{\mathrm{Î»}}_{v}|}^{k}+\underset{n=v+1}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}{|{\mathrm{Î”}}_{v}\left({\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}\right)|}^{k}\right\}}^{\frac{1}{k}}.$

Hence it follows from (28) that

${v}^{kâˆ’1}{b}_{vv}{|{\mathrm{Î»}}_{v}|}^{k}+\underset{n=v+1}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}{|{\mathrm{Î”}}_{v}{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}|}^{k}â‰¤{M}^{k}{a}_{vv}^{k}+{M}^{k}\underset{n=v+1}{\overset{\mathrm{âˆž}}{âˆ‘}}{|{\mathrm{Î”}}_{v}{\stackrel{Ë†}{a}}_{nv}|}^{k}.$

Using (17), we can find

${v}^{kâˆ’1}{b}_{vv}{|{\mathrm{Î»}}_{v}|}^{k}+\underset{n=v+1}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}{|{\mathrm{Î”}}_{v}\left({\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}\right)|}^{k}=O\left({a}_{vv}^{k}\right).$

The above inequality will be true if and only if each term on the left-hand side is $O\left({a}_{vv}^{k}\right)$. Taking the first term,

${v}^{kâˆ’1}{b}_{vv}{|{\mathrm{Î»}}_{v}|}^{k}=O\left({a}_{vv}^{k}\right)$

then

$|{\mathrm{Î»}}_{v}|=O\left({v}^{\frac{1}{k}âˆ’1}\frac{{a}_{vv}}{{b}_{vv}}\right)$

which verifies that (19) is necessary. Using the second term, we have

$\underset{n=v+1}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}{|{\mathrm{Î”}}_{v}\left({\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}\right)|}^{k}=O\left({a}_{vv}^{k}\right)$

which is condition (20). Now, if we apply (22) to ${a}_{v}={e}_{v+1}$, we have

$\stackrel{Â¯}{\mathrm{Î”}}{x}_{n}=\left\{\begin{array}{cc}0,\hfill & \text{if}nâ‰¤v,\hfill \\ {\stackrel{Ë†}{a}}_{n,v+1},\hfill & \text{if}n>v\hfill \end{array}$

and

$\stackrel{Â¯}{\mathrm{Î”}}{y}_{n}=\left\{\begin{array}{cc}0,\hfill & \text{if}nâ‰¤v,\hfill \\ {\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1},\hfill & \text{if}n>v,\hfill \end{array}$

respectively. Hence

Hence it follows from (28) that

$\underset{n=v+1}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}{|{\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1}|}^{k}â‰¤{M}^{k}{\left\{\underset{n=v+1}{\overset{\mathrm{âˆž}}{âˆ‘}}|{\stackrel{Ë†}{a}}_{n,v+1}|\right\}}^{k}.$

Using (18) we can find

$\underset{n=v+1}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}{|{\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1}|}^{k}=O\left(1\right)$

which is condition (21).

Sufficiency. We use the notations of necessity. Then

$\stackrel{Â¯}{\mathrm{Î”}}{x}_{n}=\underset{v=0}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{a}}_{nv}{a}_{v}$
(29)

which implies

${a}_{v}=\underset{r=0}{\overset{v}{âˆ‘}}{\stackrel{Ë†}{a}}_{vr}^{â€²}\stackrel{Â¯}{\mathrm{Î”}}{x}_{r}.$
(30)

In this case

$\stackrel{Â¯}{\mathrm{Î”}}{y}_{n}=\underset{v=0}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{b}}_{nv}{a}_{v}{\mathrm{Î»}}_{v}=\underset{v=0}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}\underset{r=0}{\overset{v}{âˆ‘}}{\stackrel{Ë†}{a}}_{vr}^{â€²}\stackrel{Â¯}{\mathrm{Î”}}{x}_{r}.$

On the other hand, since

${\stackrel{Ë†}{b}}_{n0}={\stackrel{Â¯}{b}}_{n0}âˆ’{\stackrel{Â¯}{b}}_{nâˆ’1,0}$

by (22), we have

$\begin{array}{rcl}\stackrel{Â¯}{\mathrm{Î”}}{y}_{n}& =& \underset{v=1}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}\left\{\underset{r=0}{\overset{v}{âˆ‘}}{\stackrel{Ë†}{a}}_{vr}^{â€²}\stackrel{Â¯}{\mathrm{Î”}}{x}_{r}\right\}\\ =& \underset{v=1}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}\left\{{\stackrel{Ë†}{a}}_{vv}^{â€²}\stackrel{Â¯}{\mathrm{Î”}}{x}_{v}+{\stackrel{Ë†}{a}}_{v,vâˆ’1}^{â€²}\stackrel{Â¯}{\mathrm{Î”}}{x}_{vâˆ’1}+\underset{r=0}{\overset{vâˆ’2}{âˆ‘}}{\stackrel{Ë†}{a}}_{vr}^{â€²}\stackrel{Â¯}{\mathrm{Î”}}{x}_{r}\right\}\\ =& \underset{v=1}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}{\stackrel{Ë†}{a}}_{vv}^{â€²}\stackrel{Â¯}{\mathrm{Î”}}{x}_{v}+\underset{v=1}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}{\stackrel{Ë†}{a}}_{v,vâˆ’1}^{â€²}\stackrel{Â¯}{\mathrm{Î”}}{x}_{vâˆ’1}+\underset{v=1}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}\underset{r=0}{\overset{vâˆ’2}{âˆ‘}}{\stackrel{Ë†}{a}}_{vr}^{â€²}\stackrel{Â¯}{\mathrm{Î”}}{x}_{r}\\ =& {\stackrel{Ë†}{b}}_{nn}{\mathrm{Î»}}_{n}{\stackrel{Ë†}{a}}_{nn}^{â€²}\stackrel{Â¯}{\mathrm{Î”}}{x}_{n}+\underset{v=1}{\overset{nâˆ’1}{âˆ‘}}\left({\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}{\stackrel{Ë†}{a}}_{vv}^{â€²}+{\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1}{\stackrel{Ë†}{a}}_{v+1,v}^{â€²}\right)\stackrel{Â¯}{\mathrm{Î”}}{x}_{v}\\ +\underset{r=0}{\overset{nâˆ’2}{âˆ‘}}\stackrel{Â¯}{\mathrm{Î”}}{x}_{r}\underset{v=r+2}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}{\stackrel{Ë†}{a}}_{vr}^{â€²}.\end{array}$
(31)

By considering the equality

$\underset{k=v}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{a}}_{nk}^{â€²}{\stackrel{Ë†}{a}}_{kv}={\mathrm{Î´}}_{nv},$

where ${\mathrm{Î´}}_{nv}$ is the Kronecker delta, we have that

$\begin{array}{rcl}{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}{\stackrel{Ë†}{a}}_{vv}^{â€²}+{\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1}{\stackrel{Ë†}{a}}_{v+1,v}^{â€²}& =& \frac{{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}}{{\stackrel{Ë†}{a}}_{vv}}+{\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1}\left(âˆ’\frac{{\stackrel{Ë†}{a}}_{v+1,v}}{{\stackrel{Ë†}{a}}_{vv}{\stackrel{Ë†}{a}}_{v+1,v+1}}\right)\\ =& \frac{{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}}{{a}_{vv}}âˆ’\frac{{\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1}\left({\stackrel{Â¯}{a}}_{v+1,v}âˆ’{\stackrel{Â¯}{a}}_{v,v}\right)}{{a}_{vv}{a}_{v+1,v+1}}\\ =& \frac{{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}}{{a}_{vv}}âˆ’\frac{{\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1}\left({a}_{v+1,v+1}+{a}_{v+1,v}âˆ’{a}_{vv}\right)}{{a}_{vv}{a}_{v+1,v+1}}\\ =& \frac{{\mathrm{Î”}}_{v}\left({\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}\right)}{{a}_{vv}}+{\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1}\frac{{a}_{vv}âˆ’{a}_{v+1,v}}{{a}_{vv}{a}_{v+1,v+1}}\end{array}$

and so

$\begin{array}{rcl}\stackrel{Â¯}{\mathrm{Î”}}{y}_{n}& =& \frac{{b}_{nn}{\mathrm{Î»}}_{n}}{{a}_{nn}}\stackrel{Â¯}{\mathrm{Î”}}{x}_{n}+\underset{v=1}{\overset{nâˆ’1}{âˆ‘}}\frac{{\mathrm{Î”}}_{v}\left({\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}\right)}{{a}_{vv}}\stackrel{Â¯}{\mathrm{Î”}}{x}_{v}+\underset{v=1}{\overset{nâˆ’1}{âˆ‘}}{\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1}\frac{{a}_{vv}âˆ’{a}_{v+1,v}}{{a}_{vv}{a}_{v+1,v+1}}\stackrel{Â¯}{\mathrm{Î”}}{x}_{v}\\ +\underset{r=0}{\overset{nâˆ’2}{âˆ‘}}\stackrel{Â¯}{\mathrm{Î”}}{x}_{r}\underset{v=r+2}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}{\stackrel{Ë†}{a}}_{vr}^{â€²}.\end{array}$

Let

Since

$|{T}_{n}\left(1\right)+{T}_{n}\left(2\right){|}^{k}â‰¤{2}^{k}\left(|{T}_{n}\left(1\right){|}^{k}+|{T}_{n}\left(2\right){|}^{k}\right)$

to complete the proof of Theorem, it is sufficient to show that

$\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}|{T}_{n}\left(i\right){|}^{k}<\mathrm{âˆž}\phantom{\rule{1em}{0ex}}\text{for}i=1,2.$

Then

$\begin{array}{rcl}\stackrel{Â¯}{{T}_{n}\left(1\right)}& =& {n}^{1âˆ’\frac{1}{k}}{T}_{n}\left(1\right)\\ =& {n}^{1âˆ’\frac{1}{k}}\frac{{b}_{nn}{\mathrm{Î»}}_{n}}{{a}_{nn}}\stackrel{Â¯}{\mathrm{Î”}}{x}_{n}+{n}^{1âˆ’\frac{1}{k}}\underset{v=1}{\overset{nâˆ’1}{âˆ‘}}\frac{{\mathrm{Î”}}_{v}\left({\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}\right)}{{a}_{vv}}\stackrel{Â¯}{\mathrm{Î”}}{x}_{v}+{n}^{1âˆ’\frac{1}{k}}\underset{v=1}{\overset{nâˆ’1}{âˆ‘}}{\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1}\frac{{a}_{vv}âˆ’{a}_{v+1,v}}{{a}_{vv}{a}_{v+1,v+1}}\stackrel{Â¯}{\mathrm{Î”}}{x}_{v}\\ =& \underset{v=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{c}_{nv}\stackrel{Â¯}{\mathrm{Î”}}{x}_{v},\end{array}$

where

${c}_{nv}=\left\{\begin{array}{cc}{n}^{1âˆ’\frac{1}{k}}\left(\frac{{\mathrm{Î”}}_{v}\left({b}_{nv}{\mathrm{Î»}}_{v}\right)}{{a}_{vv}}+{\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1}\frac{{a}_{vv}âˆ’{a}_{v+1,v}}{{a}_{vv}{a}_{v+1,v+1}}\right),\hfill & \text{if}1â‰¤vâ‰¤nâˆ’1\text{,}\hfill \\ {n}^{1âˆ’\frac{1}{k}}\frac{{b}_{nn}{\mathrm{Î»}}_{n}}{{a}_{nn}},\hfill & \text{if}v=n,\hfill \\ 0,\hfill & \text{if}v>n\text{.}\hfill \end{array}$

Now

$âˆ‘{|\stackrel{Â¯}{{T}_{n}\left(1\right)}|}^{k}<\mathrm{âˆž}\phantom{\rule{1em}{0ex}}\text{whenever}âˆ‘|\stackrel{Â¯}{\mathrm{Î”}}{x}_{n}|<\mathrm{âˆž}$

is equivalently

$\underset{v}{sup}\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{|{c}_{nv}|}^{k}<\mathrm{âˆž}$
(32)

by Lemma. But it follows from conditions (20), (21) and (23) that

$\begin{array}{rcl}\underset{n=v}{\overset{\mathrm{âˆž}}{âˆ‘}}{|{c}_{nv}|}^{k}& =& O\left(1\right)\left\{{n}^{kâˆ’1}{|\frac{{b}_{nn}{\mathrm{Î»}}_{n}}{{a}_{nn}}|}^{k}+\underset{n=v+1}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}|\frac{{\mathrm{Î”}}_{v}\left({\stackrel{Ë†}{b}}_{nv}{\mathrm{Î»}}_{v}\right)}{{a}_{vv}}+{\stackrel{Ë†}{b}}_{n,v+1}{\mathrm{Î»}}_{v+1}\frac{{a}_{vv}âˆ’{a}_{v+1,v}}{{a}_{vv}{a}_{v+1,v+1}}{|}^{k}\right\}\\ =& O\left(1\right)\phantom{\rule{1em}{0ex}}\text{as}vâ†’\mathrm{âˆž}.\end{array}$

Finally,

$\stackrel{Â¯}{{T}_{n}\left(2\right)}={n}^{1âˆ’\frac{1}{k}}{T}_{n}\left(2\right)={n}^{1âˆ’\frac{1}{k}}\underset{r=0}{\overset{nâˆ’2}{âˆ‘}}\stackrel{Â¯}{\mathrm{Î”}}{x}_{r}\underset{v=r+2}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{b}}_{nv}{\stackrel{Ë†}{a}}_{vr}^{â€²}{\mathrm{Î»}}_{v}=\underset{r=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{d}_{nr}\stackrel{Â¯}{\mathrm{Î”}}{x}_{r},$

where

${d}_{nr}=\left\{\begin{array}{cc}{n}^{1âˆ’\frac{1}{k}}{âˆ‘}_{v=r+2}^{n}{\stackrel{Ë†}{b}}_{nv}{\stackrel{Ë†}{a}}_{vr}^{â€²}{\mathrm{Î»}}_{v},\hfill & \text{if}0â‰¤râ‰¤nâˆ’2,\hfill \\ 0,\hfill & \text{if}r>nâˆ’2\text{.}\hfill \end{array}$

Now

$âˆ‘{|\stackrel{Â¯}{{T}_{n}\left(2\right)}|}^{k}<\mathrm{âˆž}\phantom{\rule{1em}{0ex}}\text{whenever}âˆ‘|\stackrel{Â¯}{\mathrm{Î”}}{x}_{n}|<\mathrm{âˆž}$

is equivalently

$\underset{r}{sup}\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{|{d}_{nr}|}^{k}<\mathrm{âˆž}$
(33)

by Lemma. But it follows from conditions (21) and (24) that

$\begin{array}{rcl}\underset{n=r+2}{\overset{\mathrm{âˆž}}{âˆ‘}}{|{d}_{nr}|}^{k}& =& O\left(1\right)\underset{n=r+2}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}{\left\{\underset{v=r+2}{\overset{\mathrm{âˆž}}{âˆ‘}}|{\stackrel{Ë†}{b}}_{nv}{\stackrel{Ë†}{a}}_{vr}^{â€²}{\mathrm{Î»}}_{v}|\right\}}^{k}\\ =& O\left(1\right)\underset{n=r+2}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}{|{\stackrel{Ë†}{b}}_{n,r+1}{\mathrm{Î»}}_{r+1}|}^{k}\\ =& O\left(1\right)\phantom{\rule{1em}{0ex}}\text{as}râ†’\mathrm{âˆž}.\end{array}$

Therefore, we have

$\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{n}^{kâˆ’1}|{T}_{n}\left(i\right){|}^{k}<\mathrm{âˆž}\phantom{\rule{1em}{0ex}}\text{for}i=1,2.$

This completes the proof of the Theorem.â€ƒâ–¡

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Ã–zarslan, H., Ari, T. A new note on generalized absolute matrix summability. J Inequal Appl 2012, 166 (2012). https://doi.org/10.1186/1029-242X-2012-166