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# A new note on generalized absolute matrix summability

Journal of Inequalities and Applications20122012:166

https://doi.org/10.1186/1029-242X-2012-166

• Received: 3 June 2012
• Accepted: 13 July 2012
• Published:

## Abstract

This paper gives necessary and sufficient conditions in order that a series $\sum {a}_{n}{\lambda }_{n}$ should be summable $|B{|}_{k}$, $k\ge 1$, whenever $\sum {a}_{n}$ is summable $|A|$. Some new results have also been obtained.

MSC:40D25, 40F05, 40G99.

## Keywords

• summability factors
• absolute matrix summability
• infinite series

## 1 Introduction

Let $\sum {a}_{n}$ 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
${P}_{n}=\sum _{v=0}^{n}{p}_{v}\to \mathrm{\infty }\phantom{\rule{1em}{0ex}}\text{as}\left(n\to \mathrm{\infty }\right),\phantom{\rule{2em}{0ex}}\left({P}_{-i}={p}_{-i}=0,i\ge 1\right).$
(1)
The sequence-to-sequence transformation
${t}_{n}=\frac{1}{{P}_{n}}\sum _{v=0}^{n}{p}_{v}{s}_{v}$
(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 ). The series $\sum {a}_{n}$ is said to be summable $|R,{p}_{n}{|}_{k}$$k\ge 1$, if (see )
$\sum _{n=1}^{\mathrm{\infty }}{n}^{k-1}|{t}_{n}-{t}_{n-1}{|}^{k}<\mathrm{\infty }.$
(3)
Let $A=\left({a}_{nv}\right)$ be a normal matrix, i.e., a lower triangular matrix of nonzero diagonal entries. Then A defines the sequence-to-sequence transformation, mapping the sequence $s=\left({s}_{n}\right)$ to $As=\left({A}_{n}\left(s\right)\right)$, where
${A}_{n}\left(s\right)=\sum _{v=0}^{n}{a}_{nv}{s}_{v},\phantom{\rule{1em}{0ex}}n=0,1,\dots .$
(4)
The series $\sum {a}_{n}$ is said to be summable $|A{|}_{k}$$k\ge 1$, if (see )
$\sum _{n=1}^{\mathrm{\infty }}{n}^{k-1}|\overline{\mathrm{\Delta }}{A}_{n}\left(s\right){|}^{k}<\mathrm{\infty },$
(5)
where
$\overline{\mathrm{\Delta }}{A}_{n}\left(s\right)={A}_{n}\left(s\right)-{A}_{n-1}\left(s\right).$

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 $\overline{A}=\left({\overline{a}}_{nv}\right)$ and $\stackrel{ˆ}{A}=\left({\stackrel{ˆ}{a}}_{nv}\right)$ as follows:
${\overline{a}}_{nv}=\sum _{i=v}^{n}{a}_{ni},\phantom{\rule{1em}{0ex}}n,v=0,1,\dots$
(6)
and
${\stackrel{ˆ}{a}}_{00}={\overline{a}}_{00}={a}_{00},\phantom{\rule{2em}{0ex}}{\stackrel{ˆ}{a}}_{nv}={\overline{a}}_{nv}-{\overline{a}}_{n-1,v},\phantom{\rule{1em}{0ex}}n=1,2,\dots .$
(7)
It may be noted that $\overline{A}$ and $\stackrel{ˆ}{A}$ are the well-known matrices of series-to-sequence and series-to-series transformations, respectively. Then, we have
${A}_{n}\left(s\right)=\sum _{v=0}^{n}{a}_{nv}{s}_{v}=\sum _{v=0}^{n}{\overline{a}}_{nv}{a}_{v}$
(8)
and
$\overline{\mathrm{\Delta }}{A}_{n}\left(s\right)=\sum _{v=0}^{n}{\stackrel{ˆ}{a}}_{nv}{a}_{v}.$
(9)

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

Sarıgöl  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${P}_{n}\to \mathrm{\infty }$and${Q}_{n}\to \mathrm{\infty }$as$n\to \mathrm{\infty }$. Then$\sum {a}_{n}{\lambda }_{n}$is summable$|R,{q}_{n}{|}_{k}$, $k\ge 1$whenever$\sum {a}_{n}$is summable$|R,{p}_{n}|$, if and only if
provided that
${W}_{n}={\left\{\sum _{v=n+1}^{\mathrm{\infty }}{v}^{k-1}{\left(\frac{{q}_{v}}{{Q}_{v}{Q}_{v-1}}\right)}^{k}\right\}}^{\frac{1}{k}}<\mathrm{\infty }.$
Theorem B The$|R,{p}_{n}|$summability implies the$|R,{q}_{n}{|}_{k}$, $k\ge 1$, summability if and only if the following conditions hold:
where
${W}_{v}={\left\{\sum _{i=v+1}^{\mathrm{\infty }}{i}^{k-1}{\left(\frac{{q}_{i}}{{Q}_{i}{Q}_{i-1}}\right)}^{k}\right\}}^{\frac{1}{k}}<\mathrm{\infty }$

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

It may be remarked that the above theorem has been proved by Orhan and Sarıgöl .

Lemma ()

$A=\left({a}_{nv}\right)\in \left({l}_{1},{l}_{k}\right)$ if and only if
$\underset{v}{sup}\sum _{n=1}^{\mathrm{\infty }}{|{a}_{nv}|}^{k}<\mathrm{\infty }$
(16)

for the cases$1\le k<\mathrm{\infty }$, where$\left({l}_{1},{l}_{k}\right)$denotes the set of all matrices A which map${l}_{1}$into${l}_{k}=\left\{x=\left({x}_{n}\right):\sum {|{x}_{n}|}^{k}<\mathrm{\infty }\right\}$.

## 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$k\ge 1$, $A=\left({a}_{nv}\right)$and$B=\left({b}_{nv}\right)$be two positive normal matrices such that
Then, in order that$\sum {a}_{n}{\lambda }_{n}$is summable$|B{|}_{k}$whenever$\sum {a}_{n}$is summable$|A|$, it is necessary that

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 ${\lambda }_{n}=1$, 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 $\sum {a}_{n}$ and $\sum {a}_{n}{\lambda }_{n}$, respectively. Then, by (8) and (9), we have
$\overline{\mathrm{\Delta }}{x}_{n}=\sum _{v=0}^{n}{\stackrel{ˆ}{a}}_{nv}{a}_{v}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\overline{\mathrm{\Delta }}{y}_{n}=\sum _{v=0}^{n}{\stackrel{ˆ}{b}}_{nv}{a}_{v}{\lambda }_{v}.$
(25)
For $k\ge 1$, we define
$A=\left\{\left({a}_{i}\right):\sum {a}_{i}\text{is summable}|A|\right\},\phantom{\rule{2em}{0ex}}B=\left\{\left({a}_{i}{\lambda }_{i}\right):\sum {a}_{i}{\lambda }_{i}\text{is summable}{|B|}_{k}\right\}.$
Then it is routine to verify that these are BK-spaces, if normed by
$\parallel X\parallel =\left\{\sum _{n=0}^{\mathrm{\infty }}|\overline{\mathrm{\Delta }}{x}_{n}|\right\}$
(26)
and
$\parallel Y\parallel ={\left\{\sum _{n=0}^{\mathrm{\infty }}{n}^{k-1}{|\overline{\mathrm{\Delta }}{y}_{n}|}^{k}\right\}}^{\frac{1}{k}}$
(27)
respectively. Since $\sum {a}_{n}$ is summable $|A|$ implies $\sum {a}_{n}{\lambda }_{n}$ is summable ${|B|}_{k}$, by the hypothesis of the theorem,
$\parallel X\parallel <\mathrm{\infty }\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}\parallel Y\parallel <\mathrm{\infty }.$
Now consider the inclusion map $c:A\to B$ 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
$\parallel Y\parallel \le M\parallel X\parallel .$
(28)
By applying (25) to ${a}_{v}={e}_{v}-{e}_{v+1}$ (${e}_{v}$ is the v th coordinate vector), we have
$\overline{\mathrm{\Delta }}{x}_{n}=\left\{\begin{array}{cc}0,\hfill & \text{if}nv\hfill \end{array}$
and
$\overline{\mathrm{\Delta }}{y}_{n}=\left\{\begin{array}{cc}0,\hfill & \text{if}nv.\hfill \end{array}$
So (26) and (27) give us
$\parallel X\parallel =\left\{{a}_{vv}+\sum _{n=v+1}^{\mathrm{\infty }}|{\mathrm{\Delta }}_{v}{\stackrel{ˆ}{a}}_{nv}|\right\}$
and
$\parallel Y\parallel ={\left\{{v}^{k-1}{b}_{vv}{|{\lambda }_{v}|}^{k}+\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}{|{\mathrm{\Delta }}_{v}\left({\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}\right)|}^{k}\right\}}^{\frac{1}{k}}.$
Hence it follows from (28) that
${v}^{k-1}{b}_{vv}{|{\lambda }_{v}|}^{k}+\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}{|{\mathrm{\Delta }}_{v}{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}|}^{k}\le {M}^{k}{a}_{vv}^{k}+{M}^{k}\sum _{n=v+1}^{\mathrm{\infty }}{|{\mathrm{\Delta }}_{v}{\stackrel{ˆ}{a}}_{nv}|}^{k}.$
Using (17), we can find
${v}^{k-1}{b}_{vv}{|{\lambda }_{v}|}^{k}+\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}{|{\mathrm{\Delta }}_{v}\left({\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}\right)|}^{k}=O\left({a}_{vv}^{k}\right).$
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,
${v}^{k-1}{b}_{vv}{|{\lambda }_{v}|}^{k}=O\left({a}_{vv}^{k}\right)$
then
$|{\lambda }_{v}|=O\left({v}^{\frac{1}{k}-1}\frac{{a}_{vv}}{{b}_{vv}}\right)$
which verifies that (19) is necessary. Using the second term, we have
$\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}{|{\mathrm{\Delta }}_{v}\left({\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}\right)|}^{k}=O\left({a}_{vv}^{k}\right)$
which is condition (20). Now, if we apply (22) to ${a}_{v}={e}_{v+1}$, we have
$\overline{\mathrm{\Delta }}{x}_{n}=\left\{\begin{array}{cc}0,\hfill & \text{if}n\le v,\hfill \\ {\stackrel{ˆ}{a}}_{n,v+1},\hfill & \text{if}n>v\hfill \end{array}$
and
$\overline{\mathrm{\Delta }}{y}_{n}=\left\{\begin{array}{cc}0,\hfill & \text{if}n\le v,\hfill \\ {\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1},\hfill & \text{if}n>v,\hfill \end{array}$
Hence it follows from (28) that
$\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}{|{\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1}|}^{k}\le {M}^{k}{\left\{\sum _{n=v+1}^{\mathrm{\infty }}|{\stackrel{ˆ}{a}}_{n,v+1}|\right\}}^{k}.$
Using (18) we can find
$\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}{|{\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1}|}^{k}=O\left(1\right)$

which is condition (21).

Sufficiency. We use the notations of necessity. Then
$\overline{\mathrm{\Delta }}{x}_{n}=\sum _{v=0}^{n}{\stackrel{ˆ}{a}}_{nv}{a}_{v}$
(29)
which implies
${a}_{v}=\sum _{r=0}^{v}{\stackrel{ˆ}{a}}_{vr}^{\prime }\overline{\mathrm{\Delta }}{x}_{r}.$
(30)
In this case
$\overline{\mathrm{\Delta }}{y}_{n}=\sum _{v=0}^{n}{\stackrel{ˆ}{b}}_{nv}{a}_{v}{\lambda }_{v}=\sum _{v=0}^{n}{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}\sum _{r=0}^{v}{\stackrel{ˆ}{a}}_{vr}^{\prime }\overline{\mathrm{\Delta }}{x}_{r}.$
On the other hand, since
${\stackrel{ˆ}{b}}_{n0}={\overline{b}}_{n0}-{\overline{b}}_{n-1,0}$
by (22), we have
$\begin{array}{rcl}\overline{\mathrm{\Delta }}{y}_{n}& =& \sum _{v=1}^{n}{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}\left\{\sum _{r=0}^{v}{\stackrel{ˆ}{a}}_{vr}^{\prime }\overline{\mathrm{\Delta }}{x}_{r}\right\}\\ =& \sum _{v=1}^{n}{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}\left\{{\stackrel{ˆ}{a}}_{vv}^{\prime }\overline{\mathrm{\Delta }}{x}_{v}+{\stackrel{ˆ}{a}}_{v,v-1}^{\prime }\overline{\mathrm{\Delta }}{x}_{v-1}+\sum _{r=0}^{v-2}{\stackrel{ˆ}{a}}_{vr}^{\prime }\overline{\mathrm{\Delta }}{x}_{r}\right\}\\ =& \sum _{v=1}^{n}{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}{\stackrel{ˆ}{a}}_{vv}^{\prime }\overline{\mathrm{\Delta }}{x}_{v}+\sum _{v=1}^{n}{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}{\stackrel{ˆ}{a}}_{v,v-1}^{\prime }\overline{\mathrm{\Delta }}{x}_{v-1}+\sum _{v=1}^{n}{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}\sum _{r=0}^{v-2}{\stackrel{ˆ}{a}}_{vr}^{\prime }\overline{\mathrm{\Delta }}{x}_{r}\\ =& {\stackrel{ˆ}{b}}_{nn}{\lambda }_{n}{\stackrel{ˆ}{a}}_{nn}^{\prime }\overline{\mathrm{\Delta }}{x}_{n}+\sum _{v=1}^{n-1}\left({\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}{\stackrel{ˆ}{a}}_{vv}^{\prime }+{\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1}{\stackrel{ˆ}{a}}_{v+1,v}^{\prime }\right)\overline{\mathrm{\Delta }}{x}_{v}\\ +\sum _{r=0}^{n-2}\overline{\mathrm{\Delta }}{x}_{r}\sum _{v=r+2}^{n}{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}{\stackrel{ˆ}{a}}_{vr}^{\prime }.\end{array}$
(31)
By considering the equality
$\sum _{k=v}^{n}{\stackrel{ˆ}{a}}_{nk}^{\prime }{\stackrel{ˆ}{a}}_{kv}={\delta }_{nv},$
where ${\delta }_{nv}$ is the Kronecker delta, we have that
$\begin{array}{rcl}{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}{\stackrel{ˆ}{a}}_{vv}^{\prime }+{\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1}{\stackrel{ˆ}{a}}_{v+1,v}^{\prime }& =& \frac{{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}}{{\stackrel{ˆ}{a}}_{vv}}+{\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1}\left(-\frac{{\stackrel{ˆ}{a}}_{v+1,v}}{{\stackrel{ˆ}{a}}_{vv}{\stackrel{ˆ}{a}}_{v+1,v+1}}\right)\\ =& \frac{{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}}{{a}_{vv}}-\frac{{\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1}\left({\overline{a}}_{v+1,v}-{\overline{a}}_{v,v}\right)}{{a}_{vv}{a}_{v+1,v+1}}\\ =& \frac{{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}}{{a}_{vv}}-\frac{{\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1}\left({a}_{v+1,v+1}+{a}_{v+1,v}-{a}_{vv}\right)}{{a}_{vv}{a}_{v+1,v+1}}\\ =& \frac{{\mathrm{\Delta }}_{v}\left({\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}\right)}{{a}_{vv}}+{\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1}\frac{{a}_{vv}-{a}_{v+1,v}}{{a}_{vv}{a}_{v+1,v+1}}\end{array}$
and so
$\begin{array}{rcl}\overline{\mathrm{\Delta }}{y}_{n}& =& \frac{{b}_{nn}{\lambda }_{n}}{{a}_{nn}}\overline{\mathrm{\Delta }}{x}_{n}+\sum _{v=1}^{n-1}\frac{{\mathrm{\Delta }}_{v}\left({\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}\right)}{{a}_{vv}}\overline{\mathrm{\Delta }}{x}_{v}+\sum _{v=1}^{n-1}{\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1}\frac{{a}_{vv}-{a}_{v+1,v}}{{a}_{vv}{a}_{v+1,v+1}}\overline{\mathrm{\Delta }}{x}_{v}\\ +\sum _{r=0}^{n-2}\overline{\mathrm{\Delta }}{x}_{r}\sum _{v=r+2}^{n}{\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}{\stackrel{ˆ}{a}}_{vr}^{\prime }.\end{array}$
Since
$|{T}_{n}\left(1\right)+{T}_{n}\left(2\right){|}^{k}\le {2}^{k}\left(|{T}_{n}\left(1\right){|}^{k}+|{T}_{n}\left(2\right){|}^{k}\right)$
to complete the proof of Theorem, it is sufficient to show that
$\sum _{n=1}^{\mathrm{\infty }}{n}^{k-1}|{T}_{n}\left(i\right){|}^{k}<\mathrm{\infty }\phantom{\rule{1em}{0ex}}\text{for}i=1,2.$
Then
$\begin{array}{rcl}\overline{{T}_{n}\left(1\right)}& =& {n}^{1-\frac{1}{k}}{T}_{n}\left(1\right)\\ =& {n}^{1-\frac{1}{k}}\frac{{b}_{nn}{\lambda }_{n}}{{a}_{nn}}\overline{\mathrm{\Delta }}{x}_{n}+{n}^{1-\frac{1}{k}}\sum _{v=1}^{n-1}\frac{{\mathrm{\Delta }}_{v}\left({\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}\right)}{{a}_{vv}}\overline{\mathrm{\Delta }}{x}_{v}+{n}^{1-\frac{1}{k}}\sum _{v=1}^{n-1}{\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1}\frac{{a}_{vv}-{a}_{v+1,v}}{{a}_{vv}{a}_{v+1,v+1}}\overline{\mathrm{\Delta }}{x}_{v}\\ =& \sum _{v=1}^{\mathrm{\infty }}{c}_{nv}\overline{\mathrm{\Delta }}{x}_{v},\end{array}$
where
${c}_{nv}=\left\{\begin{array}{cc}{n}^{1-\frac{1}{k}}\left(\frac{{\mathrm{\Delta }}_{v}\left({b}_{nv}{\lambda }_{v}\right)}{{a}_{vv}}+{\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1}\frac{{a}_{vv}-{a}_{v+1,v}}{{a}_{vv}{a}_{v+1,v+1}}\right),\hfill & \text{if}1\le v\le n-1\text{,}\hfill \\ {n}^{1-\frac{1}{k}}\frac{{b}_{nn}{\lambda }_{n}}{{a}_{nn}},\hfill & \text{if}v=n,\hfill \\ 0,\hfill & \text{if}v>n\text{.}\hfill \end{array}$
Now
$\sum {|\overline{{T}_{n}\left(1\right)}|}^{k}<\mathrm{\infty }\phantom{\rule{1em}{0ex}}\text{whenever}\sum |\overline{\mathrm{\Delta }}{x}_{n}|<\mathrm{\infty }$
is equivalently
$\underset{v}{sup}\sum _{n=1}^{\mathrm{\infty }}{|{c}_{nv}|}^{k}<\mathrm{\infty }$
(32)
by Lemma. But it follows from conditions (20), (21) and (23) that
$\begin{array}{rcl}\sum _{n=v}^{\mathrm{\infty }}{|{c}_{nv}|}^{k}& =& O\left(1\right)\left\{{n}^{k-1}{|\frac{{b}_{nn}{\lambda }_{n}}{{a}_{nn}}|}^{k}+\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}|\frac{{\mathrm{\Delta }}_{v}\left({\stackrel{ˆ}{b}}_{nv}{\lambda }_{v}\right)}{{a}_{vv}}+{\stackrel{ˆ}{b}}_{n,v+1}{\lambda }_{v+1}\frac{{a}_{vv}-{a}_{v+1,v}}{{a}_{vv}{a}_{v+1,v+1}}{|}^{k}\right\}\\ =& O\left(1\right)\phantom{\rule{1em}{0ex}}\text{as}v\to \mathrm{\infty }.\end{array}$
Finally,
$\overline{{T}_{n}\left(2\right)}={n}^{1-\frac{1}{k}}{T}_{n}\left(2\right)={n}^{1-\frac{1}{k}}\sum _{r=0}^{n-2}\overline{\mathrm{\Delta }}{x}_{r}\sum _{v=r+2}^{n}{\stackrel{ˆ}{b}}_{nv}{\stackrel{ˆ}{a}}_{vr}^{\prime }{\lambda }_{v}=\sum _{r=0}^{\mathrm{\infty }}{d}_{nr}\overline{\mathrm{\Delta }}{x}_{r},$
where
${d}_{nr}=\left\{\begin{array}{cc}{n}^{1-\frac{1}{k}}{\sum }_{v=r+2}^{n}{\stackrel{ˆ}{b}}_{nv}{\stackrel{ˆ}{a}}_{vr}^{\prime }{\lambda }_{v},\hfill & \text{if}0\le r\le n-2,\hfill \\ 0,\hfill & \text{if}r>n-2\text{.}\hfill \end{array}$
Now
$\sum {|\overline{{T}_{n}\left(2\right)}|}^{k}<\mathrm{\infty }\phantom{\rule{1em}{0ex}}\text{whenever}\sum |\overline{\mathrm{\Delta }}{x}_{n}|<\mathrm{\infty }$
is equivalently
$\underset{r}{sup}\sum _{n=1}^{\mathrm{\infty }}{|{d}_{nr}|}^{k}<\mathrm{\infty }$
(33)
by Lemma. But it follows from conditions (21) and (24) that
$\begin{array}{rcl}\sum _{n=r+2}^{\mathrm{\infty }}{|{d}_{nr}|}^{k}& =& O\left(1\right)\sum _{n=r+2}^{\mathrm{\infty }}{n}^{k-1}{\left\{\sum _{v=r+2}^{\mathrm{\infty }}|{\stackrel{ˆ}{b}}_{nv}{\stackrel{ˆ}{a}}_{vr}^{\prime }{\lambda }_{v}|\right\}}^{k}\\ =& O\left(1\right)\sum _{n=r+2}^{\mathrm{\infty }}{n}^{k-1}{|{\stackrel{ˆ}{b}}_{n,r+1}{\lambda }_{r+1}|}^{k}\\ =& O\left(1\right)\phantom{\rule{1em}{0ex}}\text{as}r\to \mathrm{\infty }.\end{array}$
Therefore, we have
$\sum _{n=1}^{\mathrm{\infty }}{n}^{k-1}|{T}_{n}\left(i\right){|}^{k}<\mathrm{\infty }\phantom{\rule{1em}{0ex}}\text{for}i=1,2.$

This completes the proof of the Theorem. □

## Authors’ Affiliations

(1)
Department of Mathematics, Erciyes University, Kayseri, 38039, Turkey

## References 