Comptes Rendus
Mathématique

Şebnem Yıldız

A new extension on the theorem of Bor

Volume 359, issue 5 (2021), p. 555-562

Published online: 13 July 2021

https://doi.org/10.5802/crmath.195

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Numerical analysis / Analyse numérique

# A new extension on the theorem of Bor

Şebnem Yıldız<sup>a</sup>

**Abstract.** In [8], Bor has obtained a main theorem dealing with Riesz summability factors of infinite series and Fourier series. In this paper, we generalized that theorem to $|A, \theta_n|_k$ summability method for taking power increasing sequence. Also some new and known results are obtained dealing with some basic summability methods.

**2020 Mathematics Subject Classification.** 26D15, 42A24, 40F05, 40G99.

Manuscript received 10th October 2020, revised 26th November 2020, accepted 1st March 2021.

## 1. Introduction

Let $\sum a_n$ be a given infinite series with partial sums $(s_n)$. By $u_n^\alpha$ and $t_n^\alpha$ we denote the nth Cesàro means of order $\alpha$, with $\alpha > -1$, of the sequence $(s_n)$ and $(na_n)$, respectively, that is (see [9])

$$u_n^\alpha = \frac{1}{A_n^\alpha} \sum_{\nu=0}^{n} A_{n-\nu}^{\alpha-1} s_\nu \quad \text{and} \quad t_n^\alpha = \frac{1}{A_n^\alpha} \sum_{\nu=1}^{n} A_{n-\nu}^{\alpha-1} v a_\nu, \qquad (1)$$

where

$$A_n^\alpha = \frac{(\alpha+1)(\alpha+2)\dots(\alpha+n)}{n!} = O(n^\alpha), \quad A_{-n}^\alpha = 0 \quad \text{for } n > 0. \qquad (2)$$

The series $\sum a_n$ is said to be summable $|C, \alpha|_k$, $k \ge 1$, if (see [11, 13])

$$\sum_{n=1}^{\infty} n^{k-1} |u_n^{\alpha} - u_{n-1}^{\alpha}|^k = \sum_{n=1}^{\infty} \frac{1}{n} |t_n^{\alpha}|^k < \infty. \qquad (3)$$

If we take $\alpha = 1$, then $|C, \alpha|_k$ summability reduces to $|C, 1|_k$ summability.

Let $(p_n)$ be a sequence of positive real numbers such that

$$P_n = \sum_{\nu=0}^{n} p_\nu \to \infty \quad \text{as} \quad n \to \infty, \quad (P_{-i} = p_{-i} = 0, \ i \ge 1). \qquad (4)$$

The sequence-to-sequence transformation

$$w_n = \frac{1}{P_n} \sum_{\nu=0}^{n} p_\nu s_\nu, \quad P_n \neq 0. \qquad (5)$$

defines the sequence $(w_n)$ of the Riesz mean or simply the $(\overline{N}, p_n)$ mean of the sequence $(s_n)$ generated by the sequence of coefficients $(p_n)$ (see [12]).

<footnote><sup>a</sup> Department of Mathematics, Ahi Evran University, Kırşehir, Turkey</footnote>

<footnote>E-mail: sebnemyildiz@ahievran.edu.tr</footnote>
Let $(\theta_n)$ be any sequence of positive constants. The series $\sum a_n$ is said to be summable $|\bar{N}, p_n; \theta_n|_k$, $k \ge 1$, if (see [18])

$$ \sum_{n=1}^{\infty} \theta_n^{k-1} |w_n - w_{n-1}|^k < \infty. \quad (6) $$

In the special case if we take $\theta_n = P_n/p_n$, then $|\bar{N}, p_n; \theta_n|_k$ summability reduces to $|\bar{N}, p_n|_k$ summability (see [2]). When $\theta_n = n$ and $p_n = 1$ for all values of $n$, then we get $|C, 1|_k$ summability. Furthermore, if we take $\theta_n = n$, then $|\bar{N}, p_n; \theta_n|_k$ summability reduces to $|R, p_n|_k$ summability (see [3]).

For any sequence $(\lambda_n)$ we write that

$$ \Delta^2 \lambda_n = \Delta \lambda_n - \Delta \lambda_{n+1} \quad \text{and} \quad \Delta \lambda_n = \lambda_n - \lambda_{n+1}. $$

A sequence $(\lambda_n)$ is said to be of bounded variation, denoted by $(\lambda_n) \in \mathcal{BV}$, if

$$ \sum_{n=1}^{\infty} |\Delta \lambda_n| < \infty. $$

Let $A = (a_{nv})$ 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 $s = (s_n)$ to $As = (A_n(s))$, where

$$ A_n(s) = \sum_{v=0}^{n} a_{nv} s_v, \quad n = 0, 1, \dots \quad (7) $$

Let $(\theta_n)$ be any sequence of positive real numbers. The series $\sum a_n$ is said to be summable $|A, \theta_n|_k$, $k \ge 1$, if (see [16, 17])

$$ \sum_{n=1}^{\infty} \theta_n^{k-1} |\bar{\Delta} A_n(s)|^k < \infty, \quad (8) $$

where

$$ \bar{\Delta} A_n(s) = A_n(s) - A_{n-1}(s). \quad (9) $$

If we take $a_{nv} = \frac{p_v}{p_n}$, then $|A, \theta_n|_k$ summability reduces to $|\bar{N}, p_n; \theta_n|_k$ summability. If we take $\theta_n = \frac{p_n}{p_n}$, then $|A, \theta_n|_k$ summability reduces to $|A, p_n|_k$ summability (see [19]). And also if we take $\theta_n = \frac{p_n}{p_n}$ and $a_{nv} = \frac{p_v}{p_n}$, then $|A, \theta_n|_k$ summability reduces to $|\bar{N}, p_n|_k$ summability. Furthermore, if we take $\theta_n = n$, $a_{nv} = \frac{p_v}{p_n}$ and $p_n = 1$ for all values of $n$, then $|A, \theta_n|_k$ summability reduces to $|C, 1|_k$ summability (see [11]). Finally, if we take $\theta_n = n$ and $a_{nv} = \frac{p_v}{p_n}$, then $|A, \theta_n|_k$ summability reduces to $|R, p_n|_k$ summability (see [3]).

**Definition 1 (cf. [1]).** A positive sequence $(b_n)$ is said to be an almost increasing sequence if there exists a positive increasing sequence $(c_n)$ and two positive constants $M$ and $N$ such that $Mc_n \le b_n \le Nc_n$.

**Definition 2 (cf. [20]).** A positive sequence $X = (X_n)$ is said to be quasi-$f$-power increasing sequence if there exists a constant $K = K(X, f) \ge 1$ such that $Kf_nX_n \ge f_mX_m$ for all $n \ge m \ge 1$, where $f = \{f_n(\sigma, \beta)\} = \{n^\sigma (\log n)^\beta, \beta \ge 0, 0 < \sigma < 1\}$.

If we take $\beta = 0$, then we have a quasi-$\sigma$-power increasing sequence (see [15]). Every almost increasing sequence is a quasi-$\sigma$-power increasing sequence for any non-negative $\sigma$, but the converse is not true for $\sigma > 0$.
## 2. The Known Results

Recently, many papers have been done for absolute matrix summability factors of infinite series and Fourier series (see [5-7, 14, 22, 23]). From these, in [14], Lee explained history of summability of infinite series and Hüseyin Bor briefly. Now we also used Bor's new theorem dealing with the Fourier series and we will extend following theorem.

**Theorem 3 (cf. [8]).** Let $(\theta_n p_n / P_n)$ be a non-increasing sequence. Let $(p_n)$ be a sequence of positive numbers such that

$$P_n = O(np_n) \quad \text{as} \quad n \to \infty. \qquad (10)$$

Let $(X_n)$ be a positive increasing sequence. If the conditions

$$\lambda_n = o(1) \quad \text{as} \quad n \to \infty, \qquad (11)$$

$$\sum_{n=1}^{m} nX_n|\Delta^2\lambda_n| = O(1) \quad \text{as} \quad m \to \infty, \qquad (12)$$

$$\sum_{n=1}^{m} \theta_{n}^{k-1} \left( \frac{p_n}{P_n} \right)^k \frac{|t_n|^k}{X_n^{k-1}} = O(X_m) \quad \text{as} \quad m \to \infty, \qquad (13)$$

are satisfied, then the series $\sum a_n \lambda_n$ is summable $|\bar{N}, p_n, \theta_n|_k, k \ge 1$.

If we take $\theta_n = P_n / p_n$, then we get a theorem dealing with $|\bar{N}, p_n|_k$ summability (see [6]).

## 3. The Main Result

Given a normal matrix $A = (a_{nv})$, we associate two lower semimatrices $\bar{A} = (\bar{a}_{nv})$ and $\hat{A} = (\hat{a}_{nv})$ as follows:

$$\bar{a}_{nv} = \sum_{i=v}^{n} a_{ni}, \quad n, v = 0, 1, \dots \qquad (14)$$

and

$$\hat{a}_{00} = \bar{a}_{00} = a_{00}, \quad \hat{a}_{nv} = \bar{a}_{nv} - \bar{a}_{n-1,v}, \quad n = 1, 2, \dots \qquad (15)$$

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

$$A_n(s) = \sum_{v=0}^{n} a_{nv} s_v = \sum_{v=0}^{n} \bar{a}_{nv} a_v \qquad (16)$$

and

$$\bar{\Delta} A_n(s) = \sum_{v=0}^{n} \hat{a}_{nv} a_v. \qquad (17)$$

By using above notations, we generalize Theorem 3 for $|A, \theta_n|_k$ summability method by taking $(X_n)$ as a quasi-$f$-power increasing sequence.

**Theorem 4.** Let $k \ge 1$ and $A = (a_{nv})$ be a positive normal matrix such that

$$\bar{a}_{n0} = 1, \quad n = 0, 1, \dots, \qquad (18)$$

$$a_{n-1,v} \ge a_{nv}, \quad \text{for } n \ge v+1, \qquad (19)$$

$$1 = O(na_{nn}), \qquad (20)$$

$$\sum_{v=1}^{n-1} a_{vv} |\hat{a}_{n,v+1}| = O(a_{nn}). \qquad (21)$$
Let $(\theta_n a_{nn})$ be a non-increasing sequence and $(X_n)$ be a quasi-$f$-power increasing sequence for some $\sigma$ $(0 < \sigma < 1)$. If the conditions (11)-(12) of Theorem 3 and $(\theta_n)$ holds for the following condition,

$$ \sum_{n=1}^{m} \theta_n^{k-1} a_{nn}^k \frac{|t_n|^k}{X_n^{k-1}} = O(X_m) \quad \text{as} \quad m \to \infty, \qquad (22) $$

are satisfied, then the series $\sum a_n \lambda_n$ is summable $|A, \theta_n|_k, k \ge 1$.

We need the following lemmas for the proof of Theorem 4.

**Lemma 5 (cf. [21]).** By using conditions (14), (15), (18) and (19), we have

$$ \sum_{v=1}^{n-1} |\Delta_v(\hat{a}_{nv})| \le a_{nn}, \qquad (23) $$

$$ \sum_{n=v+1}^{m+1} |\Delta_v(\hat{a}_{nv})| \le a_{vv}, \qquad (24) $$

$$ \sum_{n=v+1}^{m+1} |\hat{a}_{n,v+1}| = O(1). \qquad (25) $$

**Lemma 6 (cf. [4]).** Under the conditions of Theorem 3 we have the following

$$ nX_n|\Delta\lambda_n| = O(1) \quad \text{as} \quad n \to \infty, \qquad (26) $$

$$ \sum_{n=1}^{\infty} X_n |\Delta \lambda_n| < \infty, \qquad (27) $$

$$ X_n|\lambda_n| = O(1) \quad \text{as} \quad n \to \infty. \qquad (28) $$

**Proof of Theorem 4.** Let $(I_n)$ denotes the A-transform of the series $\sum_{n=1}^{\infty} a_n \lambda_n$. Then, by (16) and (17), we have

$$ \bar{\Delta} I_n = \sum_{v=1}^{n} \hat{a}_{nv} a_v \lambda_v. $$

Applying Abel's transformation to this sum, we have that

$$ \begin{aligned} \bar{\Delta} I_n &= \sum_{v=1}^{n} \hat{a}_{nv} a_v \lambda_v \frac{v}{v} = \sum_{v=1}^{n-1} \Delta \left( \frac{\hat{a}_{nv} \lambda_v}{v} \right) \sum_{r=1}^{v} r a_r + \frac{\hat{a}_{nn} \lambda_n}{n} \sum_{v=1}^{n} v a_v \\ &= \sum_{v=1}^{n-1} \Delta \left( \frac{\hat{a}_{nv} \lambda_v}{v} \right) (v+1) t_v + \hat{a}_{nn} \lambda_n \frac{n+1}{n} t_n \\ &= \sum_{v=1}^{n-1} \Delta_v (\hat{a}_{nv}) \lambda_v t_v \frac{v+1}{v} + \sum_{v=1}^{n-1} \hat{a}_{n,v+1} \Delta \lambda_v t_v \frac{v+1}{v} + \sum_{v=1}^{n-1} \hat{a}_{n,v+1} \lambda_{v+1} \frac{t_v}{v} + a_{nn} \lambda_n t_n \frac{n+1}{n} \\ &= I_{n,1} + I_{n,2} + I_{n,3} + I_{n,4}. \end{aligned} $$

To complete the proof of Theorem 4, by Minkowski's inequality, it is sufficient to show that

$$ \sum_{n=1}^{\infty} \theta_n^{k-1} |I_{n,r}|^k < \infty, \quad \text{for} \quad r = 1, 2, 3, 4. \qquad (29) $$
First, by applying Hölder's inequality with indices $k$ and $k'$, where $k > 1$ and $\frac{1}{k} + \frac{1}{k'} = 1$, we have that

$$
\begin{align*}
\sum_{n=2}^{m+1} \theta_n^{k-1} |I_{n,1}|^k &\le \sum_{n=2}^{m+1} \theta_n^{k-1} \left\{ \sum_{v=1}^{n-1} \left| \frac{v+1}{v} \right| |\Delta_v(\hat{a}_{nv})| |\lambda_v| |t_v| \right\}^k \\
&= O(1) \sum_{n=2}^{m+1} \theta_n^{k-1} \left\{ \sum_{v=1}^{n-1} |\Delta_v(\hat{a}_{nv})| |\lambda_v|^k |t_v|^k \right\} \times \left\{ \sum_{v=1}^{n-1} |\Delta_v(\hat{a}_{nv})| \right\}^{k-1} \\
&= O(1) \sum_{n=2}^{m+1} (\theta_n a_{nn})^{k-1} \left\{ \sum_{v=1}^{n-1} |\Delta_v(\hat{a}_{nv})| |\lambda_v|^k |t_v|^k \right\} \\
&= O(1) \sum_{v=1}^{m} |\lambda_v|^{k-1} |\lambda_v| |t_v|^k \sum_{n=v+1}^{m+1} (\theta_n a_{nn})^{k-1} |\Delta_v(\hat{a}_{nv})| \\
&= O(1) \sum_{v=1}^{m} (\theta_v a_{vv})^{k-1} \frac{1}{X_v^{k-1}} |\lambda_v| |t_v|^k \sum_{n=v+1}^{m+1} |\Delta_v(\hat{a}_{nv})| \\
&= O(1) \sum_{v=1}^{m} (\theta_v a_{vv})^{k-1} \frac{1}{X_v^{k-1}} |\lambda_v| |t_v|^k a_{vv} \\
&= O(1) \sum_{v=1}^{m-1} \Delta |\lambda_v| \sum_{r=1}^{v} \theta_r^{k-1} a_{rr}^k \frac{|t_r|^k}{X_r^{k-1}} + O(1) |\lambda_m| \sum_{v=1}^{m} \theta_v^{k-1} a_{vv}^k \frac{|t_v|^k}{X_v^{k-1}} \\
&= O(1) \sum_{v=1}^{m-1} |\Delta \lambda_v| X_v + O(1) |\lambda_m| X_m \\
&= O(1) \quad \text{as} \quad m \to \infty,
\end{align*}
$$

by virtue of the hypotheses of Theorem 4, Lemma 5, and Lemma 6. Now using Hölder's inequality,
we have that

$$
\begin{align*}
\sum_{n=2}^{m+1} \theta_n^{k-1} |I_{n,2}|^k &\le \sum_{n=2}^{m+1} \theta_n^{k-1} \left\{ \sum_{v=1}^{n-1} \left| \frac{v+1}{v} \right| |\hat{a}_{n,v+1}| |\Delta \lambda_v| |t_v| \right\}^k \\
&= O(1) \sum_{n=2}^{m+1} \theta_n^{k-1} \left\{ \sum_{v=1}^{n-1} |\hat{a}_{n,v+1}| |\Delta \lambda_v| |t_v| \right\}^k \\
&= O(1) \sum_{n=2}^{m+1} \theta_n^{k-1} \left\{ \sum_{v=1}^{n-1} a_{vv}^{1-k} |\hat{a}_{n,v+1}| |\Delta \lambda_v|^k |t_v|^k \right\} \times \left\{ \sum_{v=1}^{n-1} a_{vv} |\hat{a}_{n,v+1}| \right\}^{k-1} \\
&= O(1) \sum_{n=2}^{m+1} (\theta_n a_{nn})^{k-1} \sum_{v=1}^{n-1} |\Delta \lambda_v|^k a_{vv}^{1-k} |\hat{a}_{n,v+1}| |t_v|^k \\
&= O(1) \sum_{v=1}^{m} |t_v|^k a_{vv}^{1-k} |\Delta \lambda_v|^k \sum_{n=v+1}^{m+1} (\theta_n a_{nn})^{k-1} |\hat{a}_{n,v+1}| \\
&= O(1) \sum_{v=1}^{m} (\theta_v a_{vv})^{k-1} |t_v|^k a_{vv}^{1-k} |\Delta \lambda_v|^k \sum_{n=v+1}^{m+1} |\hat{a}_{n,v+1}| \\
&= O(1) \sum_{v=1}^{m} (\theta_v a_{vv})^{k-1} |t_v|^k a_{vv}^{1-k} |\Delta \lambda_v|^k \\
&= O(1) \sum_{v=1}^{m} \theta_v^{k-1} a_{vv}^k |t_v|^k (v |\Delta \lambda_v|)^{k-1} (v |\Delta \lambda_v|) \\
&= O(1) \sum_{v=1}^{m} \theta_v^{k-1} a_{vv}^k \frac{1}{X_v^{k-1}} |t_v|^k (v |\Delta \lambda_v|)
\end{align*}
$$
$$
\begin{align*}
&= O(1) \sum_{v=1}^{m-1} \Delta(v|\Delta\lambda_v|) \sum_{r=1}^{v} \theta_r^{k-1} a_{rr}^k \frac{1}{X_r^{k-1}} |t_r|^k + O(1) m |\Delta\lambda_m| \sum_{v=1}^{m} \theta_v^{k-1} a_{vv}^k \frac{1}{X_v^{k-1}} |t_v|^k \\
&= O(1) \sum_{v=1}^{m-1} |\Delta(v|\Delta\lambda_v|)| X_v + O(1) m |\Delta\lambda_m| X_m \\
&= O(1) \sum_{v=1}^{m-1} v X_v |\Delta^2\lambda_v| + O(1) \sum_{v=1}^{m-1} X_v |\Delta\lambda_v| + O(1) m |\Delta\lambda_m| X_m \\
&= O(1) \quad \text{as } m \to \infty,
\end{align*}
$$

by virtue of the hypotheses of Theorem 4, Lemma 5, and Lemma 6. Again, as in $I_{n,1}$, we have that

$$
\begin{align*}
\sum_{n=2}^{m+1} \theta_n^{k-1} |I_{n,3}|^k &= \sum_{n=2}^{m+1} \theta_n^{k-1} \left| \sum_{v=1}^{n-1} \hat{a}_{n,v+1} \lambda_{v+1} \frac{t_v}{v} \right|^k \\
&\le \sum_{n=2}^{m+1} \theta_n^{k-1} \left\{ \sum_{v=1}^{n-1} |\hat{a}_{n,v+1}| |\lambda_{v+1}| \frac{|t_v|}{v} \right\}^k \\
&= O(1) \sum_{n=2}^{m+1} \theta_n^{k-1} \left\{ \sum_{v=1}^{n-1} a_{vv}^{1-k} \frac{1}{v^k} |\hat{a}_{n,v+1}| |\lambda_{v+1}|^k |t_v|^k \right\} \times \left\{ \sum_{v=1}^{n-1} a_{vv} |\hat{a}_{n,v+1}| \right\}^{k-1} \\
&= O(1) \sum_{n=2}^{m+1} (\theta_n a_{nn})^{k-1} \sum_{v=1}^{n-1} a_{vv}^{1-k} \frac{1}{v^k} |\hat{a}_{n,v+1}| |\lambda_{v+1}|^k |t_v|^k \\
&= O(1) \sum_{v=1}^{m} a_{vv} |\lambda_{v+1}|^k |t_v|^k \sum_{n=v+1}^{m+1} (\theta_n a_{nn})^{k-1} |\hat{a}_{n,v+1}| \\
&= O(1) \sum_{v=1}^{m} (\theta_v a_{vv})^{k-1} a_{vv} |\lambda_{v+1}|^k |t_v|^k \sum_{n=v+1}^{m+1} |\hat{a}_{n,v+1}| \\
&= O(1) \sum_{v=1}^{m} \theta_v^{k-1} a_{vv}^k |\lambda_{v+1}|^k |t_v|^k \\
&= O(1) \sum_{v=1}^{m} \theta_v^{k-1} a_{vv}^k |t_v|^k |\lambda_{v+1}|^{k-1} |\lambda_{v+1}| \\
&= O(1) \sum_{v=1}^{m} \theta_v^{k-1} a_{vv}^k \frac{1}{X_v^{k-1}} |\lambda_{v+1}| |t_v|^k \\
&= O(1) \quad \text{as } m \to \infty,
\end{align*}
$$

by virtue of the hypotheses of Theorem 4, Lemma 5, and Lemma 6. Finally, as in $I_{n,1}$, we have that

$$
\begin{align*}
\sum_{n=1}^{m} \theta_n^{k-1} |I_{n,4}|^k &= O(1) \sum_{n=1}^{m} \theta_n^{k-1} a_{nn}^k |\lambda_n|^k |t_n|^k = O(1) \sum_{n=1}^{m} \theta_n^{k-1} a_{nn}^k |\lambda_n|^{k-1} |\lambda_n||t_n|^k \\
&= O(1) \sum_{n=1}^{m} \theta_n^{k-1} a_{nn}^k \frac{1}{X_n^{k-1}} |\lambda_n||t_n|^k = O(1) \quad \text{as} \quad m \to \infty,
\end{align*}
$$

by virtue of hypotheses of Theorem 4, Lemma 5, and Lemma 6. This completes the proof of
Theorem 4.
□

**4. An application of absolute matrix summability to Fourier series**

Let $f$ be a periodic function with period $2\pi$ and integrable ($L$) over $(-\pi, \pi)$. The trigonometric Fourier series of $f$ is defined as

$$
f(x) \sim \frac{1}{2} a_0 + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx) = \sum_{n=0}^{\infty} C_n(x). \quad (30)
$$
Set

$$ \phi(t) = \frac{1}{2}\{f(x+t) + f(x-t)\}, \qquad (31) $$

$$ \phi_{\alpha}(t) = \frac{\alpha}{t^{\alpha}} \int_{0}^{t} (t-u)^{\alpha-1} \phi(u) \, du, \quad (\alpha > 0). \qquad (32) $$

It is well known that if $\phi_1(t) \in \mathcal{BV}(0, \pi)$, then $t_n(x) = O(1)$, where $t_n(x)$ is the (C, 1) mean of the sequence $(nC_n(x))$ (see [10]).

The following theorem is known dealing with $|\bar{N}, p_n, \theta_n|_k$ summability factors of Fourier series.

**Theorem 7 (cf. [8]).** Let $(\frac{\theta_n p_n}{p_n})$ be a non-increasing sequence. If $\phi_1(t) \in \mathcal{BV}(0, \pi)$ and the sequences $(p_n)$, $(\lambda_n)$, and $(X_n)$ satisfy the conditions of Theorem 3, then the series $\sum C_n(x)\lambda_n$ is summable $|\bar{N}, p_n, \theta_n|_k$, $k \ge 1$.

Now, we generalize Theorem 7 for $|A, \theta_n|_k$ summability method in the following form.

**Theorem 8.** Let $(\theta_n a_{nn})$ be a non-increasing sequence, and A be a positive normal matrix as in Theorem 4, and $(X_n)$ be a quasi-f-power increasing sequence for some $\sigma$ $(0 < \sigma < 1)$. If $\phi_1(t) \in \mathcal{BV}(0, \pi)$, and the sequences $(p_n)$, $(\lambda_n)$, and $(X_n)$ satisfy the conditions of Theorem 4, then the series $\sum C_n(x)\lambda_n$ is summable $|A, \theta_n|_k$, $k \ge 1$.

It should be noted that if we take $(X_n)$ as a positive increasing sequence and $a_{nv} = \frac{p_v}{p_n}$ in Theorem 8, then we have Theorem 7.

## Applications.

(1) If we write $\sum_{v=0}^n p_v / P_v$, then $(X_n)$ is a positive increasing sequence tending to infinity as $n \to \infty$. In this case, if we take $(X_n)$ is a positive increasing sequence and $a_{nv} = \frac{p_v}{p_n}$ in Theorem 4, then we have Theorem 3.

(2) If we take $\theta_n = n$, $a_{nv} = \frac{p_v}{p_n}$ and $p_n = 1$ for all values of $n$ in Theorem 4, then we have a new result concerning $|C, 1|_k$ summability method.

(3) If we take $\theta_n = n$ and $a_{nv} = \frac{p_v}{p_n}$ in Theorem 4, then we get a new result dealing with $|R, p_n|_k$ summability method.

(4) If we take $\beta = 0$ and $a_{nv} = \frac{p_v}{p_n}$ in Theorem 4, then we have new theorem dealing with quasi-$\sigma$-power increasing sequence.

(5) If we take $\beta = 0$ and $a_{nv} = \frac{p_v}{p_n}$ in Theorem 8, then we have new theorem dealing with quasi-$\sigma$-power increasing sequence and Fourier series.

## Acknowledgments

The author wishes her sincerest thanks to the referee for invaluable suggestions for the improvement of this paper.

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