• Research
• Open Access

# Some new results on convex sequences

Journal of Inequalities and Applications20172017:165

https://doi.org/10.1186/s13660-017-1438-4

• Received: 11 May 2017
• Accepted: 26 June 2017
• Published:

## Abstract

In the present paper, we obtained a main theorem related to factored infinite series. Some new results are also deduced.

## Keywords

• absolute summability
• convex sequence
• Minkowsky inequality
• Hölder inequality

• 26D15
• 40D15
• 40F05

## 1 Introduction

Let $$\sum a_{n}$$ be a given infinite series with $$(s_{n})$$ as the sequence of partial sums. In , Borwein introduced the $$(C,\alpha ,\beta)$$ methods in the following form: Let $$\alpha+\beta\ne -1,-2,\ldots$$ . Then the $$(C,\alpha,\beta)$$ mean is defined by
$${u_{n}^{\alpha,\beta}}=\frac{1}{A_{n}^{\alpha+\beta}} \sum _{v=1}^{n}{A_{n-v}^{\alpha-1}} {A_{v}^{\beta}}s_{v},$$
(1)
where
$${A_{n}^{\alpha+\beta}}= O \bigl(n^{\alpha+\beta} \bigr),\qquad {A_{0}^{\alpha+\beta}=1}\quad \mbox{and}\quad {A_{-n}^{\alpha+\beta}=0} \quad \mbox{for } n>0.$$
(2)
The series $$\sum{a_{n}}$$ is said to be summable $${ \vert {C},\alpha,\beta ,\sigma;\delta \vert }_{k}$$, $$k\geq1$$, $$\delta\geq0$$, $$\alpha+\beta>-1$$, and $$\sigma\in {R}$$, if (see )
$$\sum_{n=1}^{\infty}n^{\sigma(\delta k+k-1)} \frac{ \vert t_{n}^{\alpha,\beta} \vert ^{k}}{{n}^{k}}< \infty,$$
(3)
where $${t_{n}^{\alpha,\beta}}$$ is the $$(C,\alpha,\beta)$$ transform of the sequence $$(na_{n})$$. It should be noted that, for $${\beta=0}$$, the $${ \vert {C},\alpha,\beta ,\sigma;\delta \vert }_{k}$$ summability method reduces to the $${ \vert {C},\alpha,\sigma;\delta \vert }_{k}$$ summability method (see ). Let us consider the sequence $$(\theta_{n}^{\alpha,\beta})$$ which is defined by (see )
\begin{aligned} \theta_{n}^{\alpha,\beta}= \textstyle\begin{cases} \vert t_{n}^{\alpha,\beta} \vert , & \alpha=1,\beta>-1, \\ \max_{1\leq v\leq n} \vert t_{v}^{\alpha,\beta} \vert , & 0< \alpha< 1, \beta>-1. \end{cases}\displaystyle \end{aligned}
(4)

## 2 The main result

Here, we shall prove the following theorem.

### Theorem

If $$(\lambda_{n})$$ is a convex sequence (see ) such that the series $$\sum\frac{\lambda_{n}}{n}$$ is convergent and let $$(\theta _{n}^{\alpha,\beta})$$ be a sequence defined as in (4). If the condition
\begin{aligned} \sum_{n=1}^{m}n^{\sigma(\delta k+k-1)} \frac{(\theta_{n}^{\alpha,\beta })^{k}}{n^{k-1}}=O(m) \quad\textit{as } {m\rightarrow\infty} \end{aligned}
(5)
holds, then the series $$\sum a_{n} \lambda_{n}$$ is summable $${ \vert {C},\alpha,\beta,\sigma;\delta \vert }_{k}$$, $$k\geq1$$, $$0\leq \delta<\alpha\leq1$$, $$\sigma\in{R}$$, and $$({\alpha+\beta +1})k-{\sigma(\delta k+k-1)}>1$$.

One should note that, if we set $$\sigma=1$$, then we obtain a well-known result of Bor (see ).

We will use the following lemmas for the proof of the theorem given above.

### Lemma 1



If $$0<\alpha\leq1$$, $$\beta>-1$$, and $$1 \leq v \leq n$$, then
\begin{aligned} \Biggl\vert {\sum_{p=0}^{v}A_{n-p}^{\alpha-1}A_{p}^{\beta}a_{p}} \Biggr\vert \leq \max_{1 \leq m \leq v} \Biggl\vert {\sum _{p=0}^{m} A_{m-p}^{\alpha-1}A_{p}^{\beta}a_{p}} \Biggr\vert . \end{aligned}
(6)

### Lemma 2



If $$(\lambda_{n})$$ is a convex sequence such that the series $$\sum\frac{\lambda_{n}}{n}$$ is convergent, then $$n{\Delta\lambda_{n}}\rightarrow0\textit{ as }n\rightarrow\infty$$ and $$\sum_{n=1}^{\infty} (n+1)\Delta^{2} {\lambda_{n}}$$ is convergent.

## 3 Proof of the theorem

Let $$(T_{n}^{\alpha,\beta})$$ be the nth $$(C,\alpha,\beta)$$ mean of the sequence $$(n{a_{n}}{\lambda_{n}})$$. Then, by (1), we have
\begin{aligned} T_{n}^{\alpha,\beta} = & \frac{1}{A_{n}^{\alpha+\beta}} \sum _{v=1}^{n}A_{n-v}^{\alpha-1}A_{v}^{\beta} va_{v}\lambda_{v} . \end{aligned}
First applying Abel’s transformation and then using Lemma 1, we have
\begin{aligned} &T_{n}^{\alpha,\beta} = \frac{1}{A_{n}^{\alpha+\beta}} \sum _{v=1}^{n-1}\Delta\lambda_{v} \sum _{p=1}^{v}A_{n-p}^{\alpha-1}A_{p}^{\beta}p a_{p}+ \frac{\lambda_{n}}{A_{n}^{\alpha+\beta}} \sum_{v=1}^{n} {A_{n-v}^{\alpha-1}}A_{v}^{\beta}va_{v}, \\ &\bigl\vert T_{n}^{\alpha,\beta} \bigr\vert \leq \frac{1}{A_{n}^{\alpha+\beta}} \sum_{v=1}^{n-1} \vert { \Delta\lambda_{v}} \vert \Biggl\vert {\sum _{p=1}^{v}A_{n-p}^{\alpha-1}A_{p}^{\beta}p a_{p}} \Biggr\vert + \frac{ \vert \lambda_{n} \vert }{A_{n}^{\alpha+\beta}} \Biggl\vert \sum _{v=1}^{n} {A_{n-v}^{\alpha-1}} {A_{v}^{\beta}}v{a_{v}} \Biggr\vert \\ &\phantom{\bigl\vert T_{n}^{\alpha,\beta} \bigr\vert }\leq \frac{1}{A_{n}^{\alpha+\beta}} \sum_{v=1}^{n-1} A_{v}^{\alpha}A_{v}^{\beta} \theta_{v}^{\alpha,\beta} \vert {\Delta \lambda_{v}} \vert + \vert {\lambda_{n}} \vert \theta_{n}^{\alpha,\beta} \\ &\phantom{\bigl\vert T_{n}^{\alpha,\beta} \bigr\vert }= T_{n,1}^{\alpha,\beta} + T_{n,2}^{\alpha,\beta}. \end{aligned}
In order to complete the proof of the theorem by using Minkowski’s inequality, it is sufficient to show that
$$\sum_{n=1}^{\infty}n^{\sigma(\delta k+k-1)} \frac{ \vert T_{n,r}^{\alpha,\beta} \vert ^{k}}{n^{k}} < \infty,\quad \mbox{for } r=1,2.$$
For $$k>1$$, we can apply Hölder’s inequality with indices k and $${k'}$$, where $$\frac{1}{k}+\frac{1}{k'}=1$$, and we obtain
\begin{aligned} \sum_{n=2}^{m+1}n^{\sigma(\delta k+k-1)} \frac{ \vert {T_{n,1}^{\alpha,\beta}} \vert ^{k}}{n^{k}} \leq{}&\sum_{n=2}^{m+1} n^{\sigma(\delta k+k-1)-k} \Biggl\vert { \frac{1}{A_{n}^{\alpha+\beta}} \sum _{v=1}^{n-1}A_{v}^{\alpha}A_{v}^{\beta} \theta_{v}^{\alpha,\beta}\Delta \lambda_{v}} \Biggr\vert ^{k} \\ = {}& O(1) \sum_{n=2}^{m+1} \frac{1}{n^{(\alpha+ \beta +1)k-\sigma(\delta k+k-1)}} \Biggl\{ \sum_{v=1}^{n-1} v^{\alpha k}v^{\beta k} \Delta\lambda_{v} \bigl( \theta_{v}^{\alpha,\beta} \bigr)^{k} \Biggr\} \\ &{}\times \Biggl\{ \sum_{v=1}^{n-1}\Delta \lambda_{v} \Biggr\} ^{k-1} \\ = {}& O(1)\sum_{v=1}^{m} v^{(\alpha+\beta) k} \Delta\lambda_{v} \bigl(\theta_{v}^{\alpha,\beta} \bigr)^{k} \sum_{n=v+1}^{m+1} \frac{1}{n^{(\alpha+ \beta+1)k-\sigma(\delta k+k-1)}} \\ = {}& O(1)\sum_{v=1}^{m}v^{(\alpha+\beta) k} \Delta\lambda_{v} \bigl(\theta_{v}^{\alpha,\beta} \bigr)^{k} \int_{v}^{\infty} \frac{dx}{ x^{(\alpha+ \beta+1)k-\sigma(\delta k+k-1)}} \\ = {}& O(1)\sum_{v=1}^{m}\Delta \lambda_{v}v^{\sigma(\delta k+k-1)} \frac{(\theta_{v}^{\alpha,\beta})^{k}}{v^{k-1}} \\ = {}& O(1)\sum_{v=1}^{m-1}\Delta(\Delta \lambda_{v})\sum_{p=1}^{v} p^{\sigma(\delta k+k-1)}\frac{(\theta_{p}^{\alpha,\beta})^{k}}{p^{k-1}} \\ &{}+ O(1)\Delta\lambda_{m}\sum_{v=1}^{m} v^{\sigma(\delta k+k-1)}\frac {(\theta_{v}^{\alpha,\beta})^{k}}{v^{k-1}} \\ ={} & O(1)\sum_{v=1}^{m} v \Delta^{2}\lambda_{v} + O(1)m {\Delta\lambda_{m}} = O(1) \quad\mbox{as } {m\rightarrow\infty}, \end{aligned}
by virtue of hypotheses of the theorem and Lemma 2. Similarly, we have
\begin{aligned} \sum_{n=1}^{m}n^{\sigma(\delta k+k-1)} \frac{ \vert {T_{n,2}^{\alpha,\beta}} \vert ^{k}}{n^{k}} = {}& O(1)\sum_{n=1}^{m} \frac {\lambda_{n}}{n}n^{\sigma(\delta k+k-1)}\frac{(\theta_{n}^{\alpha,\beta })^{k}}{n^{k-1}} \\ = {}& O(1)\sum_{n=1}^{m-1} \Delta \biggl( \frac{\lambda_{n}}{n} \biggr)\sum_{v=1}^{n}v^{\sigma(\delta k+k-1)} \frac{({\theta_{v}^{\alpha,\beta}})^{k}}{v^{k-1}} \\ &{} + O(1)\frac{\lambda_{m}}{m}\sum_{n=1}^{m}n^{\sigma(\delta k+k-1)} \frac{({\theta_{n}^{\alpha,\beta}})^{k}}{n^{k-1}} \\ = {}& O(1)\sum_{n=1}^{m-1}\Delta \lambda_{n} + O(1)\sum_{n=1}^{m-1} \frac{\lambda_{n+1}}{n+1} + O(1)\lambda_{m} \\ ={} & O(1)\sum_{n=1}^{m-1}\Delta \lambda_{n} + O(1)\sum_{n=2}^{m-1} \frac{\lambda_{n}}{n} + O(1)\lambda_{m} \\ = {}& O(1) (\lambda_{1}-\lambda_{m})+ O(1)\sum _{n=1}^{m-1}\frac{\lambda_{n}}{n} + O(1) \lambda_{m} \\ = {}&O(1) \quad\mbox{as } {m\rightarrow\infty} \end{aligned}
in view of hypotheses of the theorem and Lemma 2. This completes the proof of the theorem.

## 4 Conclusions

By selecting proper values for α, β, δ, and σ, we have some new results concerning the $${ \vert {C,1} \vert }_{k}$$, $${ \vert {C},\alpha \vert }_{k}$$, and $${ \vert {C},\alpha ;\delta \vert }_{k}$$ summability methods.

## Declarations 