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Some new results on convex sequences

Journal of Inequalities and Applications20172017:165

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


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


  • 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 [1], 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
$$ {u_{n}^{\alpha,\beta}}=\frac{1}{A_{n}^{\alpha+\beta}} \sum _{v=1}^{n}{A_{n-v}^{\alpha-1}} {A_{v}^{\beta}}s_{v}, $$
$$ {A_{n}^{\alpha+\beta}}= O \bigl(n^{\alpha+\beta} \bigr),\qquad {A_{0}^{\alpha+\beta}=1}\quad \mbox{and}\quad {A_{-n}^{\alpha+\beta}=0} \quad \mbox{for } n>0. $$
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 [2])
$$ \sum_{n=1}^{\infty}n^{\sigma(\delta k+k-1)} \frac{ \vert t_{n}^{\alpha,\beta} \vert ^{k}}{{n}^{k}}< \infty, $$
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 [3]). Let us consider the sequence \((\theta_{n}^{\alpha,\beta})\) which is defined by (see [4])
$$\begin{aligned} \theta_{n}^{\alpha,\beta}= \textstyle\begin{cases} \vert t_{n}^{\alpha,\beta} \vert , & \alpha=1,\beta>-1, \\ \max_{1\leq v\leq n} \vert t_{v}^{\alpha,\beta} \vert , & 0< \alpha< 1, \beta>-1. \end{cases}\displaystyle \end{aligned}$$

2 The main result

Here, we shall prove the following theorem.


If \((\lambda_{n})\) is a convex sequence (see [5]) 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
$$\begin{aligned} \sum_{n=1}^{m}n^{\sigma(\delta k+k-1)} \frac{(\theta_{n}^{\alpha,\beta })^{k}}{n^{k-1}}=O(m) \quad\textit{as } {m\rightarrow\infty} \end{aligned}$$
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 [6]).

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
$$\begin{aligned} \Biggl\vert {\sum_{p=0}^{v}A_{n-p}^{\alpha-1}A_{p}^{\beta}a_{p}} \Biggr\vert \leq \max_{1 \leq m \leq v} \Biggl\vert {\sum _{p=0}^{m} A_{m-p}^{\alpha-1}A_{p}^{\beta}a_{p}} \Biggr\vert . \end{aligned}$$

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.


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Authors’ Affiliations

Independent Consultant, P.O. Box 121, TR-06502 Bahçelievler, Ankara, Turkey


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