# Power Increasing Sequences and Their Some New Applications

Authors: Hüseyin Bor
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

In the work of Bor (2008), we have proved a result dealing
with|N−,pn,θn|k summability factors by using a quasi-β-power increasing sequence. In this paper, we prove that result under less and more weaker conditions. Some new results
have also been obtained.


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## Body

## 1. Introduction

A positive sequence(bn) is said to be almost increasing if there exists a positive increasing sequence (cn) and two positive constants A and B such that Acn≤bn≤Bcn (see [1]). We write ℬ𝒱𝒪=ℬ𝒱∩𝒞𝒪, where 𝒞𝒪={x=(xk)∈Ω:limk|xk|=0}, ℬ𝒱={x=(xk)∈Ω:∑k‍|xk-xk+1|<∞} and Ω being the space of all real or complex-valued sequences. A positive sequence X=(Xn) is said to be a quasi-β-power increasing sequence if there exists a constant K=K(β,X)≥1 such that KnβXn≥mβXm holds for all n≥m≥1. It should be noted that every almost increasing sequence is a quasi-β-power increasing sequence for any nonnegative β, but the converse is not true for β>0. Moreover, for any positive β there exists a quasi-β-power increasing sequence tending to infinity, but it is not almost increasing (see [2]). Let ∑‍an be a given infinite series with partial sums (sn). Let (pn) be a sequence of positive numbers such that
(1.1)Pn=∑v=0n‍pv→∞asn→∞,(P-i=p-i=0,i≥1).
Let (θn) be any sequence of positive real constants. The series ∑‍an is said to be summable |N-,pn|k,k≥1, if (see [3])
(1.2)∑n=1∞‍(Pnpn)k-1|Vn-Vn-1|k<∞,
and it is said to be summable |N-,pn,θn|k,k≥1, if (see [4])
(1.3)∑n=1∞‍θnk-1|Vn-Vn-1|k<∞,
where
(1.4)Vn=1Pn∑v=0n‍pvsv.
If we take θn=Pn/pn, then |N-,pn,θn|k summability reduces to |N-,pn|k summability. Also if we take θn=n and pn=1 for all values of n, then we get |C,1|k summability (see [5]). Furthermore, if we take θn=n, then |N-,pn,θn|k summability reduces to |R,pn|k summability (see [6]).


## 2. Known Result

In [7], we have proved the following theorem dealing with |N-,pn,θn|k summability factors of infinite series.Theorem 2.1.
Let(λn)∈ℬ𝒱𝒪, (Xn) be a quasi-β-power increasing sequence for some β(0<β<1), and let (θnpn/Pn) be a nonincreasing sequence. Suppose also there exists sequences (λn) and (pn) such that
(2.1)|λm|Xm=O(1)asm→∞,∑n=1m‍nXn|Δ2λn|=O(1),∑n=1m‍Pnn=O(Pm)asm→∞.
If
(2.2)∑n=1m‍|tn|kn=O(Xm)asm→∞,(2.3)∑n=1m‍θnk-1(pnPn)k|tn|k=O(Xm)asm→∞,
are satisfied, then the series ∑‍anλn is summable |N-,pn,θn|k,k≥1, where (tn) is the nth (C,1) mean of the sequence (nan).Remark 2.2.
It should be noticed that, if we take(Xn) as an almost increasing sequence and θn=Pn/pn, then we obtain a theorem of Mazhar (see [8]), in this case the condition “(λn)∈ℬ𝒱𝒪” is not needed.


## 3. The Main Result

The aim of this paper is to prove Theorem2.1 under less and more weaker conditions. Now, we prove the following theorem.Theorem 3.1.
Let(Xn) be a quasi-β-power increasing sequence for some β(0<β<1), and let (θnpn/Pn) be a nonincreasing sequence. Suppose also there exists sequences (λn) and (pn) such that conditions (2.1) of Theorem 2.1 are satisfied. If
(3.1)∑n=1m‍|tn|knXnk-1=O(Xm)asm→∞,(3.2)∑n=1m‍θnk-1(pnPn)k|tn|kXnk-1=O(Xm)asm→∞
are satisfied, then the series ∑‍anλn is summable |N-,pn,θn|k,  k≥1.Remark 3.2.
It should be noted that conditions (3.1) and (3.2) are the same as conditions (2.2) and (2.3), respectively, when k=1. When k>1, conditions (3.1) and (3.2) are weaker than conditions (2.2) and (2.3), respectively. But the converses are not true. In fact, if (2.2) is satisfied, then we get that
(3.3)∑n=1m‍|tn|knXnk-1=O(1X1k-1)∑n=1m‍|tn|kn=O(Xm).
If (3.1) is satisfied, then for k>1, we obtain that
(3.4)∑n=1m‍|tn|kn=∑n=1m‍Xnk-1|tn|knXnk-1=O(Xmk-1)∑n=1m‍|tn|knXnk-1=O(Xmk)≠O(Xm).
The similar argument is also valid for the conditions (2.3) and (3.2). Also it should be noted that condition “(λn)∈ℬ𝒱𝒪” has been removed.
We need following lemma for the proof of our theorem.Lemma 3.3 (see [9]).
Under the conditions on the sequences(Xn) and (λn) as expressed in the statement of the theorem, one has the following:
(3.5)nXn|Δλn|=O(1),∑n=1∞‍Xn|Δλn|<∞.


## 4. Proof of the Theorem

Let(Tn) denote the (N-,pn) mean of the series ∑‍anλn. Then, for n≥1, we have
(4.1)Tn-Tn-1=pnPnPn-1∑v=1n‍Pv-1avλv=pnPnPn-1∑v=1n‍Pv-1λvvvav.
By Abel's transformation, we have
(4.2)Tn-Tn-1=n+1nPnpntnλn-pnPnPn-1∑v=1n-1‍pvtvλvv+1v+pnPnPn-1∑v=1n-1‍PvΔλvtvv+1v+pnPnPn-1∑v=1n-1‍Pvtvλv+11v=Tn,1+Tn,2+Tn,3+Tn,4.
To complete the proof of the theorem, by Minkowski's inequality, it is enough to show that
(4.3)∑n=1∞‍θnk-1|Tn,r|k<∞,forr=1,2,3,4.
Firstly, we have that
(4.4)∑n=1m‍θnk-1|Tn,1|k=∑n=1m‍θnk-1|λn|k-1|λn|(pnPn)k|tn|k=O(1)∑n=1m‍|λn|θnk-1(1Xv)k-1(pnPn)k|tn|k=O(1)∑n=1m-1‍Δ|λn|∑v=1n‍θvk-1(pvPv)k|tv|kXvk-1+O(1)|λm|∑n=1m‍θnk-1(pnPn)k|tn|kXnk-1=O(1)∑n=1m-1‍|Δλn|Xn+O(1)|λm|Xm=O(1)asm→∞,
by virtue of the hypotheses of the theorem and lemma. Now, when k>1 applying Hölder's inequality with indices k and k', where (1/k)+(1/k′)=1, as in Tn,1, we have that
(4.5)∑n=2m+1θnk-1|Tn,2|k=O(1)∑n=2m+1‍θnk-1(pnPn)k1Pn-1{∑v=1n-1‍pv|λv|k|tv|k}×{1Pn-1∑v=1n-1‍pv}k-1=O(1)∑v=1m‍pv|λv|k-1|λv||tv|k∑n=v+1m+1‍(θnpnPn)k-1pnPnPn-1=O(1)∑v=1m‍(θvpvPv)k-1pv|tv|k|λv|(1Xv)k-1∑n=v+1m+1‍pnPnPn-1=O(1)∑v=1m‍θvk-1(pvPv)k|λv|(1Xv)k-1|tv|k=O(1)∑v=1m‍|λv|θvk-1(pvPv)k|tv|kXvk-1=O(1)asm→∞.
Again we have that
(4.6)∑n=2m+1‍θnk-1|Tn,3|k=O(1)∑n=2m+1‍θnk-1(pnPn)k1Pn-1{∑v=1n-1‍Pvv|Δλv|kvk|tv|k}×{1Pn-1∑v=1n-1‍Pvv}k-1=O(1)∑v=1m‍Pvv|tv|kvk|Δλv|k-1|Δλv|∑n=v+1m+1‍(θnpnPn)k-1pnPnPn-1=O(1)∑v=1m‍(θvpvPv)k-1vk-1(1vXv)k-1|Δλv||tv|k=O(1)(θ1p1P1)k-1∑v=1m‍v|Δλv||tv|kvXvk-1=O(1)∑v=1m-1‍Δ(v|Δλv|)∑i=1v‍|ti|kiXik-1+O(1)m|Δλm|∑v=1m‍|tv|kvXvk-1=O(1)∑v=1m-1‍|Δ(v|Δλv|)|Xv+O(1)m|Δλm|Xm=O(1)∑v=1m-1‍|(v+1)|Δ2λv|-|Δλv||Xv+O(1)m|Δλm|Xm=O(1)∑v=1m-1‍v|Δ2λv|Xv+O(1)∑v=1m-1‍|Δλv|Xv+O(1)m|Δλm|Xm=O(1)asm→∞,
by virtue of the hypotheses of the theorem and lemma. Finally, we have that
(4.7)∑n=2m+1‍θnk-1|Tn,4|k≤∑n=2m+1‍θnk-1(pnPn)k1Pn-1∑v=1n-1‍Pv|λv+1|k|tv|k1v×{1Pn-1∑v=1n-1‍Pvv}k-1=O(1)∑v=1m‍Pv|λv+1|k-1|λv+1||tv|k1v∑n=v+1m+1‍(θnpnPn)k-1pnPnPn-1=O(1)∑v=1m‍Pv(1Xv)k-1|λv+1||tv|k1v∑n=v+1m+1‍(θnpnPn)k-1pnPnPn-1=O(1)∑v=1m‍|λv+1|(θvpvPv)k-1|tv|kvXvk-1=O(1)(θ1p1P1)k-1∑v=1m‍|λv+1||tv|kvXvk-1=O(1)∑v=1m-1‍Δ|λv+1|∑r=1v‍|tr|krXrk-1+O(1)|λm+1|∑v=1m‍|tv|kvXvk-1=O(1)∑v=1m-1‍|Δλv+1|Xv+1+O(1)|λm+1|Xm+1=O(1)asm→∞,
by virtue of the hypotheses of the theorem and lemma. This completes the proof of the theorem. If we take pn=1 for all values of n and θn=n, then we get a result dealing with |C,1|k summability factors. Also, if we take pn=1 for all values of n, then we have a new result for |C,1,θn|k summability. Finally, if we take θn=n, then we have another new result for |R,pn|k summability factors.



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*Source: 436094-2012-10-15.xml*