# Note on Product Summability of an Infinite Series

Authors: W. T. Sulaiman
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

New results concerning product summability of an infinite series are given. Some special cases are also deduced.


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## Body

## 1. Introduction

Let∑an be  a given infinite series with partial
sums sn. Let unα denote the nth Cesaro mean of order α>−1 of the sequence (sn). The
series ∑an is
summable  |C,α|k,k≥1 if(1.1)∑n=1∞nk−1|unα−un−1α|k<∞(Flett [1]). For α=1,|C,α|k reduces  to |C,1|k summability.Let(pn) be a sequence of positive real constants such
that Pn=p0+⋯+pn→∞ as n→∞(P−1=p−1=0). The (N,p) transform ϕn of (sn) generated by (pn) is defined by(1.2)ϕn=1Pn∑v=0npn−vsv. The sequence-to-sequence transformation
(1.3)Φn=1Pn∑v=0npvsv defines the sequence (Φn) of (N¯,pn) transform of (sn) generated by (pn). The series ∑an is summable |R,pn|k,k≥1 if(1.4)∑n=1∞nk−1|Φn−Φn−1|k<∞. In the special case when pn=1 for all n (resp., k = 1), |R,pn|k summability reduces to |C,1|k (resp., |R,pn|) summability.The series∑an is said to be summable |(N,p)(N,q)|, when the (N,p) transform of the (N,q) transform of (sn) is a sequence of bounded variation (see Das [2]).We give the following new
definition.Let(Tn) define the sequence of the (N¯,qn) transform of the (N¯,pn) transform of (sn) generated by the sequences (qn) and (pn), respectively. The series ∑an is said to be summable |(R,qn)(R,pn)|k,k≥1 if(1.5)∑n=1∞nk−1|Tn−Tn−1|k<∞. We may assume through the paper
that Qn=q0+⋯+qn→∞, as n→∞; Rn=r0+⋯+rn→∞, as n→∞.


## 2. New Results

We state and prove the following.Theorem 2.1.
Letk≥1,(λn) be a sequence of constants. Define
(2.1)fv=∑r=vnqrPr,Fv=∑r=vnprfr. Let (2.2)pnQn=O(Pn),(2.3)∑n=v+1∞nk−1qnkQnkQn−1=O((vqv)k−1Qvk).  Then, sufficient conditions for the implication
(2.4)∑an⁢is summable|R,rn|k⟹∑anλn⁢is summable|(R,qn)(R,pn)|k are (2.5)|λv|Fv=O(Qv),(2.6)|λn|=O(Qn),(2.7)pvRv|λv|=O(Qv),(2.8)pvqvRv|λv|=O(QvQv−1rv),(2.9)pnqnRn|λn|=O(PnQnrn),(2.10)Rv−1|Δλv|Fv+1=O(Qvrv),(2.11)Rv−1|Δλv|=O(Qvrv).Proof.
Let(Sn) be the sequence of partial sums of ∑anλn. Let vn,Vn be the (N¯,rn), (N¯,qn)(N¯,pn) transforms of the sequences (sn),(Sn), respectively. We write tn=vn−vn−1,Tn=Vn−Vn−1. Therefore,
(2.12)tn=rnRnRn−1∑v=1nRv−1av,(2.13)Vn=1Qn∑r=0nqr1Pr∑v=0rpvSv=1Qn∑v=0npvSv∑r=vnqrPr=1Qn∑v=0npvSvfv.
Also,
(2.14)Tn=Vn−Vn−1=qnQnQn−1∑r=0nprSrfr+pnSnfnQn−1=qnQnQn−1∑r=0vprfr∑v=0ravλv+pnqnPnQn−1∑v=0navλv=qnQnQn−1∑v=0navλv∑r=vnprfr+pnqnPnQn−1∑v=0navλv=qnQnQn−1∑v=1nRv−1avλvRv−1∑r=vnprfr+pnqnPnQn−1∑v=1nRv−1avλvRv−1=qnQnQn−1(∑v=1n−1(∑r=1vRr−1ar)Δv(λvRv−1∑r=vnprfr)+(∑v=1nRv−1av)λnpnfnRn−1)+pnqnPnQn−1(∑v=1n−1(∑r=1vRr−1ar)Δ(λvRv−1)+(∑v=1nRv−1av)λnRn−1)=qnQnQn−1(∑v=1n−1(tvλvFv+Rv−1rvpvtvλvfv+Rv−1rvtvΔλvFv+1))+pnqnRnQnQn−1rntnλnfn+pnqnPnQn−1(∑v=1n−1(tvλv+Rv−1rvtvΔλv))+pnqnRnPnQn−1rntnλn,=∑j=17Tnj.  In order to complete the proof, it is sufficient to show that
(2.15)∑n=1∞nk−1|Tnj|k<∞,j=1,2,3,4,5,6,7. Applying Holder's inequality,
(2.16)∑n=2m+1nk−1|Tn1|k=∑n=2m+1nk−1|qnQnQn−1∑v=1n−1tvλvFv|k≤∑n=2m+1nk−1qnkQnkQn−1∑v=1n−11qvk−1|tv|k|λv|kFvk(∑v=1n−1qvQn−1)k−1=O(1)∑v=1m1qvk−1|tv|k|λv|kFvk∑n=v+1m+1nk−1qnkQnkQn−1=O(1)∑v=1mvk−1|tv|k|λv|kFvkQvk=O(1),∑n=2m+1nk−1|Tn2|k=∑n=2m+1nk−1|qnQnQn−1∑v=1n−1Rv−1pvrvtvλvfv|k≤∑n=2m+1nk−1qnkQnkQn−1∑v=1n−1Rvkpvkqvk−1|tv|k|λv|kfvk(∑v=1n−1qvQn−1)k−1=O(1)∑v=1mRvkpvkqvk−1|tv|k|λv|kfvk∑n=v+1m+1nk−1qnkQnkQn−1=O(1)∑v=1mvk−1|tv|kRvkpvkQvk|λv|kfvk=O(1)∑v=1mvk−1|tv|k=O(1),∑n=2m+1nk−1|Tn3|k=∑n=2m+1nk−1|qnQnQn−1∑v=1n−1Rv−1rvtvΔλvFv+1|k≤∑n=1m+1nk−1qnkQnkQn−1∑v=1n−1Rv−1kqvk−1rvk|tv|k|Δλv|kFv+1k(∑v=1n−1qvQn−1)k=O(1)∑v=1mRv−1kqvk−1rvk|tv|k|Δλv|kFv+1k∑n=v+1m+1nk−1qnkQnkQn−1=O(1)∑v=1mvk−1|tv|kRv−1kQvkrvk|Δλv|kFv+1k=O(1),∑n=1mnk−1|Tn4|k=∑n=1mnk−1|pnqnRnQnQn−1rntnλnfn|k=O(1)∑n=1mnk−1|tn|k|λn|kpnkqnkRnkQnkQn−1krnk=O(1),∑n=2m+1nk−1|Tn5|k=∑n=2m+1nk−1|pnqnPnQn−1∑v=1n−1tvλv|k≤∑n=1m+1nk−1pnkqnkPnkQn−1∑v=1n−1|tv|k|λv|k1qvk−1(∑v=1n−1qvQn−1)k−1=O(1)∑v=1m|tv|k|λv|k1qvk−1∑n=v+1m+1nk−1pnkqnkPnkQn−1=O(1)∑v=1m|tv|k|λv|k1qvk−1∑n=v+1m+1nk−1qnkQnkQn−1=O(1)∑v=1mvk−1|tv|k|λv|k1Qvk=O(1),∑n=2m+1nk−1|Tn6|k=∑n=2m+1nk−1|pnqnPnQn−1∑v=1n−1Rv−1rvtvΔλv|k≤∑n=2m+1nk−1pnkqnkPnkQn−1∑v=1n−1Rv−1kqk−1rvk|tv|k|Δλv|k(∑v=1n−1qvQn−1)k−1=O(1)∑v=1mRv−1kqvk−1rvk|tv|k|Δλv|k∑n=v+1m+1nk−1pnkqnkPnkQn−1=O(1)∑v=1mvk−1|tv|k|Δλv|kRv−1kQvkrvk=O(1). Finally,
(2.17)∑n=1mnk−1|Tn7|k=∑n=1mnk−1|pnqnRnPnQn−1rntnλn|k=O(1)∑n=1mnk−1|tn|k|λn|k(pnqnRnPnQnrn)k=O(1).
This completes the
proof of the theorem.Theorem 2.2.
Let (2.3) be
satisfied and
(2.18)Pv=O(pvQv),(2.19)Qn=O(nqn). Then, necessary conditions for the implication (2.4) to be satisfied are
(2.20)|λv|=O(QvQv−1rv(1+Fv)qvRv),|λv|=O(v1−1/krvQvpvfvRv),|Δλv|=O(v1−1/krvQv(1+Fv+1)Rv).Proof.
Fork≥1 define
(2.21)A*={(aj):∑aj⁢is summable|R,rn|k},B*={(bj):∑bjλj⁢is summable|(R,qn)(R,pn)|k}. From (2.14), we have
(2.22)Tn=∑v=1n(qnFvQnQn−1+pnqnPnQn−1)avλv. With tn and Tn as defined by (2.12) and (2.22), the spaces A* and B* are BK-spaces with norms defined by
(2.23)‖c‖1={|t0|k+∑n=1∞nk−1|tn|k}1/k,‖c‖2={|T0|k+∑n=1∞nk−1|Tn|k}1/k, respectively. By the hypothesis
of the theorem,
(2.24)‖c‖1<∞⟹‖c‖2<∞. The inclusion map i:A*→B* defined by i(a)=a is continuous since A* and B* are BK-spaces. By the closed graph
theorem, there exists
a constant K>0 such that
(2.25)‖c‖2≤K‖c‖1. Let en denote the nth coordinate vector. From (2.12) and (2.22) with (an) defined by an=en−en+1,n=v,an=0, otherwise, we  have
(2.26)tn={0,n<v,rvRv,n=v,−rnrvRnRn−1,n>v,Tn={0,n<v,(qvFvQvQv−1+pvqvPvQv−1)λv,n=v,Δv((qnFvQnQn−1+pnqnPnQn−1)λv),n>v. From (2.23), we have
(2.27)‖c‖1={vk−1(qvQv)k+∑n=v+1∞nk−1(qnqvQnQn−1)k}1/k,‖c‖2={vk−1|(qvFvQvQv−1+pvqvPvQv−1)λv|k+∑n=v+1∞nk−1|Δv((qnFvQnQn−1+pnqnPnQn−1)λv)|k}1/k. Applying (2.25), we obtain
(2.28)vk−1|(qvFvQvQv−1+pvqvPvQv−1)λv|k+∑n=v+1∞nk−1|Δv((qnFvQnQn−1+pnqnPnQn−1)λv)|k=O(1)(vk−1(rvRv)k+∑n=v+1∞nk−1(rnrvRnRn−1)k). As the right-hand side of (2.28), by (2.3), is
(2.29)=O(1)(vk−1(rvRv)k+rvkRvk−1∑n=v+1∞nk−1rnkRnkRn−1)=O(1)(vk−1(rvRv)k+(rvRv)k−1vk−1(rvRv)k)=O(vk−1(rvRv)k), and the fact that each term of
the left-hand side of (2.28) is O(vk−1(rv/Rv)k), we obtain
(2.30)vk−1(qvFvQvQv−1+pvqvPvQv−1)k|λv|k=O(vk−1(rvRv)k), which implies by (2.18)
(2.31)(qvQvQv−1)k(1+Fv)k|λv|k=O(rvRv)k, that is,
(2.32)|λv|=O(QvQv−1rv(1+Fv)qvRv). Also, we  have, by (2.28),
(2.33)∑n=v+1∞nk−1|(qnpvfvQnQn−1)λv+(qnFv+1QnQn−1+pnqnPnQn−1)Δλv|k=O(vk−1(rvRv)k). The above, via the linear independence of λv and Δλv, implies
(2.34)∑n=v+1∞nk−1(qnFv+1QnQn−1+pnqnPnQn−1)k|Δλv|k=O(vk−1(qvQv)k)|Δλv|k(1+Fv+1)k∑n=v+1∞nk−1(qnQnQn−1)k=O(vk−1(qvQv)k) by (2.18). As by (2.19), via the mean value theorem,
(2.35)1Qvk=∑n=v+1∞Δ(1Qn−1k)=O(1)∑n=v+1∞|ΔQn−1k|QnkQn−1k=O(1)∑n=v+1∞Qn−1k−1qnQnkQn−1k=O(1)∑n=v+1∞nk−1(qnQnQn−1)k. Then,
(2.36)|Δλv|k(1+Fv+1)k1Qvk=O(vk−1(rvRv)k), which implies
(2.37)Δλv=O(v1−1/krvQv(1+Fv+1)Rv). Also, by (2.28),
(2.38)∑n=v+1∞nk−1|qnpvfvQnQn−1λv|k=O(vk−1(rvRv)k),pvkfvk|λv|k∑n=v+1∞nk−1(qnQnQn−1)k=O(vk−1(rvRv)k),pvkfvk|λv|k1Qvk=O(vk−1(rvRv)k), which implies
(2.39)λv=O(v1−1/krvQvpvfvRv).


## 3. Applications

Corollary 3.1.
Letk≥1. Define
(3.1)fv=∑r=vnqrr,Fv=∑r=vnfr. Let
(3.2)n=O(Qn). Then, sufficient conditions for
the implication
(3.3)∑an⁢is summable|C,1|k⟹∑anλn⁢is summable|(R,qn)(C,1)|k are (2.5), (2.6), and the  following:
(3.4)v|λv|=O(Qv),vqv|λv|=O(QvQv−1),nqn|λn|=O(nQn),v|Δλv|Fv+1=O(Qv),|Δλv|=O(qv),v|Δλv|=O(Qv).Proof.
The proof follows from Theorem2.1 by putting pn=rn=1 for all n.Corollary 3.2.
Letk≥1. Define
(3.5)fv=∑r=vn1Pr,Fv=∑r=vnprfr. Let  (2.2) be satisfied. Then, sufficient conditions for the implication
(3.6)∑an⁢is summable|C,1|k⟹∑anλn⁢is summable|(C,1)(R,pn)|k are
(3.7)|λv|Fv=O(v),|λn|=O(n),pv|λv|=O(1),|Δλv|Fv+1=O(1),|Δλv|=O(1).Proof.
The proof follows from Theorem2.1, by putting qn=rn=1, for all n, noticing that (2.3) is satisfied as
(3.8)∑n=v+1∞1n(n−1)=∑n=v+1∞(1n−1−1n)=1v.Corollary 3.3.
Letfv,Fv be as defined in (3.1). Let (2.3) and (3.2) be satisfied. Then, sufficient conditions for
the implication
(3.9)∑an⁢is summable|R,rn|k⟹∑anλn⁢is summable|(R,qn)(C,1)|k are (2.5), (2.6), (2.10), (2.11), and the following:
(3.10)Rv|λv|=O(Qv),qvRv|λv|=O(QvQv−1rv),qnRn|λn|=O(nQnrn).Proof.
The proof follows from Theorem2.1, by outing pn=1 for all n.Corollary 3.4.
Letfv,Fv be as defined in (3.1). Let (2.3), (2.19) be satisfied and
(3.11)v=O(Qv). Then, necessary conditions for
the implication (3.3) are
(3.12)λv=O(QvQv−1(1+Fv)vqv),λv=O(Qvv1/kfv),Δλv=O(Qvv1/k(1+Fv+1)).Proof.
The proof follows from Theorem2.2 by putting pn=rn=1 for all n.Corollary 3.5.
Letfv,Fv be as defined in (3.5). Let
(3.13)Pv=O(vpv). Then, necessary conditions for
the implication (3.5) to be satisfied are
(3.14)λv=O(v1+Fv),λv=O(v1−1/kpvfv),Δλv=O(v1−1/k1+Fv+1).Proof.
The proof follows from Theorem2.2, by putting qn=rn=1, keeping in mind that (2.3) is satisfied as in the case of (3.8).Corollary 3.6.
Letfv,Fv be as defined in (3.1). Let (2.3), (2.19), and (3.2) be all
satisfied. Then, necessary conditions
for the implication (3.9) to be satisfied
are
(3.15)λv=O(QvQv−1rv(1+Fv)qvRv),λv=O(v1−1/krvQvfvRv),Δλv=O(v1−1/krvQv(1+Fv+1)Rv).Proof.
The proof follows from Theorem2.2, by putting pn=1 for all n.



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*Source: 372604-2008-05-22.xml*