Gen. Math. Notes, Vol. 2, No. 2, February 2011, pp. 7-13
ISSN 2219-7184; Copyright © ICSRS Publication, 2011
www.i-csrs.org
Available free online at http://www.geman.in
On Some Absolute Summability Factors of
Infinite Series
W.T. Sulaiman
Department of Computer Engineering
College of Engineering
University of Mosul, Iraq
E-mail: waadsulaiman@hotmail.com
(Received: 4-11-10 /Accepted: 11-11-10)
Abstract
ϕ − N , pn k
In this paper, a general theorem concerning
factors of infinite series has been
proved. The presented result giving improvement as well as generalization of some known
results.
Keywords: Absolute Summability, Infinite series
1
Introduction
Let (ϕ n ) be a sequence of positive real numbers, let
∑a
n
be an infinite series with the
sequence of partial sums (s n ) . Let (t n ) denote the n-th (C, 1) means of the sequence (na n ) .
The series
∑a
n
is said to be summable C ,1 k , k ≥ 1, if (see [1])
∞
1
∑nt
(1.1)
n =1
k
n
< ∞.
and it is said to be summable ϕ − C ,1 k , k ≥ 1 , if (see [5])
∞
∑
(1.2)
ϕ nk −1
k
tn < ∞ .
k
n
If we are taking ϕ n = n, ϕ − C ,1 k reduces to C ,1 k summability.
n =1
Let
( pn )
be a sequence of positive numbers such that
8
W.T. Sulaiman
n
Pn = ∑ pv → ∞ as n → ∞
v =1
( P−1 = P−1 = 0 ) .
The sequence-to-sequence transformation
1 n
(1.3)
un =
∑ pv sv
Pn v =0
defines the sequence (u n ) of the Riesz mean or simply the (N , p n ) mean of the sequence
generated by the sequence of coefficients ( p n ) (see[2]) . The series
(s n )
∑a
n
is said to be summable R, p n k , k ≥ 1 if
∞
∑n
(1.4)
k −1
n =1
k
u n − u n−1
< ∞.
In the special case when p n = 1 for all n, then R, p n
∑a
C ,1 k summability. The series
∞
∑ϕ
n =1
For ϕ n = n, ϕ − R, p n
k −1
n
n
k
summability is the same as
is summable ϕ − N , p n k , k ≥ 1, if
u n − u n −1 < ∞ .
k
summability is the same as R, p n
k
k
summability .
Concerning C ,1 k summability, Mazhar [3] has proved the following
Theorem 1.1. If
(1.5)
λm = o (1),
as m → ∞ ,
m
(1.6)
∑ n log n ∆ λ
2
n =1
m
∑
tv
n
= Ο (1) ,
as m → ∞ ,
k
= Ο (log m) as m → ∞ ,
v
then the series ∑ a n λ n is summable C ,1 k , k ≥ 1.
Ozarslan [4], on the other hand , generalized the previous result by giving the following
(1.7)
v =1
Theorem 1.2. Let (ϕ n ) be a sequence of positive real numbers and the conditions (1.5)(1.6) of Theorem (1.1) are satisfied . If
m
(1.8)
∑
v =1
ϕ vk −1
v
k
tv
k
= Ο (log m)
as m → ∞ ,
 ϕ vk −1 
 k  ,
=
Ο
∑
k +1
n =v n
 v 
then the series ∑ a n λ n is summable ϕ − C ,1 k , k ≥ 1.
∞
(1.9)
ϕ nk −1
It should be mentioned that on taking ϕ n = n in Theorem (1.2), we get Theorem 1.1.
The aim of this paper is that to give three improvements to Theorem 1.2. Firstly by
weakining the conditions and secondly by generalizing the result replacing log m by χ m ,
and thirdly by adding new parameter. In fact, we present the following
On Some Absolute Summability Factors of…
2
9
Main Result
Theorem 2.1. Let (ϕ n ) , (χ n ) be sequences of positive real numbers such that (χ n ) is
nondecreasing and the condition (1.5), is satisfied. If
(2.1)
np n = Ο (Pn ) , Pn = Ο (np n ) , as n → ∞ ,
(2.2)
β n +1 = Ο (β n ) ,
(
)
∆β n = Ο n −1 β n , as n → ∞ ,
(2.3)
∞
∑ nχ
(2.4)
n =1
n
∆2 λ n = Ο (1) ,
ϕ nk −1 β v s n
∑
n k χ nk −1
n =1
k
m
(2.5)
ϕ nk −1
m
∑v
(2.6)
n =v
then the series
∑a λ β
n
n
n
k
Pn −1
k
= Ο (χ m ) ,
 ϕ k −1
= Ο  k −v1
 v Pv
as m → ∞,

 ,

is summable ϕ − N , p n k , k ≥ 1 .
Remark 1 On putting p n = 1, β n = 1, χ n = log n , we obtain an improvement to Theorem 1.2
.
Remark 2 It may be noted thet condition (2.5) is weaker than condition
k
k
m ϕ k −1 β
sn
n
n
(2.5)′
= Ο (χ m ) ,
∑
nk
n =1
for we have (2.5)′ ⇒ (2.5) , but not conversely. In fact if (2.5)′ holds, then
ϕ nk −1 β n s n
∑
n k χ nk −1
n =1
k
m
k
= Ο (1)
ϕ nk −1 β n s n
k
m
∑
k
nk
n =1
= Ο (χ m ),
(
2
. 4
)
′
while, if (2.5) is satisfied, we have
since by the mean value theorem,
∆χ nk −1 = Ο (1) χ nk −2 ∆χ n ,
then, we have
m
∑
n =1
ϕ nk −1 β n s n
k
n
ϕ nk −1 β n s n
=∑
n k χ nk −1
n =1
k
 n ϕ vk −1 β v k s v
= ∑∑

v k χ vk −1
n =1 v =1

k
k
k
k
m
m −1
= Ο (1)
m −1
∑χ
n =1
k −1
n
χ nk −1
k

 m k −1
 ∆ χ k −1 +  ϕ n β n s n
n

∑
n k χ nk −1

 n=1
(
∆χ n + Ο (1) χ mk
m
= Ο (1) χ mk −1 ∑ ∆χ n + Ο (1) χ mk
v =1
= Ο (1) χ + Ο (1) χ mk
( )
k
m
( )
= Ο χ mk ≠ Ο χ m , for k>1.
)
k

 χ k −1
 m

10
3
W.T. Sulaiman
Lemma
The following Lemma is needed
Lemma 3.1. The conditions (1.5) and (2.4) implies
∞
∑χ
(3.1)
n =1
n
∆λ n = Ο(1) ,
(3.2)
n χ n ∆λ n = Ο (1) ,
as n → ∞ ,
(3.3)
χ n λn = Ο (1) ,
as n → ∞ .
Proof. By virtue of (1.5),
∞
∑ χ n ∆λn =
n =1
≤
=
∞
∞
n =1
∞
v =n
∞
∑ χ n ∑ ∆ ∆λv
∑χ ∑ ∆ λ
2
n
n =1
∞
v= n
∑ ∆2 λv
v =1
v
∑χ
n =1
∞
∑v χ
= Ο (1)
v =1
= Ο(1) .
v
v
∆2 λv
v
∞
∑ ∆ ∆λ
n χ n ∆λ n = n χ n
v=n
v
∞
≤ n χ n ∑ ∆ ∆λv
v=n
∞
≤ n χ n ∑ ∆2 λv
= Ο (1)
v =n
∞
∑v χ
n =v
= Ο (1) .
v
∆2 λv
∞
χ n λ n = χ n ∑ ∆ λv
v=n
∞
≤ χ n ∑ ∆λv
v =n
= Ο (1)
∞
∑χ
v =n
v
∆λv
= Ο (1) , by the first part .
On Some Absolute Summability Factors of…
4
11
Proof of Theorem 2.1
Let Tn be the ( N , p n ) mean of the series
Tn =
1
Pn
n
v
v =1
r =1
∞
n =1
n
∑ (P
1
Pn
∑ pv ∑ a r λr β r =
∑
n
v =1
a n λ n β n . By definition, we have
− Pv −1 ) a v λv β v ,
and hence
pn n
∑ Pv−1av λv β v , n ≥ 1.
Pn Pn −1 v =1
Using Abel’s transformation, we have
p n n −1
p
Tn − Tn −1 =
s v ∆(Pv −1λv β v ) + n λ n β n s n
∑
Pn Pn−1 v =1
Pn
n −1
pn
(− p v λv β v sv + Pv λv ∆β v s v + Pv β v +1∆λv s v ) + p n λn β n s n
=
∑
Pn Pn −1 v =1
Pn
= Tn1 + Tn 2 + Tn 3 + Tn 4 .
Tn − Tn −1 =
(
)
Since Tn1 + Tn 2 + Tn3 + Tn 4 ≤ 4 k Tn1 + Tn 2 + Tn3 + Tn 4 ,
proof, it is sufficient to show that
k
∞
∑ϕ
n =1
k −1
n
Tnr
k
< ∞,
k
k
k
in order to complete the
r = 1, 2, 3, 4 .
Applying Holder's inequality, we have via Lemma 3.1
m
∑ϕ
n=2
k −1
n
Tn1
k
=
m
∑ϕ
n=2
k −1
n
m
p n n −1
∑ p v λv β v s v
Pn Pn −1 v =1
ϕ nk −1 p nk  n −1
∑P
= Ο (1)
k
n
n=2
= Ο (1)
m
n=2
∑p
v =1
= Ο (1)
m
m
v =1
v
λv β v s v
k
pv ϕ
ϕ
v =1
λv β v s v
k
λv β v s v
v
∑v
∑
∑p
k
v
k
v =1
= Ο (1)
n −1
Pn−1
m
∑p

 ∑ p v λv β v s v 
 v =1

k
n
m
v =1
= Ο (1)
Pnk−1
ϕ nk −1 p nk
∑P
= Ο (1)
k
k −1
k −1
v
Pv
k −1
v
k
v
λv
k
k
k
m
k
∑P
n=v
k
n
m
ϕ
∑n
λv β v s v
k −1
k
P
n −1
k −1
n
k
Pn−1
k
β v s v λv
k
k
∞
ϕ vk −1
k
k
β
s
∑
v
v ∑ ∆ λn
k
k −1
v =1 v χ v
n =v
k −1
m
∞
ϕ
k
k
= Ο (1) ∑ k v k −1 β v s v ∑ ∆λ n
v =1 v χ v
n=v
= Ο (1)
m
k
ϕ nk −1 p nk
n =v
k
k
 n −1 p v 
 ∑

P
v
=
1
n
−
1


k −1
12
W.T. Sulaiman
∞
n
∑ ∆λ ∑ v
= Ο (1)
n
n =1
∞
∑χ
= Ο (1)
= Ο (1) .
m
∑ϕ
n=2
k −1
n
Tn 2
k
=
m
∑ϕ
n= 2
n =1
v =1
∆λ n
n
ϕ nk −1 p nk  n −1
m
∑P
k
n
n =1
∑P
k
n
n =1
∑P
= Ο (1)
λv β v s v
k
v
∑ p v λv β v s v
k
∑p
v =1
k
m
= Ο (1)
−k
Pn −1
m
= Ο (1)
k
k
k
v =1
= Ο (1)
ϕ vk −1 pv
m
∑v
v =1
= Ο (1)
m
∑
k −1

k
k
 λv β v s v

k −1
m
ϕ n p nk
∑
k
n = v Pn Pn −1
∑v
k
n −v
k
λv β v s v
k
k
k
k
 n−1 p v 
 ∑

 v =1 Pn −1 
ϕ nk −1
m
λv β v s v
Pv
ϕ vk −1
k
P
v p v  v
∑
v =1
 pv
n −1
k
−1
k
n
n =1
k

 ∑ v Pv λv β v s v 
 v =1

Pnk−1
ϕ nk −1 p nk
m



 ∑ Pv λv ∆β v s v
Pnk−1  v =1
ϕ nk −1 p nk  n −1
m
= Ο (1)
k
p n n −1
∑ Pv λv ∆β v sv
Pn Pn −1 v =1
k −1
n
= Ο (1)
ϕ vk −1
k
k
β v sv
k
k −1
χv
k
Pn −1
k
k
v
= Ο (1) , as in the case of Tn1 .
v =1
m +1
∑ϕ
n =1
k −1
n
Tn 3
k
= Ο (1)
m +1
∑ϕ
n =1
= Ο (1)
m +1
k −1
n
ϕ nk −1 p nk  n−1
∑P
k
n
n =1
m +1
p n n −1
∑ Pv β v+1∆λv sv
Pn Pn −1 v =1

 ∑ Pv β v ∆λv s v 
 v =1

Pnk−1
ϕ nk −1 p nk
n −1
k
k
 n −1

= Ο (1) ∑ k k ∑ k −1 β v ∆λv s v  ∑ χ v ∆λv 
n =1 Pn Pn −1 v =1 χ v
 v =1

k
k −1
k
m
m +1
P
ϕ p
k
k
= Ο (1) ∑ kv −1 β v ∆λv s v ∑ nk k n
v =1 χ v
n = v +1 Pn Pn −1
= Ο (1)
m
Pvk
∑χ
v =1
= Ο (1)
m
∑χ
v =1
= Ο (1)
m
∑v
v =1
β v ∆λv s v
k
k −1
v
Pv
Pvk
β
v
k −1
v
k −1
v
k −1 k −1
v
ϕ
χ
k
∆λv s v
k
k
∑v
n = v +1
k
m +1
k
k
Pnk−1
ϕ nk −1
∑v
n = v +1
β v ∆λv s v
k
ϕ nk −1
m +1
k
k
Pn −1
k −1
k −1
On Some Absolute Summability Factors of…
13
ϕ vk −1 β v sv
= Ο (1) ∑ v ∆λv
v k χ vk −1
v =1
k
m
m −1
∑ ∆(v ∆λ )∑
= Ο (1)
v
v
v =1
r =1
k
ϕ rk −1 β r s r
r k χ rk −1
k
k
+ Ο (1) m ∆λ m
m −1
∑v ∆ λ
= Ο (1)
2
v =1
v
m −1
∑ ∆λ
χ v + Ο (1)
v =1
= Ο (1) .
m
∑ϕ
n =1
k −1
n
Tn 4
k
=
m
∑ϕ
n =1
k −1
n
pn
β n λn s n
Pn
m
= Ο (1) ∑ ϕ
n =1
 pn 
  β n
 Pn 
k
m
v
∑ ∆λ ∑
v
v =1
m
∑χ
v =1
v =1
k
k
χ v + Ο (1) m ∆λ m χ m
k
ϕ nk −1 β n s n
= Ο (1) ∑
n k χ nk −1
n =1
= Ο (1)
∑
ϕ vk −1 β v s v
v k χ vk −1
k
k −1
n
m
= Ο (1)
v
m
v
k
k
s n λn
k
λn
∞
∑ ∆λ
v=n
ϕ nk −1 β n s n
n =1
k −1
k
v
k
n k χ nk −1
∆λv
= Ο (1) .
This completes the proof of the Theorem.
References
[1]
[2]
[3]
[4]
[5]
T.M. Flett, On an extension of absolute summability and some theorems of
Littlewood and Paley, Proc. London Math. Soc., 7 (1957), 113-141.
G.H. Hardy, Divergent Series, Oxford Univ. Press, Oxford, (1949).
S.M. Mazhar, On C ,1 k summability factors of infinite series, Indian J. Math., 14
(1972), 45-48.
H.S. Ozarslan, On absolute Cesaro summability factors of infinite series,
Communications in Mathematical Analysis, 3 (2007), 53-56.
H. Seyhan, The absolute summability methods, Ph.D. Thesis, Kayseri, (1995), 1-57.