IJMMS 25:6 (2001) 389–395
PII. S016117120100309X
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
GENERALIZATION OF SOME ABSOLUTE
SUMMABILITY METHODS
W. T. SULAIMAN
(Received 6 January 1999)
Abstract. Mazhar (1971) gave the characterization for the series an n to be summable
|N, pn | whenever an is summable |C, α|k , α ≥ 0, k ≥ 1. Here we prove two theorems, the
first concerns the sufficient conditions and the second the necessary conditions satisfied
by n in order to have an n summable |N̄, pn |k whenever an is summable |C, α|k ,
k ≥ 1.
2000 Mathematics Subject Classification. Primary 40G.
Let an be a given infinite series with partial sums sn . Let σnδ and rnδ denote the nth
Cesàro mean of order δ (δ > −1) of the sequences {sn } and {nan }, respectively. The
series an is said to be summable |C, δ|k , k ≥ 1 if
∞
n=1
δ k
nk−1 σnδ − σn−1
< ∞,
(1)
or equivalently
∞
n=1
k
n−1 rnδ < ∞.
(2)
Let Pn be a sequence of positive real constants such that
P n = p0 + p 1 + · · · + p n ,
A series
Pn → ∞ as n → ∞, P−1 = p−1 = 0 .
(3)
an is said to be summable |N̄, pn |k , k ≥ 1, (see [1]), if
∞ Pn k−1 tn − tn−1 k < ∞,
p
n
n=1
(4)
where
tn =
1
Pn
∞
ν=0
pν s ν .
(5)
390
W. T. SULAIMAN
For pn = 1/(n+1), the summability |N̄, pn |k reduces to the well-known summability
|R, log n, 1|k . In general, the two summabilities |N̄, pn |k and |C, α|k are independent
of each other, but for pn = 1 and α = 1 they are equivalent. For any real α and integer
n ≥ 0, we define
∆α n =
∞
ν=n
A−α−1
ν−n ν ,
(6)
whenever the series is convergent.
Theorem A (see [2]). The necessary and sufficient conditions for the series an n
to be summable |N̄, pn | whenever an is summable |C, α|k , α ≥ 0, k ≥ 1 are
(i) nα+1−(1/k ) ∆α (n /n) ∈ k , 1/k + 1/k = 1;
(ii) n−1/k n ∈ k , 0 ≤ α ≤ 1;
(iii) nα−(1/k ) (pn /Pn )n ∈ k , α > 1;
where (a) pn = O(pn+1 ), (b) (n + 1)pn = O(Pn ), and (c) Pn = O(nα pn ) (α > 1).
Theorem A includes as special cases k = 1 and pn = 1, the result of Mohapatra [4]
and Mehdi [3], respectively.
We need the following lemmas.
Lemma 1 (see [5]). If σ > δ > 0, then
∞
(n − ν)δ−1
= O ν δ−σ .
σ
n
n=ν+1
(7)
Lemma 2. If σ > δ ≥ 0, then
∞
δ−σ Aδ−1
n−ν
.
σ =O ν
A
n
n=ν+1
(8)
Proof. If δ > 0, then by Lemma 1,
∞
∞
Aδ−1
(n − ν)δ−1
n−ν
=
O(1)
= O ν δ−σ .
σ
σ
An
n
n=ν+1
n=ν+1
(9)
If δ = 0, then by Lemma 1,
∞
A−1
1
n−ν
σ = O(1) σ
A
A
n
n
n=ν+1
∞
−1 −σ A
.
n−ν = O n
n=ν+1
(10)
GENERALIZATION OF SOME ABSOLUTE SUMMABILITY METHODS
391
Lemma 3 (see [6]). Let k 1, Pn = p0 + p1 + · · · + pn → ∞ as n → ∞, pn > 0. Then
for all ν ≥ 1,
1
k
kPν−1
∞
pn
n=ν
k
Pn Pn−1
1
k
Pν−1
.
(11)
We prove the following theorems.
Theorem 4. Let {n } be monotonic nonincreasing sequence of constants. Suppose
that {Pn n /n} is nonincreasing, and
npn = O Pn .
(12)
Then the following are sufficient conditions for an n to be summable |N̄, pn |k when
ever an is summable |C, α|k , 0 < α ≤ αk ≤ 1:
(i) n = O(npn /Pn ),
(ii) ∆n = O(pn /Pn ).
Proof of Theorem 4. Define
Tn =
Tn =
=
=
=
1/k
pn /Pn
Pn−1
1/k
pn /Pn
Pn−1
1/k
pn /Pn
Pn−1
1/k
pn /Pn
Pn−1
1/k
pn /Pn
Pn−1
=
1/k
pn /Pn
Pn−1
n
Pν−1 aν ν ,
ν=1
α
tn
=
n
1 α−1
A
ναν ,
α
An ν=1 n−ν
n
n
Pν−1 ν −α−1 α α
Aν−r Ar tr
ν
ν=1
r =1
n
r =1
n
α
Aα
r tr
n
Pν−1 ν −α−1
Aν−r
ν
ν=r
n−r
Pν−r −1 ν+r −α−1
Aν
ν +r
r =1
ν=0
 n−r −1  ν
n
Pν+r −1 ν+r
α α
−α−1 

Ar tr
Aµ
∆ν
ν +r
r =1
ν=0
µ=0


n−r
 Pn−1 n
A−α−1
+
µ
n
µ=0
n
n−r
−1
pν+r ν+r
Pν+r −1 ν+r
α
−
Aα
t
A−α
r r
ν
(ν + r )(ν + r + 1)
ν +r +1
r =1
ν=0
Pn−1
Pν+r ∆ν ν+r
+ A−α
+
n ,
n−r
ν +r +1
n
α
Aα
r tr
= Tn,1 + Tn,2 + Tn,3 + Tn,4 .
(13)
In order to prove sufficiency, by Minkowski’s inequality, it is sufficient to show that
m
Tn,r k < ∞,
r =1
r = 1, 2, 3, 4.
(14)
392
W. T. SULAIMAN
n=1
k
Pn Pn−1
k
n
n−r
−1
Pν+r −1 ν+r
α α
−α
Ar tr
Aν
(ν + r )(ν + r + 1) m
pn
n−r
−1
n=1
k
Pn Pn−1
m
m
Tn,1 k =
n=1
≤
pn
r =1
ν=0
n
α Pr −1 r
Aα
r tr
r
r =1
ν=0
k
A−α
ν
ν +r +1
Pn n
being nonincreasing
n
k
n
m
n
pn α α Pr −1 r (ν − r )−α A
t
=
r r
k
r
ν
ν=r
n=1 Pn Pn−1 r =1
n
k
m
pn α α Pr −1 r −α r
= O(1)
A
t
r r
k
r
n=1 Pn Pn−1 r =1
n
k
m
pn α Pr
≤ O(1)
p
t
r r
r
k
r
p
P
P
r
n
n−1 r =1
n=1
n
k−1
n
m
k Pr k k pn
pr
r ≤ O(1)
pr trα P P
r pr
P
n=1 n n−1 r =1
r =1 n−1
m
m
k
k Pr
k pn
r = O(1)
pr trα r
p
P
r
n=r n Pn−1
r =1
m
k
pr t α k Pr
r k
= O(1)
r
P
r pr
r =1 r
as
= O(1)
m
1
t α k .
r
r
r =1
m
m
Tn,2 k =
n=1
≤
≤
(15)
pn
n=1
k
Pn Pn−1
m
pn
k
n=1 Pn Pn−1
m
pn
k
n=1 Pn Pn−1
k
n
n−r
−1
α α
−α −pν+r ν+r Ar tr
Aν
ν +r +1 r =1
ν=0
k
n
n−r
−1 A−α ν
α α
Ar tr pr r
ν +r +1
r =1
n
k
α
tr pr r ν=0
r =1
n
α k k
pn
t pr r ≤ O(1)
r
P
P
n
n−1
n=1
r =1
m
m
n
α k k t pr r = O(1)
r =1
r
n=r
n
pr
P
r =1 n−1
k−1
pn
Pn Pn−1
m
= O(1)
pr t α k r k
r
P
r =1 r
= O(1)
m
1
t α k r k
r
r
r =1
= O(1)
m
1
t α k
r
r
r =1
as r = O(1) being nonincreasing.
(16)
GENERALIZATION OF SOME ABSOLUTE SUMMABILITY METHODS
m
m
Tn,3 k =
n=1
=
pn
n=1
k
Pn Pn−1
m
pn
k
n=1 Pn Pn−1
≤ O(1)
m
k
n
n−r
−1
α α
−α Pν+r ∆ν ν+r Ar tr
Aν
ν +r +1 r =1
ν=0
k
n
n−r
−1 A−α ν
α α
Ar tr Pr ∆r
ν +r +1
r =1
pn
k
n=1 Pn Pn−1
ν=0
n
k
α
tr pr ∆r r =1
n
k−1
α k Pr k k pr
pr tr
∆r
pr
P
r =1
r =1 n−1
m
m
k Pr k pn
∆r k
= O(1)
pr trα p
P
Pn−1
r
n
n=r
r =1
m
k
pr k Pr
t α ∆r k
= O(1)
r
P
p
r
r =1 r
m
pn
= O(1)
P
Pn−1
n
n=1
= O(1)
393
n
(17)
m
1
t α k .
r
r
r =1
m
m
Tn,4 k =
n=1
n=1
pn
k
Pn Pn−1
n
k
α −α Pn−1
Aα
t
A
n
r r
n−r
n
r =1
n
m
k k
pn |n |k α k =
Ar trα A−α
n−r
k
Pn n r =1
n=1
n
k−1
1
r =1
m
n
k k
pn |n |k α k =
Ar trα A−α
n−r
P
n
n
n=1
r =1
=
m
r =1
Aα
r
≤ O(1)
m
k α k pn |n |k −α k
t An−r
r
nPn
n=r
m
r =1
Aα
r
= O(1)
m
r =1
= O(1)
Aα
r
(18)
m
k α k k A−αk
n−r
t r r
2
n=r n
k α k k −1−αk
t r r
r
m
1
t α k .
r
r
r =1
Theorem 5. The necessary conditions for
an is summable |C, α|k , k > 1, are
(i) n = O{(Pn /pn )1/k ν 1−α−(1/k) },
(ii) ∆n = O(ν 1−α−(1/k) ).
an n to be summable |N̄, pn |k whenever
Proof of Theorem 5. For k 1, we define
A = ai : ai is summable |C, α|k ,
B = ai : ai i is summable N̄, pn k .
(19)
394
W. T. SULAIMAN
These are BK-spaces, if normed by
∞
∞
1 k 1/k
k 1/k
(20)
tn
,
a2 =
,
Tn
a1 =
n
n=1
n=1
respectively. Since (1/n)|tn |k < ∞ ⇒ |Tn |k < ∞ by the hypothesis of Theorem 5,
then
(21)
a1 < ∞ ⇒ a2 < ∞.
Consider the inclusion map i : A → B defined by i(x) = x, i is continuous which follows
as A and B are BK-spaces. Therefore, there exists a constant M such that
a2 ≤ Ma1 .
(22)
Applying the values of tn and Tn , stated in Theorem 5, to a = eν − eν+1 (where eν is
the νth coordinate vector), we have



0,
if n < ν,




 ν

,
if n = ν,
t n = Aα
(23)
ν


α−1 


A
∆
 ν n−ν


, if n > ν,
Aα
n


if n < ν,

0,




1/k
 pν

,
if n = ν,
(24)
Tn =
Pν



1/k



p /P


 n n
∆ Pν−1 ν , if n > ν.
Pn−1
Equalities (20) give
α−1 k 1/k
∞
1
1 ν k
∆ν An−ν ν +
,
a1 =
α
A
ν Aα
ν
ν
n
n=ν+1
(25)
1/k
∞
k
k
pν p
n
ν +
∆ pν−1 ν .
a2 =
k
Pν
n=ν+1 Pn Pn−1
By inequality (22),
α−1 k k
∞
∞
pn 1
pν ∆ν νAn−ν .
ν k +
∆ Pν−1 ν k ≤ M k 1 ν
+
α
α
k
An
Pν
ν Aν
ν
n=ν+1 Pn Pn−1
n=ν+1
(26)
α−1
α−1
Since ∆ νAα−1
n−ν = −An−ν + ν∆ν An−ν , then by Minkowski’s inequality
α−1 k ∞
∞
k−αk−1 ∆ν νAn−ν ν k 1
1 (n − ν)αk−k
k 1
+
+O
M
=O v
Aα
ν Aα
ν
ν n=ν+1
Aαk
ν
n
n
n=ν+1
∞
k αk−2k
(n − ν)
ν
+O
ν n=ν+1
nαk
−k k−αk−1 +O ν
+ O ν −k , by Lemma 1
=O ν
= O ν k−αk−1 .
(27)
GENERALIZATION OF SOME ABSOLUTE SUMMABILITY METHODS
395
As each part of the left-hand side of (26) is O(ν k−αk−1 ), we have
ν = O
∆ Pν−1 ν k
Pv
pv
∞
1/k
n=ν+1
ν 1−α−(1/k) ,
pn
k
Pn Pn−1
= O ν k−αk−1 ,
(28)
which, by Lemma 3, implies
∆ Pν−1 ν k = O P k ν k−αk−1 .
ν−1
(29)
Now, as ∆(Pν−1 ν = −pν ν + Pν ∆ν ,
pν
1 ν +
∆ Pν−1 ν
Pν
Pν
Pν 1/k 1−α−(1/k)
1 pν
O
ν
O Pν−1 ν 1−α−(1/k)
+
=
Pν
pν
Pν
1−(1/k)
1 pν
=O
ν 1−α−(1/k) +
O Pν−1 ν 1−α−(1/k)
Pν
Pν
1−α−(1/k) .
=O v
∆ν =
(30)
This completes the proof of the theorem.
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[1]
[2]
[3]
[4]
[5]
[6]
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W. T. Sulaiman: Department of Mathematics, College of Science, University of Qatar,
Doha, Qatar