Volume 8 (2007), Issue 3, Article 82, 7 pp.
A GENERAL NOTE ON INCREASING SEQUENCES
HÜSEYİN BOR
D EPARTMENT OF M ATHEMATICS
E RCIYES U NIVERSITY
38039 K AYSERI , TURKEY
bor@erciyes.edu.tr
URL: http://fef.erciyes.edu.tr/math/hbor.htm
Received 23 December, 2006; accepted 08 August, 2007
Communicated by S.S. Dragomir
A BSTRACT. In the present paper, a general theorem on | N̄ , pn |k summability factors of infinite
series has been proved under more weaker conditions. Also we have obtained a new result
concerning the | C, 1 |k summability factors.
Key words and phrases: Absolute summability, Summability factors, Almost and power increasing sequences, Infinite series.
2000 Mathematics Subject Classification. 40D15, 40F05, 40G99.
1. I NTRODUCTION
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
denote by BVO the expression BV ∩ CO , where CO and BV are the
Pset of all null sequences and
the set of all sequences with bounded variation, respectively. Let an be a given infinite series
with partial sums (sn ). We denote by uαn and tαn the n-th Cesàro means of order α, with α > −1,
of the sequences (sn ) and (nan ), respectively, i.e.,
n
1 X α−1
α
A sv ,
(1.1)
un = α
An v=0 n−v
tαn
(1.2)
n
1 X α−1
= α
A vav ,
An v=1 n−v
where
Aαn = O(nα ),
(1.3)
The series
(1.4)
009-07
P
α > −1,
Aα0 = 1 and Aα−n = 0 for n > 0.
an is said to be summable |C, α|k , k ≥ 1, if (see [6, 8])
∞
∞
X
k X
|tαn |k
nk−1 uαn − uαn−1 =
< ∞.
n
n=1
n=1
2
H ÜSEYIN B OR
If we take α = 1, then we get |C, 1|k summability.
Let (pn ) be a sequence of positive numbers such that
n
X
(1.5)
Pn =
pv → ∞ as n → ∞, (P−i = p−i = 0, i ≥ 1).
v=0
The sequence-to-sequence transformation
σn =
(1.6)
n
1 X
p v sv
Pn v=0
defines the sequence (σn ) of the Riesz mean or simply the (N̄ , pn )P
mean of the sequence (sn ),
generated
by the sequence of coefficients (pn ) (see [7]). The series an is said to be summable
N̄ , pn , k ≥ 1, if (see [2, 3])
k
(1.7)
∞
X
(Pn /pn )k−1 |∆σn−1 |k < ∞,
n=1
where
n
pn X
(1.8)
∆σn−1 = −
Pv−1 av , n ≥ 1.
Pn Pn−1 v=1
In the special case pn = 1 for all values of n, N̄ , pn k summability is the same as |C, 1|k
summability.
2. K NOWN R ESULTS
Mishra and Srivastava [10] have proved the following theorem concerning the N̄ , pn summability factors.
Theorem A. Let (Xn ) be a positive non-decreasing sequence and let there be sequences (βn )
and (λn ) such that
|∆λn | ≤ βn ,
(2.1)
βn → 0 as
(2.2)
∞
X
(2.3)
n → ∞,
n |∆βn | Xn < ∞,
n=1
|λn | Xn = O(1).
(2.4)
If
(2.5)
n
X
|sv |
v=1
v
= O(Xn ) as
n→∞
and (pn ) is a sequence such that
(2.6)
Pn = O(npn ),
(2.7)
Pn ∆pn = O(pn pn+1 ),
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 82, 7 pp.
http://jipam.vu.edu.au/
A G ENERAL N OTE ON I NCREASING S EQUENCES
3
P∞
n λn
an Pnp
is summable N̄ , pn .
n
Later on Bor [4] generalized Theorem A for N̄ , pn k summability in the following form.
then the series
n=1
Theorem B. Let (Xn ) be a positive non-decreasing sequence and the sequences (βn ) and (λn )
are such that conditions (2.1) – (2.7) of Theorem A are satisfied with the condition (2.5) replaced
by:
n
X
|sv |k
(2.8)
v=1
Then the series
P∞
Pn λn
n=1 an npn
v
= O(Xn ) as
n → ∞.
is summable N̄ , pn k , k ≥ 1.
It may be noticed that if we take k = 1, then we get Theorem A.
Quite recently Bor [5] has proved Theorem B under weaker conditions by taking an almost
increasing sequence instead of a positive non-decreasing sequence.
Theorem C. Let (Xn ) be an almost increasing
sequence. If the conditions
(2.1) – (2.4) and
P
Pn λn
(2.6) – (2.8) are satisfied, then the series ∞
a
is
summable
N̄
,
p
n k , k ≥ 1.
n=1 n npn
Remark 2.1. It should be noted that, under the conditions of Theorem B, (λn ) is bounded and
∆λn = O(1/n) (see [4]).
3. M AIN R ESULT
The aim of this paper is to prove Theorem C under weaker conditions. For this we need the
concept of quasi β-power increasing sequences. A positive sequence (γn ) is said to be a quasi
β-power increasing sequence if there exists a constant K = K(β, γ) ≥ 1 such that
Knβ γn ≥ mβ γm
(3.1)
holds for all n ≥ m ≥ 1. It should be noted that almost every increasing sequence is a quasi
β-power increasing sequence for any nonnegative β, but the converse need not be true as can be
seen by taking the example, say γn = n−β for β > 0.
Now we shall prove the following theorem.
Theorem 3.1. Let (Xn ) be a quasi β-power increasing sequence for some 0 < β < 1. If the
conditions (2.1) – (2.4), (2.6) – (2.8) and
(λn ) ∈ BVO
(3.2)
are satisfied, then the series
P∞
n=1
n λn
an Pnp
is summable N̄ , pn k , k ≥ 1.
n
It should be noted that if we take (Xn ) as an almost increasing sequence, then we get Theorem
C. In this case, condition (3.2) is not needed.
We require the following lemma for the proof of Theorem 3.1.
Lemma 3.2 ([9]). Except for the condition (3.2), under the conditions on (Xn ), (βn ) and (λn )
as taken in the statement of Theorem 3.1, the following conditions hold, when (2.3) is satisfied:
(3.3)
(3.4)
nXn βn = O(1),
∞
X
βn Xn < ∞.
n=1
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 82, 7 pp.
http://jipam.vu.edu.au/
4
H ÜSEYIN B OR
Proof of Theorem 3.1. Let (Tn ) denote the (N̄ , pn ) mean of the series
definition, we have
Tn =
(3.5)
P∞
n=1
an Pn λn
.
npn
Then, by
n
v
n
1 X X ar P r λ r
1 X
av P v λ v
pv
=
(Pn − Pv−1 )
,
Pn v=1 r=1 rpr
Pn v=1
vpv
and thus
Tn − Tn−1
(3.6)
n
pn X Pv−1 Pv av λv
,
=
Pn Pn−1 v=1
vpv
n ≥ 1.
Using Abel’s transformation, we get
Tn − Tn−1
n
Pv−1 Pv λv
λn s n
pn X
=
sv ∆
+
Pn Pn−1 v=1
vpv
n
n−1
pn X Pv+1 Pv ∆λv
sn λn
=
+
sv
n
Pn Pn−1 v=1 (v + 1)pv+1
n−1
n−1
pn X
Pv
pn X
1
+
P v s v λv ∆
−
s v Pv λv
Pn Pn−1 v 1
vpv
Pn Pn−1 v=1
v
=
= Tn,1 + Tn,2 + Tn,3 + Tn,4 ,
say.
To prove Theorem 3.1, by Minkowski’s inequality, it is sufficient to show that
k−1
∞ X
Pn
(3.7)
pn
n=1
|Tn,r |k < ∞,
for
r = 1, 2, 3, 4.
Firstly by using Abel’s transformation, we have
k−1
m X
Pn
n=1
pn
k−1
m X
Pn
|sn |k
|Tn,1 | =
|λn |k−1 |λn |
npn
n
n=1
k
= O(1)
m
X
n=1
= O(1)
m−1
X
n=1
= O(1)
m−1
X
|λn |
| |sn | |k
n
∆ |λn |
n
X
|sv |k
v=1
v
+ O(1) |λm |
m
X
|sn |k
n=1
n
|∆λn | Xn + O(1) |λm | Xm
n=1
= O(1)
m−1
X
βn Xn + O(1) |λm | Xm = O(1)
as
m → ∞,
n=1
by virtue of the hypotheses of Theorem 3.1 and Lemma 3.2.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 82, 7 pp.
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A G ENERAL N OTE ON I NCREASING S EQUENCES
5
Now, using the fact that Pv+1 = O((v + 1)pv+1 ), by (2.6), and then applying Hölder’s inequality, we have
k
m+1
n−1
m+1
X pn X
X Pn k−1
k
|Tn,2 | = O(1)
Pv sv ∆λv k
pn
P P
n=2 n n−1 v=1
n=2
(
)k
m+1
n−1
X pn
X
Pv
= O(1)
|sv | pv |∆λv |
p
P Pk
n=2 n n−1
v=1 v
!k−1
k
m+1
n−1 n−1
X pn X
X
Pv
1
= O(1)
|sv |k pv |∆λv |k
pv
P
P
p
P
n
n−1
v
n−1
n=2
v=1
v=1
m
m+1
k
X Pv
X
pn
= O(1)
|sv |k pv |∆λv |k
pv
P P
v=1
n=v+1 n n−1
m
k−1
X
Pv |∆λv |
= O(1)
|sv |k |∆λv |
pv
v=1
k−1
m
X
Pv
k
= O(1)
|sv | |∆λv |
vpv
v=1
= O(1)
m
X
v=1
= O(1)
m−1
X
vβv
|sv |k
v
∆(vβv )
v=1
= O(1)
m−1
X
v
X
|sr |k
r=1
r
+ O(1)mβm
m
X
|sv |k
v=1
v
|∆(vβv )| Xv + O(1)mβm Xm
v=1
= O(1)
m−1
X
v |∆βv | Xv + O(1)
v=1
m−1
X
|βv | Xv + O(1)mβm Xm = O(1)
v=1
as m → ∞, in view of the hypotheses of Theorem 3.1 and Lemma 3.2.
Pv
Again, since ∆( vp
) = O( v1 ), by (2.6) and (2.7) (see [10]), as in Tn,1 we have
v
( n−1
)k
m+1
m+1
X
X Pn k−1
X
1
p
n
|Tn,3 |k = O(1)
Pv |sv | |λv |
k
p
v
P
P
n
n
n−1
n=2
n=2
v=1
( n−1 )k
m+1
X pn
X Pv
1
= O(1)
pv |sv | |λv |
pv
v
P Pk
n=2 n n−1
v=1
(
)k−1
k
m+1
n−1 n−1
X pn X
Pv
1 X
k
k
= O(1)
pv |sv | |λv |
pv
P P
vpv
Pn−1 v=1
n=2 n n−1 v=1
k
m m+1
X
X
Pv
pn
k
k
= O(1)
|sv | pv |λv |
vpv
P P
v=1
n=v+1 n n−1
k
m X
Pv
1 v
= O(1)
pv |sv |k |λv |k .
vpv
Pv v
v=1
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 82, 7 pp.
http://jipam.vu.edu.au/
6
H ÜSEYIN B OR
= O(1)
= O(1)
k−1
m X
Pv
vpv
v=1
m
X
|λv |
v=1
= O(1)
m−1
X
k−1
|λv |
|sv |k
|λv |
v
|sv |k
v
Xv βv + O(1)Xm |λm | = O(1)
as
m → ∞.
v=1
Finally, using Hölder’s inequality, as in Tn,3 we have
k
m+1
m+1
n−1
X Pn k−1
X pn X
P
v
| Tn,4 |k =
s
λ
v
v
k
p
v
P
P
n
n=2 n n−1 v=1
n=2
k
m+1
n−1
X pn X
Pv
=
s
p
λ
v
v
v
k
vpv
P P
n=2 n n−1 v=1
m+1
X
k
n−1
Pv
pn X
k
|sv |
pv |λv |k
≤
P P
vpv
n=2 n n−1 v=1
m
k
X
Pv
1 v
= O(1)
|sv |k pv |λv |k
·
vpv
Pv v
v=1
= O(1)
m
X
|λv |
v=1
= O(1)
m−1
X
1
n−1
X
Pn−1
v=1
!k−1
pv
|sv |k
v
Xv βv + O(1)Xm |λm | = O(1)
as
m → ∞.
v=1
Therefore we get
k−1
m X
Pn
n=1
pn
|Tn,r |k = O(1)
as
m → ∞,
for
r = 1, 2, 3, 4.
This completes the proof of Theorem 3.1.
Finally if we take pn = 1 for all values of n in the theorem, then we obtain a new result
concerning the |C, 1|k summability factors.
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AND
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A G ENERAL N OTE ON I NCREASING S EQUENCES
7
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http://jipam.vu.edu.au/