PORTUGALIAE MATHEMATICA
Vol. 54 Fasc. 4 – 1997
ON ϕ − |N , pn ; δ|k SUMMABILITY FACTORS
H. Seyhan and A. Sönmez
Abstract: In this paper a general theorem on ϕ − |N , pn ; δ|k summability factors,
which generalizes a result of Bor [2] on |N , pn |k summability factors, has been proved.
1 – Introduction
Let (ϕn ) be a sequence of positive real numbers and let an be a given infinite
series with the sequence of partial sums (sn ). Let (pn ) be a sequence of positive
real constants such that
P
(1.1)
Pn =
n
X
pv → ∞ as n → ∞ ,
(P−i = p−i = 0, i ≥ 1) .
v=0
The sequence-to-sequence transformation
(1.2)
Tn =
n
1 X
pv s v
Pn v=0
(Pn 6= 0)
defines the sequence (Tn ) of the (N , pn ) means of the sequence (sn ) generated by
the sequence of coefficients (pn ) (see [3]).
P
The series an is said to be summable |N , pn |k , k ≥ 1, if (see [1])
(1.3)
¶
∞ µ
X
Pn k−1
n=1
pn
|Tn − Tn−1 |k < ∞ .
In the special case when pn = 1 for all values of n (resp. k = 1), then |N , pn |k
summability is the same as |C, 1|k (resp. |N , pn |) summability.
Received : April 27, 1996; Revised : July 5, 1996.
394
The series
(see [5])
H. SEYHAN and A. SÖNMEZ
P
an is said to be summable ϕ − |N , pn ; δ|k , k ≥ 1 and δ ≥ 0, if
∞
X
(1.4)
ϕnδk+k−1 |Tn − Tn−1 |k < ∞ .
n=1
If we take δ = 0 and ϕn =
|N , pn |k summability.
Pn
pn ,
then ϕ − |N , pn ; δk | summability is the same as
2 – The following theorem is known
Theorem A ([2]). Let (pn ) be a sequence of positive numbers such that
(2.1)
Pn = O(n pn )
as n → ∞ .
Let (Xn ) be a positive non-decreasing sequence and let there be sequences
(βn ) and (λn ) such that
(2.2)
|∆λn | ≤ βn ,
(2.3)
βn → 0
(2.4)
λm Xm = O(1)
∞
X
(2.5)
as n → ∞ ,
as m → ∞ ,
n Xn |∆βn | < ∞ .
n=1
If
m
X
pn
(2.6)
then the series
P
n=1 n
P
|tn |k = O(Xm )
as m → ∞ ,
an λn is summable |N , pn |k , k ≥ 1.
3. The object of this paper is to generalize above theorem in the following
form.
Theorem. Let (pn ) be a sequence of positive numbers such that condition
(2.1) of Theorem A is satisfied and let (ϕn ) be a sequence of positive real numbers
395
ON ϕ − |N , pn ; δ|k SUMMABILITY FACTORS
such that
(3.1)
ϕn pn = O(Pn ) ,
∞
X
(3.2)
ϕδk−1
n
n=v+1
1
Pn−1
=O
µ
ϕδk
v
1
Pv
¶
.
If (Xn ) is a positive non-decreasing sequence and suppose that there exist sequences (λn ) and (βn ) such that conditions (2.2)–(2.5) of Theorem A are satisfied.
If
m
X
(3.3)
ϕδk−1
|tn |k = O(Xm )
n
as m → ∞ ,
n=1
then the series
P
an λn is summable ϕ − |N , pn ; δ|k , k ≥ 1 and 0 ≤ δk < 1.
If we take δ = 0 and ϕn = Ppnn in this theorem, then we get Theorem A.
We need the following lemma for the proof of our theorem.
Lemma ([4]). If (Xn ) is a positive non-decreasing sequence and (βn ) is a
positive sequence such that (2.3) and (2.5) hold, then
(3.4)
n Xn βn = o(1)
∞
X
(3.5)
as n → ∞ ,
Xn β n < ∞ .
n=1
4 – Proof of the Theorem
Let (Tn ) be the sequence of (N , pn ) means of the series
definition, we have
Tn =
P
an λn . Then, by
n
n
v
X
1 X
1 X
pv
(Pn − Pv−1 ) av λv .
a i λi =
Pn v=0 i=0
Pn v=0
Then, for n ≥ 1, we have
Tn − Tn−1 =
n
n
X
X
pn
Pv−1 λv
pn
Pv−1 av λv =
v av .
Pn Pn−1 v=1
Pn Pn−1 v=1
v
396
H. SEYHAN and A. SÖNMEZ
Using Abel’s transformation, we get
Tn − Tn−1 =
n−1
X
v+1
(n + 1)
pn
pv t v λv
pn t n λn −
nPn
Pn Pn−1 v=1
v
n−1
n−1
X
X
v+1
1
pn
pn
Pv ∆λv tv
pv tv λv+1
+
−
Pn Pn−1 v=1
v
Pn Pn−1 v=1
v
= Tn,1 + Tn,2 + Tn,3 + Tn,4 ,
say .
By Minkowski’s inequality it is sufficient to show that
∞
X
ϕnδk+k−1 |Tn,r |k < ∞,
for r = 1, 2, 3, 4 .
n=1
Since λn = O(1/Xn ) = O(1), by (2.4), we get that
m
X
n=1
ϕnδk+k−1 |Tn,1 |k
=
m
X
|λn |
k−1
|λn | ϕδk−1
|tn |k
n
= O(1)
m
X
|λn | ϕδk−1
|tn |k
n
n=1
n=1
= O(1)
= O(1)
m−1
X
n=1
m−1
X
∆|λn |
n
X
ϕδk−1
|tv |k
v
+ O(1) |λm |
v=1
m
X
ϕδk−1
|tn |k
n
n=1
βn Xn + O(1) |λm | Xm = O(1)
as m → ∞ ,
n=1
by virtue of the hypotheses and the Lemma.
1
k
Now, when k > 1, applying Hölder’s inequality with indices k and k 0 where
+ k10 = 1, as in Tn,1 , we have that
m+1
X
ϕnδk+k−1 |Tn,2 |k = O(1)
n=2
= O(1)
= O(1)
m+1
X
n=2
m
X
ϕδk−1
n
X
1 nn−1
Pn−1
pv |tv |k |λv |k
v=1
|λv |k−1 |λv | pv |tv |k
v=1
m
X
v=1
m+1
X
n=v+1
|λv | ϕδk−1
|tv |k = O(1)
v
o½
ϕδk−1
n
1
Pn−1
1
Pn−1
as m → ∞ .
n−1
X
v=1
pv
¾k−1
397
ON ϕ − |N , pn ; δ|k SUMMABILITY FACTORS
Since v βv = o(1/Xv ) = O(1), by (3.4), using the fact that Pv = O(v pv ), by
(2.1), we have that
m+1
X
ϕnδk+k−1 |Tn,3 |k
= O(1)
n=2
= O(1)
= O(1)
m+1
X
ϕδk−1
n
n=2
m+1
X
ϕδk−1
n
n=2
m
X
(v βv )
X
1 nn−1
k
Pn−1
1
Pn−1
k−1
v pv βv |tv |
v=1
nn−1
X
k
ok
(vβv ) pv |tv |
k
v=1
v βv pv |tv |
k
m+1
X
o½
1
Pn−1
ϕδk−1
n
n=v+1
v=1
m
X
n−1
X
pv
v=1
¾k−1
1
Pn−1
m
X
pv δk
v βv
= O(1)
v βv ϕδk−1
|tv |k
ϕv |tv |k = O(1)
v
P
v
v=1
v=1
= O(1)
= O(1)
= O(1)
m−1
X
v=1
m−1
X
v=1
m−1
X
∆(v βv )
v
X
|tr |k ϕδk−1
+ O(1) m βv
r
r=1
ϕδk−1
|tv |k
v
v=1
|∆(v βv )| Xv + O(1) m βm Xm
v|∆βv | Xv + O(1)
v=1
= O(1)
m
X
m−1
X
βv+1 Xv + O(1) m βm Xm
v=1
as m → ∞ ,
by virtue of the hypotheses and the Lemma.
Finally, using the fact that Pv = O(v pv ), by (2.1), as in Tn,1 and Tn,2 , we
have that
m+1
X
ϕnδk+k−1 |Tn,4 |k = O(1)
n=2
m+1
X
ϕδk−1
n
n=2
= O(1)
m+1
X
n=2
= O(1)
k
Pn−1
1
Pn−1
pv |tv | |λv+1 |
v=1
nn−1
X
v=1
m+1
X
pv |tv |k |λv+1 |k−1 |λv+1 |
m
X
ϕδk−1
|tv |k |λv+1 | = O(1)
v
v=1
ok
pv |tv |k |λv+1 |k
m
X
v=1
= O(1)
ϕδk−1
n
X
1 nn−1
n=v+1
o½
ϕδk−1
n
1
Pn−1
1
Pn−1
as m → ∞ .
n−1
X
v=1
pv
¾k−1
398
H. SEYHAN and A. SÖNMEZ
Therefore, we get that
m+1
X
ϕnδk+k−1 |Tn,r |k = O(1)
as m → ∞, for r = 1, 2, 3, 4 .
n=1
This completes the proof of the theorem.
REFERENCES
[1] Bor, H. – On two summability methods, Math. Proc. Cambridge Phil. Soc., 97
(1985), 147–149.
[2] Bor, H. – On |N , pn |k summability factors, Kuwait Jnl. Sci. & Eng., 23 (1996),
1–5.
[3] Hardy, G.H. – Divergent Series, Oxford Univ. Press, Oxford, 1949.
[4] Mishra, K.N. – On the absolute Nörlund summability factors of infinite series,
Indian J. Pure Appl. Math., (1983), 40–43.
[5] Seyhan, H. – Ph.D. Thesis, Erciyes University, Kayseri, 1995.
H. Seyhan and A. Sönmez,
Department of Mathematics, Erciyes University,
Kayseri 38039 – TURKEY