Bor [2] extended this result to absolute summability of orderk ≥1.. Unfortunately, an incorrect definition of absolute summability was used.. In this note, we establish the corresponding
Trang 1Volume 2007, Article ID 86095, 8 pages
doi:10.1155/2007/86095
Research Article
Ekrem Savas¸ and B E Rhoades
Received 9 November 2006; Accepted 29 March 2007
Recommended by Martin J Bohner
We obtain sufficient conditions on a nonnegative lower triangular matrix A and a se-quenceλ nfor the series
a n λ n /na nnto be absolutely summable of orderk ≥1 byA.
Copyright © 2007 E Savas¸ and B E Rhoades This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, dis-tribution, and reproduction in any medium, provided the original work is properly cited
A weighted mean matrix, denoted by (N, p n), is a lower triangular matrix with entries
p k /P n, where{ p k }is a nonnegative sequence withp0> 0, and P n:=n k =0p k
Mishra and Srivastava [1] obtained sufficient conditions on a sequence { p k }and a sequence{ λ n }for the series
a n P n λ n /np n to be absolutely summable by the weighted mean matrix (N, p n) Bor [2] extended this result to absolute summability of orderk ≥1 Unfortunately, an incorrect definition of absolute summability was used
In this note, we establish the corresponding result for a nonnegative triangle, using the correct definition of absolute summability of orderk ≥1, (see [3]) As a corollary, we obtain the corrected version of Bor’s result
LetA be an infinite lower triangular matrix We may associate with A two lower
trian-gular matricesA and A, whose entries are defined by
a nk =n
i = k a ni, ank = a nk − a n −1,k, (1) respectively The motivation for these definitions will become clear as we proceed LetA be an infinite matrix The seriesa kis said to be absolutely summable byA, of
orderk ≥1, written as| A | k, if
∞
k =0
n k −1 Δt n −1k < ∞, (2)
Trang 2whereΔ is the forward difference operator and t n denotes thenth term of the matrix
transform of the sequence{ s n }, wheres n:=n k =0a k
Thus
t n =n
k =0
a nk s k =n
k =0
a nk k
ν =0
a ν =n
ν =0
a ν n
k = ν a nk =n
ν =0
a nν a ν,
t n − t n −1=
n
ν =0
a nν a ν −
n−1
ν =0
a n −1,ν a ν =
n
ν =0
a nν a ν,
(3)
sincea n −1,n =0
The result to be proved is the following
Theorem 1 Let A be a triangle with nonnegative entries satisfying
(i)a n0 = 1, n =0, 1, ,
(ii)a n −1,ν ≥ a nν for n ≥ ν + 1,
(iii)na nn O(1),
(iv)Δ(1/a nn)= O(1),
(v)n
ν =0a νν | a n,ν+1 | = O(a nn ).
If { X n } is a positive nondecreasing sequence and the sequences { λ n } and { β n } satisfy
(vi)| Δλ n | ≤ β n ,
(vii) limβ n = 0,
(viii)| λ n | X n = O(1),
(ix)∞
n =1nX n | Δβ n | < ∞ ,
(x)T n:=n ν =1(| s ν | k /ν) = O(X n ),
then the series∞
ν =1a n λ n /na nn is summable | A | k , k ≥ 1.
The proof of the theorem requires the following lemma
Lemma 2 (see Mishra and Srivastava [1]) Let { X n } be a positive nondecreasing sequence and the sequences { β n } , { λ n } satisfy conditions (vi)–(ix) of Theorem 1 Then
∞
n =1
Since{ X n }is nondecreasing,X n ≥ X0, which is a positive constant Hence condition (viii) implies thatλ nis bounded It also follows from (4) thatβ n = O(1/n), and hence that
Δλ n = O(1/n) by condition (iv).
Proof Let T ndenote thenth term of the A-transform of the series(a n λ n)/(na nn) Then
we may write
T n =n
ν =0
a nν ν
i =0
a i λ i
a ii i =
m
i =0
a i λ i
a ii i
n
ν = i a nν =n
i =0
a ni a i λ i
Trang 3T n − T n −1=n
i =0
a ni a i λ i
a ii i −
n−1
i =0
a n −1,i a i λ i
a ii i =
n
i =0
a ni − a n −1,ia i λ i
a ii i =
n
i =0
a ni a i λ i
a ii i
=
n
i =0
a ni λ i
a ii i
s i − s i −1
=
n−1
i =0
a ni λ i
a ii is i+a nn a λ nn n ns n −
n
i =0
a ni λ i s i −1
a ii i
=
n−1
i =0
a ni λ i
a ii is i+a nn a λ nn n ns n −
n−1
i =0
a n,i+1 λ i+1 s i
(i + 1)a i+1,i+1
=n
i =0
a ni λ i
a ii i − a n,i+1(i + 1)a λ i+1 i+1,i+1
s i+a nn λ n
na nn
(7)
We may write
a ni λ i
ia ii − an,i+1 λ i+1
(i + 1)a i+1,i+1 = ani λ i
ia ii − an,i+1 λ i+1
(i + 1)a i+1,i+1+ an,i+1 λ i
(i + 1)a i+1,i+1 − an,i+1 λ i
(i + 1)a i+1,i+1
=Δi
a ni
ia ii
λ i+ an,i+1
(i + 1)a i+1,i+1Δλ i
.
(8)
Also we may write
Δi
a ni
ia ii
λ i = ani
ia ii λ i − an,i+1
(i + 1)a i+1,i+1 λ i − an,i+1
ia ii λ i+an,i+1
ia ii λ i
=Δi
a ni
λ i
ia ii +a n,i+1 λ i
1
ia ii − 1
(i + 1)a i+1,i+1
.
(9)
Hence,
T n − T n −1=
n−1
i =0
Δi
a ni
ia ii λ i s i+
n−1
i =0
a n,i+1 λ i
1
ia ii − 1
(i + 1)a i+1,i+1
s i
+
n−1
i =0
a n,i+1
(i + 1)a i+1,i+1Δi(λ i)s i+λ n
n s n
= T n1+T n2+T n3+T n4, say.
(10)
To finish the proof of the theorem, it will be sufficient to show that
∞
n =1
n k −1 T nrk < ∞, forr =1, 2, 3, 4. (11)
Trang 4Using H¨older’s inequality and (iii),
I1=
m+1
n =1
n k −1 T n1k
≤
m+1
n =1
n k −1
n −1
i =0
Δi
a ni
ia ii λ i s i
k
= O(1) m+1
n =1
n k −1
n−1
i =0
Δi
a ni
λ i s i k
= O(1) m+1
n =1
n k −1
n −1
i =0
Δi
a niλ iks ik n −1
i =0
Δi
a ni k −1.
(12)
But using (ii),
Δi
a ni
= a ni − a n,i+1 = a ni − a n −1,i − a n,i+1+a n −1,i+1 = a ni − a n −1,i ≤0. (13) Thus using (i),
n−1
i =0
Δi
a ni = n−1
i =0
a n −1,i − a ni =1−1 +a nn = a nn (14)
From (viii), it follows that λ n = O(1) Using (iii), (vi), (x), and property (5) of
Lemma 2,
I1= O(1) m+1
n =1
na nnk −1n−1
i =0
λ iks ikΔi
a ni
= O(1) m+1
n =1
na nnk −1
n−1
i =0
λ ik −1 λ iΔi
a nis ik
= O(1)m
i =0
λ is ik m+1
n = i+1
na nnk −1 Δi
a ni
= O(1)m
i =0
λ is ik a ii =λ0s0k a00+O(1)m
i =1
λ is ik i
= O(1) + O(1)m
i =1
λ i i
r =1
s rk
r −
i −1
r =1
s rk r
= O(1)
m
i =1
λ i i
r =1
s rk
r −
m−1
j =0
λ j+1 j
r =1
s rk r
= O(1) m −
1
i =1
Δλ i i
r =1
1
rs rk
+O(1)λ m m
i =1
s ik i
Trang 5= O(1) m−1
i =1
Δλ iX i+O(1)λ mX m
= O(1)m
i =1
β i X i+O(1)λ mX m = O(1),
I2=
m+1
n =1
n k −1 T n2k
=
m+1
n =1
n k −1
n−1
i =0
a n,i+1 λ iΔ
1
ia ii
s i
k
= O(1) m+1
n =1
n k −1 n−1
i =0
a n,i+1λ iΔ 1
ia ii
s ik .
(15) Now
Δ
1
ia ii
ia ii − 1
(i + 1)a i+1,i+1
ia ii − 1
(i + 1)a i+1,i+1+ 1
(i + 1)a ii − 1
(i + 1)a ii
(i + 1)
1
a ii − a i+1,i+11
+ 1
a ii
1
i −
1
i + 1
(i + 1)
Δ
1
a ii
ia ii
.
(16)
Thus using (iv) and (ii),
Δ
1
ia ii
=
i + 11
Δ
1
a ii
ia ii
≤ i + 11 a i+1,i+1 − a ii
a ii a i+1,i+1 + 1
ia ii
i + 1
O(1) + O(1).
(17)
Hence, using H¨older’s inequality, (v) and (iii),
I2= O(1) m+1
n =1
n k −1 n−1
i =0
a n,i+1λ i 1
i + 1s ik
= O(1) m+1
n =1
n k −1 n−1
i =0
a n,i+1a iiλ is ik
= O(1) m+1
n =1
n k −1
n −1
i =0
a n,i+1a iiλ iks ik n −1
i =0
a iia n,i+1 k −1
= O(1) m+1
n =1
na nnk −1n−1
i =0
a n,i+1a iiλ iks ik
Trang 6= O(1)m
i =0
λ iks ik a ii
m+1
n = i+1
na nnk −1 a n,i+1
= O(1)m
i =0
λ iks ik a ii
m+1
n = i+1a n,i+1.
(18) From [4],
m+1
Hence,
I2= O(1)m
i =1
λ iks ik a ii = O(1)m
i =1
λ iλ ik −1 s ik1
i =
m
i =1
λ is ik
i = O(1), (20)
as in the proof ofI1
Using (iii), H¨older’s inequality, and (v),
I3=
m+1
n =1
n k −1 T n3k
=
m+1
n =1
n k −1
n−1
i =0
a n,i+1
Δλ i
s i
(i + 1)a i+1,i+1
k
= O(1) m+1
n =1
n k −1
n−1
i =0
a n,i+1Δλ is i k
= O(1) m+1
n =1
n k −1 n−1
i =0
a ii
a iia n,i+1Δλ is ik
= O(1) m+1
n =1
n k −1 n−1
i =0
a iia n,i+1
a k ii
Δλ iks ik n−1
i =0
a iia n,i+1k −1
= O(1) m+1
n =1
na nnk −1n−1
i =0
a iia n,i+1
a k ii
Δλ iks ik
= O(1) m+1
n =1
n−1
i =0
a n,i+1Δλ iks ik 1
a k
ii a ii
= O(1)m
i =0
a ii
a k ii
Δλ iks ik m+1
n = i+1
a n,i+1
= O(1)m
i =0
Δλ i
a ii
k −1
Δλ is ik
= O(1)m
i =0
Δλ is ik = O(1)m
i =0
s ik β i
(21)
Trang 7Since| s i | k = i(T i − T i −1) by (x), we have
I3= O(1)m
i =1
iT i − T i −1
Using Abel’s transformation, (vi), and (5),
I3= O(1) m−1
i =1
T iΔiβ i
+O(1)mT n β n
= O(1) m−1
i =1
iΔβ iX i+O(1) m−1
i =1
X i β i+O(1)mX n β n = O(1).
(23)
Using (viii) and (x),
I4=
m+1
n =1
n k −1 T n4k
=
m+1
n =1
n k −1
s n λ n n
k
=
m+1
n =1
s nkλ nk1
n
= m+1
n =1
s nk
n λ nλ nk −1
= O(1),
(24)
Corollary 3 Let { p n } be a positive sequence such that P n =n k =0p k → ∞ and satisfies
(i)np n O(P n );
(ii)Δ(P n /p n)= O(1).
If { X n } is a positive nondecreasing sequence and the sequences { λ n } and { β n } are such that
(iii)| Δλ n | ≤ β n ,
(iv)β n → 0 as n → ∞ ,
(v)| λ n | X n = O(1) as n → ∞ ,
(vi)∞
n =1nX n | Δβ n | < ∞ ,
(vii)T n =n ν =1| s ν | k /ν = O(X n ),
then the series
(a n P n λ n)/(np n ) is summable | N, p n | k , k ≥ 1.
Proof Conditions (iii)–(vii) ofCorollary 3are, respectively, conditions (vi)–(x) of Theo-rem 1
Conditions (i), (ii), and (v) ofTheorem 1are automatically satisfied for any weighted mean method Conditions (iii) and (iv) ofTheorem 1become, respectively, conditions
Acknowledgment
The first author received support from the Scientific and Technical Research Council of Turkey
Trang 8[1] K N Mishra and R S L Srivastava, “On| N, p–– n | summability factors of infinite series,” Indian
Journal of Pure and Applied Mathematics, vol 15, no 6, pp 651–656, 1984.
[2] H Bor, “A note on| N, p–– n | k summability factors of infinite series,” Indian Journal of Pure and
Applied Mathematics, vol 18, no 4, pp 330–336, 1987.
[3] T M Flett, “On an extension of absolute summability and some theorems of Littlewood and
Paley,” Proceedings of the London Mathematical Society Third Series, vol 7, pp 113–141, 1957 [4] B E Rhoades and E Savas¸, “A note on absolute summability factors,” Periodica Mathematica
Hungarica, vol 51, no 1, pp 53–60, 2005.
Ekrem Savas¸: Department of Mathematics, Faculty of Sciences and Arts, Istanbul Ticaret University, Uskudar, 34672 Istanbul, Turkey
Email addresses:ekremsavas@yahoo.com ; esavas@iticu.edu.tr
B E Rhoades: Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA
Email address:rhoades@indiana.edu
... 8[1] K N Mishra and R S L Srivastava, ? ?On< /small>| N, p–– n | summability factors of infinite. .. class="text_page_counter">Trang 7
Since| s i | k = i(T i − T i −1)... ik i
Trang 5= O(1) m−1
i