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Some Tauberian theorems for (A)(C,α) summability method
Mathematical and Computer Modelling 52 (2010) 738–743
Contents lists available at ScienceDirect
Mathematical and Computer Modelling journal homepage: www.elsevier.com/locate/mcm
Some Tauberian theorems for (A)(C , α) summability method İbrahim Çanak a,∗ , Yılmaz Erdem b , Ümit Totur b a
Ege University, Faculty of Sciences, Department of Mathematics, İzmir, 35100, Turkey
b
Adnan Menderes University, Faculty of Sciences and Art, Department of Mathematics, Aydın, 09010, Turkey
article
abstract
info
In this paper we give some Tauberian conditions in order to obtain (C , α) summability from (A)(C , α) summability. © 2010 Elsevier Ltd. All rights reserved.
Article history: Received 26 March 2010 Accepted 1 May 2010 Keywords: Abel summability Cesàro summability Tauberian conditions Tauberian theorems
1. Introduction Let an be an infinite series of real numbers with sequence of nth partial sums (sn ). Let Aαn be defined by the generating P∞ α n function (1 − x)−α−1 = n=0 An x (|x| < 1), where α > −1. If
P
sαn =
Snα Aαn
=
n 1 X α−1 A sk → s Aαn k=0 n−k
as n → ∞, we say that an is (C , α) summable to s, where α > −1, and we write sn → s (C , α). We write τn = nan (n = 0P , 1, 2, . . .) and define τnα as (C , α) mean of (τn ). Note that (C , 0) summability is ordinary P convergence. ∞ If the series n=0 an xn is convergent for 0 ≤ x < 1 and tends to s as x → 1− , then we say that an is Abel summable to s, and sn → s (A). It is well known that, for β > α > −1, (C , α) ⊂ (C , β) ⊂ (A) (see [1]). P∞we write P If n=0 (sαn − sαn−1 )xn is convergent for 0 ≤ x < 1 and tends to s as x → 1− , then we say that an is (A)(C , α) summable to s and we write sn → s (A)(C , α). (A)(C , 0) summability P is Abel summability. P A well known result due to Abel [2] states that if an is convergent to s, then P an is Abel summable to s. That the converse of Abel’s result is not generally true follows from the example of the series (−1)n , which is Abel summable to 1/2, but not convergent. However, the converse of Abel’s theorem is valid under some conditions called Tauberian conditions. Any theorem which states that convergence of series follows from a summability method and some Tauberian condition(s) is said to be a Tauberian theorem. A first partial converse of Abel’s theorem is obtained by Tauber [3].
P
an is (A) summable to s and nan = o(1), then
Theorem 1.1 ([3]). If
P
Theorem 1.2 ([3]). If
P
∗
1 an is (A) summable to s and τn1 = n+ 1
Pn
k=1
P
an is convergent to s.
kak = o(1), then
P
an is convergent to s.
Corresponding author. Tel.: +90 232 388 4000; fax: +90 232 342 6951. E-mail addresses: [email protected], [email protected] (İ. Çanak), [email protected] (Y. Erdem), [email protected] (Ü. Totur).
0895-7177/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2010.05.001
İ. Çanak et al. / Mathematical and Computer Modelling 52 (2010) 738–743
739
Theorem 1.1 is strengthened by Littlewood [4] as follows. Theorem 1.3 ([4]). If
P
an is (A) summable to s and nan = O(1), then
P
an is convergent to s.
Hardy and Littlewood [5] replaced the condition nan = O(1) by one-sided Tauberian condition nan > −H, where H is a positive constant. Theorem 1.4 ([5]). If
P
an is (A) summable to s and nan > −H , where H is a positive constant, then
P
an is convergent to s.
Hardy and Littlewood [5] proved that Abel and (C , 1) summability methods are equivalent for the series sn ≥ 0 for all n. Theorem 1.5 ([5]). If sn ≥ 0 for all n and Theorem 1.6 ([4]). If
P
P
an is (A) summable to s, then
an is (A) summable to s and sn = O(1), then
P
P
an is (C , 1) summable to s.
an is (C , 1) summable to s.
Szàsz [6] obtained the following theorem which generalizes Theorem 1.6.
P
Theorem 1.7 ([6]). If to s.
an is (A) summable to s and τn1 > −H, where H is a positive constant, then
P
an is (C , 1) summable
Dik [7] introduced the general control modulo of the oscillatory behavior of the integer order and obtained some Tauberian conditions such as n∆τn1 > −H and 1
n X (k∆)2 τk1 > −H ,
n + 1 k =1
where H is a positive constant, for Abel summability method. Pati [8] proved some Tauberian theorems, which improve classical Tauberian theorems listed above, for (A)(C , α) summability method and gave short proofs of some classical Tauberian theorems as special cases of some of his results. The aim of this paper is to introduce some new Tauberian conditions for (A)(C , α) summability method using the concept of the general control modulo of the oscillatory behavior of the integer order defined by Dik [7] and to extend and generalize some of the results of Pati [8]. We need his following results to prove our results. Theorem 1.8 ([8]). If
P
an is (A)(C , α) summable to s for some α ≥ 0, and
α
τn > −H ,
(1)
where H is a positive constant, then
P
an is (C , α) summable to s.
Theorem 1.9 ([8]). TheP necessary and sufficient condition that the (A)(C , α + 1) summability of (C , α) summability of an to s, is that
P
τnα+1 = o(1).
an to s, α > −1, implies the (2)
Theorem 1.10 ([8]). If Snα+β ≥ 0,
α ≥ 0, β > 0,
(3)
and
τnα+1 > −H ,
(4)
then the (A)(C , α + β) summability of
P
an to s implies the (C , α + 1) summability of
P
an to s.
2. Theorems We prove the following theorems. Theorem 2.1. If
P
an is (A)(C , α) summable to s for some α ≥ 0, and
n∆τnα+1 > −H , where H is a positive constant, then
(5)
P
an is (C , α) summable to s.
740
İ. Çanak et al. / Mathematical and Computer Modelling 52 (2010) 738–743
Taking α = 0 in Theorem 2.1, we obtain the following theorem. Corollary 2.2 ([7]). If
P
an is (A) summable to s, and
n∆τn1 > −H ,
(6)
where H is a positive constant, then Theorem 2.3. If
P
P
an is convergent to s.
an is (A)(C , α) summable to s for some α ≥ 0, and
(n∆)2 τnα+2 > −H ,
(7)
where H is a positive constant, then Theorem 2.4. If
P
P
an is (C , α) summable to s.
an is (A)(C , α + 1) summable to s for some α > −1, and
n∆τnα+2 = o(1), then
P
(8)
an is (C , α) summable to s.
We observe that the condition (8) is more general than the condition (2). Theorem 2.5. If
P
an is (A)(C , α + 1) summable to s for some α > −1, and
α+3
(n∆)2 τn = o(1), P then an is (C , α) summable to s.
(9)
Theorem 2.6. If Snα+β ≥ 0,
α ≥ 0, β > 0,
(10)
and n∆τnα+1 > −H ,
(11)
then the (A)(C , α + β) summability of
P
an to s implies the (C , α + 1) summability of
P
an to s.
Theorem 2.7. If Snα+β ≥ 0,
α ≥ 0, β > 0,
(12)
and
(n∆)2 τnα+2 > −H ,
(13)
then the (A)(C , α + β) summability of
P
an to s implies the (C , α + 1) summability of
3. Lemmas For the proof of our theorems, we need the following lemmas. Lemma 3.1 ([9,10]). For α > −1, τnα = n∆sαn = n(sαn − sαn−1 ). 1 ). Lemma 3.2 ([10,11]). For α > −1, τnα+1 = (α + 1)(sαn − sα+ n
Lemma 3.3. For α > −1, n∆τnα+1 = (α + 1)(τnα − τnα+1 ) Proof. By Lemma 3.2, n∆sαn − n∆snα+1 = n∆
1
α+1
τnα+1 =
1 By Lemma 3.1, τnα − τnα+1 = α+ n∆τnα+1 . 1
1
α+1
n∆τnα+1 .
Lemma 3.4. For α > −1,
(n∆)2 τnα+2 = (α + 2)(α + 1)(τnα − τnα+1 ) − (α + 2)2 (τnα+1 − τnα+2 ).
P
an to s.
İ. Çanak et al. / Mathematical and Computer Modelling 52 (2010) 738–743
741
Proof. Replacing α by α + 1 in Lemma 3.3, we have
τnα+1 − τnα+2 =
1
α+2
n∆τnα+2 .
(14)
Taking the backward differences of both sides of (14) and then multiplying the resulting equality by n, we obtain
(n∆)2 τnα+2 = (α + 2) n∆τnα+1 − n∆τnα+2 . By Lemma 3.3,
(n∆)2 τnα+2 = (α + 2)(α + 1)(τnα − τnα+1 ) − (α + 2)2 (τnα+1 − τnα+2 ). Lemma 3.5 ([1]). For β > α > −1, (A)(C , α) ⊂ (A)(C , β). 4. Proofs 1 Proof of Theorem 2.1. By hypothesis, sαn → s (A). It then follows sα+ → s (A) by Lemma 3.5. Hence, n
sαn − snα+1 → 0 (A).
(15) α
α+1
By Lemma 3.3 and hypothesis (5), τn − τn
> −H1 , where H1 is a positive constant. Since
1 τnα − τnα+1 = n∆(sαn − sα+ ) > −H1 , n
(16)
we have sαn − snα+1 = o(1) by Theorem 1.4. By Lemma 3.2, we get τnα+1 = o(1). By (16) and the fact that every convergent sequence is bounded, we have τnα > −H2 , where H2 is a positive constant. The condition (1) of Theorem 1.8 is now satisfied. This completes the proof of Theorem 2.1. 1 2 Proof of Theorem 2.3. By hypothesis, sαn → s (A). It then follows sα+ → s (A) and sα+ → s (A) by Lemma 3.5. Hence, n n
sαn − snα+1 → 0 (A),
(17)
2 snα+1 − sα+ → 0 (A). n
(18)
and
By Lemma 3.4 and hypothesis (7),
(α + 2)(α + 1)(τnα − τnα+1 ) − (α + 2)2 (τnα+1 − τnα+2 ) > −H , α
α+1
where H is a positive constant. Since τn − τn that
α
α+1
= n∆(sn − sn
(19) α+1
) and τn
α+2
− τn
α+1
= n∆(sn
α+2
− sn
), it follows from (19)
1 1 n∆[(α + 2)(α + 1)(sαn − sα+ ) − (α + 2)2 (sα+ − snα+2 )] > −H . n n
From (17), (18) and (19), we have 1 1 2 (α + 2)(α + 1)(sαn − sα+ ) − (α + 2)2 (sα+ − sα+ ) = o(1), n n n
by Theorem 1.4. Hence, by Lemma 3.2, (α + 2)(τnα+1 − τnα+2 ) = o(1), which implies that 1 2 τnα+1 − τnα+2 = n∆(sα+ − sα+ ) = o(1). n n
(20)
1 2 From (18) and (20), we have sα+ − sα+ = o(1) by Theorem 1.1. By Lemma 3.2, τnα+1 = o(1). By (19) and the fact that every n n convergent sequence is bounded, we have τnα > −H1 , where H1 is a positive constant. The condition (1) of Theorem 1.8 is now satisfied. This completes the proof of Theorem 2.3. 1 Proof of Theorem 2.4. By hypothesis, sα+ → s (A). It then follows snα+2 → s (A). Hence, n 2 snα+1 − sα+ → 0(A). n
(21)
By Lemma 3.3 and hypothesis (8), we have 2 n∆(snα+1 − sα+ ) = O(1). n
α+1
(22) α+2
α+2
From (21) and (22), we get sn − sn = o(1) by Theorem 1.3. By Lemma 3.2, τn = o(1). By Lemma 3.3 and hypothesis (8), n∆τnα+2 = (α + 2)(τnα+1 − τnα+2 ) = o(1). Hence, τnα+1 = o(1). The condition (2) of Theorem 1.9 is now satisfied. This completes the proof of Theorem 2.4.
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İ. Çanak et al. / Mathematical and Computer Modelling 52 (2010) 738–743
1 2 3 Proof of Theorem 2.5. By hypothesis, sα+ → s (A). It then follows sα+ → s (A) and sα+ → s (A). Hence, n n n 1 sα+ − snα+2 → 0 (A), n
(23)
2 sα+ − snα+3 → 0 (A). n
(24)
and
By Lemma 3.4 and hypothesis (9), we have
(α + 2)(α + 3)(τnα+1 − τnα+2 ) − (α + 3)2 (τnα+2 − τnα+3 ) = o(1).
(25)
By Lemma 3.2, 1 2 τnα+1 − τnα+2 = n∆(sα+ − sα+ ) n n
and 2 3 τnα+2 − τnα+3 = n∆(sα+ − sα+ ). n n
Using the equalities above in (25), we have 1 2 2 3 n∆ (α + 2)(α + 3)(sα+ − sα+ ) − (α + 3)2 (sα+ − sα+ ) = o(1). n n n n
(26)
From (23), (24) and (26), we get 1 2 2 (α + 2)(α + 3)(sα+ − sα+ ) − (α + 3)2 (sα+ − snα+3 ) = o(1). n n n 1 2 2 3 Since sα+ − sα+ = α+1 2 τnα+2 and sα+ − sα+ = α+1 3 τnα+3 by Lemma 3.2, we have (α + 3)(τnα+2 − τnα+3 ) = o(1), which n n n n implies that 2 3 τnα+2 − τnα+3 = n∆(sα+ − sα+ ) = o(1). n n
α+2
(27) α+2
α+3
α+1
From (24) and (27), we get sn − sn = o(1) by Theorem 1.1. By Lemma 3.2, τn = o(1). Hence, τn The condition (2) of Theorem 1.9 is now satisfied. This completes the proof of Theorem 2.5. α+β
α+β
α+β
Proof of Theorem 2.6. Since Sn ≥ 0 for all n, we have sn ≥ 0 for all n by the definition of sn α+β is summable (A) to s. Hence by Theorem 1.5, sn → s (C , 1), which is equivalent to
= o(1) by (25). α+β
. By hypothesis, (sn
sαn → s(C , β + 1).
)
(28) α
α+1
It follows from (28) that sn → s (A) and sn
α
α+1
→ s (A). Hence, sn − sn
α+1
→ 0 (A). Applying Theorem 1.4 to τn
τnα+1 = o(1).
, we have (29)
α
Now, by (29) and hypothesis (11), we have τn > −H2 , where H2 is a positive constant. The condition (4) of Theorem 1.10 is now satisfied. This completes the proof of Theorem 2.6. Proof of Theorem 2.7. We know from Theorem 2.6 that sαn → s (C , β + 1). It follows from (28) that snα+k → s (A) for k = 1, 2, 3. Hence, 1 sαn − sα+ → 0(A), n
(30)
1 sα+ − snα+2 → 0(A). n
(31)
and
By Lemma 3.4 and hypothesis (13),
(α + 1)(α + 2)(τnα − τnα+1 ) − (α + 2)2 (τnα+1 − τnα+2 ) > −H , α
α+1
where H is a positive constant. Since τn − τn
α
α+1
= n∆(sn − sn
(32) α+1
) and τn
α+2
− τn
1 1 2 n∆[(α + 1)(α + 2)(sαn − sα+ ) − (α + 2)2 (sα+ − sα+ )] > −H n n n
α+1
= n∆(sn
α+2
− sn
), we have (33)
from (32). From (30), (31) and (33), we have 1 1 (α + 1)(α + 2)(sαn − sα+ ) − (α + 2)2 (sα+ − snα+2 ) = o(1) n n
α
α+1
by Theorem 1.4. Since (α + 1)(sn − sn
α+1
) = τn
1 2 τnα+1 − τnα+2 = n∆(sα+ − sα+ ) = o(1). n n
α+1
and (α + 2)(sn
(34) α+2
− sn
α+2
) = τn
by Lemma 3.2, it follows from (34) that (35)
İ. Çanak et al. / Mathematical and Computer Modelling 52 (2010) 738–743
743
From (31) and (33) we have 2 snα+1 − sα+ = o(1) n
(36)
by Theorem 1.1. By (34) and (36),
τnα+2 = o(1).
(37)
If we put (37) in (35), we get
τnα+1 = o(1).
(38) α
If we put (35) and (37) in (32), we have τn > −H1 , where H1 is a positive constant. The condition (4) of Theorem 1.10 is now satisfied. This completes the proof of Theorem 2.7. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
G.H. Hardy, Divergent Series, 2nd ed., Chelsea, New York, NY, 1991. m(m−1) x + 1.2 x2 + · · ·, J. für Math. 1 (1826) 311–339. N.H. Abel, Recherches sur la série 1 + m 1 A. Tauber, Ein satz der Theorie der unendlichen Reihen, Monatsh. f. Math. 8 (1897) 273–277. J.E. Littlewood, The converse of Abel’s theorem on power series, Proc. Lond. Math. Soc. 9 (2) (1911) 434–448. G.H. Hardy, J.E. Littlewood, Tauberian theorems concerning power and Dirichlet’s series whose coefficients are positive, Proc. Lond. Math. Soc. 13 (2) (1913) 174–191. O. Szàsz, Generalization of two theorems of Hardy and Littlewood on power series, Duke Math. J. 1 (1935) 105–111. M. Dik, Tauberian theorems for sequences with moderately oscillatory control modulo, Math. Morav. 5 (2001) 57–94. T. Pati, On tauberian theorems, in: D. Rath, S. Nanda (Eds.), Sequences, Summability and Fourier Analysis, Narosa Publishing House, 2005, pp. 84–96. E. Kogbetliantz, Sur le séries absolument sommables par la méthode des moyennes arihtmétiques, Bull. Soc. Math. 49 (2) (1925) 234–251. E. Kogbetliantz, Sommation des séries et intégrals divergents par les moyennes arithmétiques et typiques, Memorial Sci. Math. 51 (1931) 1–84. G.H. Hardy, Theorems relating to the summability and convergence of slowly oscillating series, Proc. Lond. Math. Soc. 8 (2) (1910) 301–320.
Contents lists available at ScienceDirect
Mathematical and Computer Modelling journal homepage: www.elsevier.com/locate/mcm
Some Tauberian theorems for (A)(C , α) summability method İbrahim Çanak a,∗ , Yılmaz Erdem b , Ümit Totur b a
Ege University, Faculty of Sciences, Department of Mathematics, İzmir, 35100, Turkey
b
Adnan Menderes University, Faculty of Sciences and Art, Department of Mathematics, Aydın, 09010, Turkey
article
abstract
info
In this paper we give some Tauberian conditions in order to obtain (C , α) summability from (A)(C , α) summability. © 2010 Elsevier Ltd. All rights reserved.
Article history: Received 26 March 2010 Accepted 1 May 2010 Keywords: Abel summability Cesàro summability Tauberian conditions Tauberian theorems
1. Introduction Let an be an infinite series of real numbers with sequence of nth partial sums (sn ). Let Aαn be defined by the generating P∞ α n function (1 − x)−α−1 = n=0 An x (|x| < 1), where α > −1. If
P
sαn =
Snα Aαn
=
n 1 X α−1 A sk → s Aαn k=0 n−k
as n → ∞, we say that an is (C , α) summable to s, where α > −1, and we write sn → s (C , α). We write τn = nan (n = 0P , 1, 2, . . .) and define τnα as (C , α) mean of (τn ). Note that (C , 0) summability is ordinary P convergence. ∞ If the series n=0 an xn is convergent for 0 ≤ x < 1 and tends to s as x → 1− , then we say that an is Abel summable to s, and sn → s (A). It is well known that, for β > α > −1, (C , α) ⊂ (C , β) ⊂ (A) (see [1]). P∞we write P If n=0 (sαn − sαn−1 )xn is convergent for 0 ≤ x < 1 and tends to s as x → 1− , then we say that an is (A)(C , α) summable to s and we write sn → s (A)(C , α). (A)(C , 0) summability P is Abel summability. P A well known result due to Abel [2] states that if an is convergent to s, then P an is Abel summable to s. That the converse of Abel’s result is not generally true follows from the example of the series (−1)n , which is Abel summable to 1/2, but not convergent. However, the converse of Abel’s theorem is valid under some conditions called Tauberian conditions. Any theorem which states that convergence of series follows from a summability method and some Tauberian condition(s) is said to be a Tauberian theorem. A first partial converse of Abel’s theorem is obtained by Tauber [3].
P
an is (A) summable to s and nan = o(1), then
Theorem 1.1 ([3]). If
P
Theorem 1.2 ([3]). If
P
∗
1 an is (A) summable to s and τn1 = n+ 1
Pn
k=1
P
an is convergent to s.
kak = o(1), then
P
an is convergent to s.
Corresponding author. Tel.: +90 232 388 4000; fax: +90 232 342 6951. E-mail addresses: [email protected], [email protected] (İ. Çanak), [email protected] (Y. Erdem), [email protected] (Ü. Totur).
0895-7177/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2010.05.001
İ. Çanak et al. / Mathematical and Computer Modelling 52 (2010) 738–743
739
Theorem 1.1 is strengthened by Littlewood [4] as follows. Theorem 1.3 ([4]). If
P
an is (A) summable to s and nan = O(1), then
P
an is convergent to s.
Hardy and Littlewood [5] replaced the condition nan = O(1) by one-sided Tauberian condition nan > −H, where H is a positive constant. Theorem 1.4 ([5]). If
P
an is (A) summable to s and nan > −H , where H is a positive constant, then
P
an is convergent to s.
Hardy and Littlewood [5] proved that Abel and (C , 1) summability methods are equivalent for the series sn ≥ 0 for all n. Theorem 1.5 ([5]). If sn ≥ 0 for all n and Theorem 1.6 ([4]). If
P
P
an is (A) summable to s, then
an is (A) summable to s and sn = O(1), then
P
P
an is (C , 1) summable to s.
an is (C , 1) summable to s.
Szàsz [6] obtained the following theorem which generalizes Theorem 1.6.
P
Theorem 1.7 ([6]). If to s.
an is (A) summable to s and τn1 > −H, where H is a positive constant, then
P
an is (C , 1) summable
Dik [7] introduced the general control modulo of the oscillatory behavior of the integer order and obtained some Tauberian conditions such as n∆τn1 > −H and 1
n X (k∆)2 τk1 > −H ,
n + 1 k =1
where H is a positive constant, for Abel summability method. Pati [8] proved some Tauberian theorems, which improve classical Tauberian theorems listed above, for (A)(C , α) summability method and gave short proofs of some classical Tauberian theorems as special cases of some of his results. The aim of this paper is to introduce some new Tauberian conditions for (A)(C , α) summability method using the concept of the general control modulo of the oscillatory behavior of the integer order defined by Dik [7] and to extend and generalize some of the results of Pati [8]. We need his following results to prove our results. Theorem 1.8 ([8]). If
P
an is (A)(C , α) summable to s for some α ≥ 0, and
α
τn > −H ,
(1)
where H is a positive constant, then
P
an is (C , α) summable to s.
Theorem 1.9 ([8]). TheP necessary and sufficient condition that the (A)(C , α + 1) summability of (C , α) summability of an to s, is that
P
τnα+1 = o(1).
an to s, α > −1, implies the (2)
Theorem 1.10 ([8]). If Snα+β ≥ 0,
α ≥ 0, β > 0,
(3)
and
τnα+1 > −H ,
(4)
then the (A)(C , α + β) summability of
P
an to s implies the (C , α + 1) summability of
P
an to s.
2. Theorems We prove the following theorems. Theorem 2.1. If
P
an is (A)(C , α) summable to s for some α ≥ 0, and
n∆τnα+1 > −H , where H is a positive constant, then
(5)
P
an is (C , α) summable to s.
740
İ. Çanak et al. / Mathematical and Computer Modelling 52 (2010) 738–743
Taking α = 0 in Theorem 2.1, we obtain the following theorem. Corollary 2.2 ([7]). If
P
an is (A) summable to s, and
n∆τn1 > −H ,
(6)
where H is a positive constant, then Theorem 2.3. If
P
P
an is convergent to s.
an is (A)(C , α) summable to s for some α ≥ 0, and
(n∆)2 τnα+2 > −H ,
(7)
where H is a positive constant, then Theorem 2.4. If
P
P
an is (C , α) summable to s.
an is (A)(C , α + 1) summable to s for some α > −1, and
n∆τnα+2 = o(1), then
P
(8)
an is (C , α) summable to s.
We observe that the condition (8) is more general than the condition (2). Theorem 2.5. If
P
an is (A)(C , α + 1) summable to s for some α > −1, and
α+3
(n∆)2 τn = o(1), P then an is (C , α) summable to s.
(9)
Theorem 2.6. If Snα+β ≥ 0,
α ≥ 0, β > 0,
(10)
and n∆τnα+1 > −H ,
(11)
then the (A)(C , α + β) summability of
P
an to s implies the (C , α + 1) summability of
P
an to s.
Theorem 2.7. If Snα+β ≥ 0,
α ≥ 0, β > 0,
(12)
and
(n∆)2 τnα+2 > −H ,
(13)
then the (A)(C , α + β) summability of
P
an to s implies the (C , α + 1) summability of
3. Lemmas For the proof of our theorems, we need the following lemmas. Lemma 3.1 ([9,10]). For α > −1, τnα = n∆sαn = n(sαn − sαn−1 ). 1 ). Lemma 3.2 ([10,11]). For α > −1, τnα+1 = (α + 1)(sαn − sα+ n
Lemma 3.3. For α > −1, n∆τnα+1 = (α + 1)(τnα − τnα+1 ) Proof. By Lemma 3.2, n∆sαn − n∆snα+1 = n∆
1
α+1
τnα+1 =
1 By Lemma 3.1, τnα − τnα+1 = α+ n∆τnα+1 . 1
1
α+1
n∆τnα+1 .
Lemma 3.4. For α > −1,
(n∆)2 τnα+2 = (α + 2)(α + 1)(τnα − τnα+1 ) − (α + 2)2 (τnα+1 − τnα+2 ).
P
an to s.
İ. Çanak et al. / Mathematical and Computer Modelling 52 (2010) 738–743
741
Proof. Replacing α by α + 1 in Lemma 3.3, we have
τnα+1 − τnα+2 =
1
α+2
n∆τnα+2 .
(14)
Taking the backward differences of both sides of (14) and then multiplying the resulting equality by n, we obtain
(n∆)2 τnα+2 = (α + 2) n∆τnα+1 − n∆τnα+2 . By Lemma 3.3,
(n∆)2 τnα+2 = (α + 2)(α + 1)(τnα − τnα+1 ) − (α + 2)2 (τnα+1 − τnα+2 ). Lemma 3.5 ([1]). For β > α > −1, (A)(C , α) ⊂ (A)(C , β). 4. Proofs 1 Proof of Theorem 2.1. By hypothesis, sαn → s (A). It then follows sα+ → s (A) by Lemma 3.5. Hence, n
sαn − snα+1 → 0 (A).
(15) α
α+1
By Lemma 3.3 and hypothesis (5), τn − τn
> −H1 , where H1 is a positive constant. Since
1 τnα − τnα+1 = n∆(sαn − sα+ ) > −H1 , n
(16)
we have sαn − snα+1 = o(1) by Theorem 1.4. By Lemma 3.2, we get τnα+1 = o(1). By (16) and the fact that every convergent sequence is bounded, we have τnα > −H2 , where H2 is a positive constant. The condition (1) of Theorem 1.8 is now satisfied. This completes the proof of Theorem 2.1. 1 2 Proof of Theorem 2.3. By hypothesis, sαn → s (A). It then follows sα+ → s (A) and sα+ → s (A) by Lemma 3.5. Hence, n n
sαn − snα+1 → 0 (A),
(17)
2 snα+1 − sα+ → 0 (A). n
(18)
and
By Lemma 3.4 and hypothesis (7),
(α + 2)(α + 1)(τnα − τnα+1 ) − (α + 2)2 (τnα+1 − τnα+2 ) > −H , α
α+1
where H is a positive constant. Since τn − τn that
α
α+1
= n∆(sn − sn
(19) α+1
) and τn
α+2
− τn
α+1
= n∆(sn
α+2
− sn
), it follows from (19)
1 1 n∆[(α + 2)(α + 1)(sαn − sα+ ) − (α + 2)2 (sα+ − snα+2 )] > −H . n n
From (17), (18) and (19), we have 1 1 2 (α + 2)(α + 1)(sαn − sα+ ) − (α + 2)2 (sα+ − sα+ ) = o(1), n n n
by Theorem 1.4. Hence, by Lemma 3.2, (α + 2)(τnα+1 − τnα+2 ) = o(1), which implies that 1 2 τnα+1 − τnα+2 = n∆(sα+ − sα+ ) = o(1). n n
(20)
1 2 From (18) and (20), we have sα+ − sα+ = o(1) by Theorem 1.1. By Lemma 3.2, τnα+1 = o(1). By (19) and the fact that every n n convergent sequence is bounded, we have τnα > −H1 , where H1 is a positive constant. The condition (1) of Theorem 1.8 is now satisfied. This completes the proof of Theorem 2.3. 1 Proof of Theorem 2.4. By hypothesis, sα+ → s (A). It then follows snα+2 → s (A). Hence, n 2 snα+1 − sα+ → 0(A). n
(21)
By Lemma 3.3 and hypothesis (8), we have 2 n∆(snα+1 − sα+ ) = O(1). n
α+1
(22) α+2
α+2
From (21) and (22), we get sn − sn = o(1) by Theorem 1.3. By Lemma 3.2, τn = o(1). By Lemma 3.3 and hypothesis (8), n∆τnα+2 = (α + 2)(τnα+1 − τnα+2 ) = o(1). Hence, τnα+1 = o(1). The condition (2) of Theorem 1.9 is now satisfied. This completes the proof of Theorem 2.4.
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1 2 3 Proof of Theorem 2.5. By hypothesis, sα+ → s (A). It then follows sα+ → s (A) and sα+ → s (A). Hence, n n n 1 sα+ − snα+2 → 0 (A), n
(23)
2 sα+ − snα+3 → 0 (A). n
(24)
and
By Lemma 3.4 and hypothesis (9), we have
(α + 2)(α + 3)(τnα+1 − τnα+2 ) − (α + 3)2 (τnα+2 − τnα+3 ) = o(1).
(25)
By Lemma 3.2, 1 2 τnα+1 − τnα+2 = n∆(sα+ − sα+ ) n n
and 2 3 τnα+2 − τnα+3 = n∆(sα+ − sα+ ). n n
Using the equalities above in (25), we have 1 2 2 3 n∆ (α + 2)(α + 3)(sα+ − sα+ ) − (α + 3)2 (sα+ − sα+ ) = o(1). n n n n
(26)
From (23), (24) and (26), we get 1 2 2 (α + 2)(α + 3)(sα+ − sα+ ) − (α + 3)2 (sα+ − snα+3 ) = o(1). n n n 1 2 2 3 Since sα+ − sα+ = α+1 2 τnα+2 and sα+ − sα+ = α+1 3 τnα+3 by Lemma 3.2, we have (α + 3)(τnα+2 − τnα+3 ) = o(1), which n n n n implies that 2 3 τnα+2 − τnα+3 = n∆(sα+ − sα+ ) = o(1). n n
α+2
(27) α+2
α+3
α+1
From (24) and (27), we get sn − sn = o(1) by Theorem 1.1. By Lemma 3.2, τn = o(1). Hence, τn The condition (2) of Theorem 1.9 is now satisfied. This completes the proof of Theorem 2.5. α+β
α+β
α+β
Proof of Theorem 2.6. Since Sn ≥ 0 for all n, we have sn ≥ 0 for all n by the definition of sn α+β is summable (A) to s. Hence by Theorem 1.5, sn → s (C , 1), which is equivalent to
= o(1) by (25). α+β
. By hypothesis, (sn
sαn → s(C , β + 1).
)
(28) α
α+1
It follows from (28) that sn → s (A) and sn
α
α+1
→ s (A). Hence, sn − sn
α+1
→ 0 (A). Applying Theorem 1.4 to τn
τnα+1 = o(1).
, we have (29)
α
Now, by (29) and hypothesis (11), we have τn > −H2 , where H2 is a positive constant. The condition (4) of Theorem 1.10 is now satisfied. This completes the proof of Theorem 2.6. Proof of Theorem 2.7. We know from Theorem 2.6 that sαn → s (C , β + 1). It follows from (28) that snα+k → s (A) for k = 1, 2, 3. Hence, 1 sαn − sα+ → 0(A), n
(30)
1 sα+ − snα+2 → 0(A). n
(31)
and
By Lemma 3.4 and hypothesis (13),
(α + 1)(α + 2)(τnα − τnα+1 ) − (α + 2)2 (τnα+1 − τnα+2 ) > −H , α
α+1
where H is a positive constant. Since τn − τn
α
α+1
= n∆(sn − sn
(32) α+1
) and τn
α+2
− τn
1 1 2 n∆[(α + 1)(α + 2)(sαn − sα+ ) − (α + 2)2 (sα+ − sα+ )] > −H n n n
α+1
= n∆(sn
α+2
− sn
), we have (33)
from (32). From (30), (31) and (33), we have 1 1 (α + 1)(α + 2)(sαn − sα+ ) − (α + 2)2 (sα+ − snα+2 ) = o(1) n n
α
α+1
by Theorem 1.4. Since (α + 1)(sn − sn
α+1
) = τn
1 2 τnα+1 − τnα+2 = n∆(sα+ − sα+ ) = o(1). n n
α+1
and (α + 2)(sn
(34) α+2
− sn
α+2
) = τn
by Lemma 3.2, it follows from (34) that (35)
İ. Çanak et al. / Mathematical and Computer Modelling 52 (2010) 738–743
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From (31) and (33) we have 2 snα+1 − sα+ = o(1) n
(36)
by Theorem 1.1. By (34) and (36),
τnα+2 = o(1).
(37)
If we put (37) in (35), we get
τnα+1 = o(1).
(38) α
If we put (35) and (37) in (32), we have τn > −H1 , where H1 is a positive constant. The condition (4) of Theorem 1.10 is now satisfied. This completes the proof of Theorem 2.7. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
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