L 1 -Convergence Of The Sine Series Whose Coe¢ cients Belong To Some Generalized Classes Of Sequences

L ¹ -Convergence Of The Sine Series Whose Coe¢ cients Belong To Some Generalized Classes Of Sequences

Xhevat Zahir Krasniqi

Received 7 March 2019

Abstract

In this paper, we have introduced three generalized classes of sequences. In addition, we have studied the L¹-convergence of sine series whose coe¢ cients belong to them. Finally, we show that our results covers some results proved previously by others.

1 Introduction

We consider trigonometric sine series

g(x) = X1 k=1

aksinkx;

with its partial sums

S^s_n(x) = Xn k=1

a_ksinkx;

and

nlim!1S_n^s(x) =g(x):

It is a well-known fact that if a trigonometric series converges inL¹-norm, then it is a Fourier series. In general, the converse of this statement is not always true. So, the question to be considered is how to make possible that the converse statement to be true? For this purpose a lot of researchers have introduced the so-called modi…ed trigonometric cosine sums or modi…ed trigonometric sine sums or both as well as some classes of sequences to which the coe¢ cients of a trigonometric series belong to.

The most famous modi…ed trigonometric sums appearing in literature are f_n(x) =1

2 Xn k=0

a_k+ Xn k=1

Xn j=k

a_jcoskx;

introduced in [4], and then the modi…ed cosine and sine sums g_n^c(x) =a0

2 + Xn k=1

Xn j=k

aj

j kcoskx and

g_n^s(x) = Xn k=1

Xn j=k

a_j

j ksinkx;

Mathematics Sub ject Classi…cations: 42A10, 42A16, 42A32.

yUniversity of Prishtina "Hasan Prishtina", Faculty of Education, Department of Mathematics and Informatics, Avenue

"Mother Theresa " no. 5, Prishtinë 10000, Kosovo

96

introduced in [11], where a_i:=a_i a_i+1. We do not recall here all modi…ed trigonometric sums introduced by others, however we suggest the interested reader to consult the references [5]-[9] of the present paper and references therein to …nd other modi…ed trigonometric sums.

Seemingly, it was S. A. Telyakovskii [13] who introduced the classSeof sequences and F. Móricz [10] who introduced the classesBVg andCe of sequences.

Very recently, the authors of [2] introduced the following modi…ed sine sums

z_n^s(x) = Xn k=1

2 4ak+1

k+ 1+ Xn j=k

2 aj

j 3

5ksinkx;

where ²ai:= ( ai) =ai 2ai+1+ai+2, and the following classes of sequences:

De…nition 1 A sequence (ak) tending to zero belongs to the class Cer, r = 0;1;2; : : :, if for every " > 0, there exists >0 independent onn and such that for alln,

Z

0

X1 k=n

b_kD_k^(r+1)(x) dx ";

wherebk= ^a_k^k andD_k^(r+1)(x)denotes the (r+ 1)-th derivative of the Dirichlet’s kernel Dk(x) =1

2 + Xk j=1

cosjx= sin k+¹₂ x 2 sin^x₂ :

De…nition 2 A sequence (a_k)tending to zero belongs to the class Se_r,r= 0;1;2; : : :, if there exists a non- increasing sequence (B_k) of numbers so that,j b_kj B_k, 8k 2 f1;2; : : :g, and P₁

k=1k^r+1B_k <1, where bk =^a_k^k.

De…nition 3 A sequence(ak)tending to zero belongs to the classBVgr,r= 0;1;2; : : :, if X1

k=1

k^r+1j bkj<1; whereb_k= ^a_k^k.

In the same paper there has been proved the following results.

Theorem 1 ([2]) Se_r Ce_r\BVg_r for eachr2 f0;1;2; : : :g.

Theorem 2 ([2]) Let (an)2Ce\BVg andlimn!1anlogn= 0. Then

nlim!1kz^s_n gk= 0:

Theorem 3 ([2]) Let (an)2Ce\BVg andlimn!1anlogn= 0. Then

nlim!1kS^s_n gk= 0:

Theorem 4 ([2]) Let (an)2Cer\BVgr andlimn!1n^ranlogn= 0,r2 f0;1;2; : : :g. Then

nlim!1k(z^s_n)^(r) g^(r)k= 0:

Theorem 5 ([2]) Let (a_n)2Ce_r\BVg_r andlim_n!1n^ra_nlogn= 0,r2 f0;1;2; : : :g. Then

nlim!1k(S_n^s)^(r) g^(r)k= 0:

Corollary 1 ([2]) Let (an)2Ser andlimn!1n^ranlogn= 0,r2 f0;1;2; : : :g. Then (i) lim_n!1k(z_n^s)^(r) g^(r)k= 0:

(ii) lim_n!1k(S_n^s)^(r) g^(r)k= 0:

Now we introduce the following generalized modi…ed sine sums

z^s_n;m(x) = Xn k=1

2 4 ak+1

(k+ 1)^m+ Xn j=k

2 aj

j^m 3

5k^msinkx; m2 f1;2; : : :g;

where again ²ci:= ( ci) :=ci 2ci+1+ci+2.

Remark 1 Note that z^s_n;1(x) z_n^s(x)which have been introduced for the …st time in [3].

Further we generalize the classes Cer,Ser, andBVgr, (r2 f0;1;2; : : :g) as follows:

De…nition 4 A sequence (ak)tending to zero belongs to the class Cer;m, (r= 0;1;2; : : :; m= 1;2; : : :), if for every" >0, there exists >0 independent onnand such that for all n,

Z

0

X1 k=n

bk;mD_k^(r+m)(x) dx ";

wherebk= _k^am^k,D_k^(r+m)(x)denotes the (r+m)-th derivative of the Dirichlet kernel D_k(x) =1

2 + Xk j=1

cosjx= sin k+¹₂ x 2 sin^x₂ :

Remark 2 It is clear that Ce_r+1;m Ce_r;m,(r= 0;1;2; : : :; m= 1;2; : : :), however, the converse inclusion need not be true in general as shown in the next example.

Example 1 De…neb_n;m=P₁

k=n 1

k^r+m+2,(r= 0;1;2; : : :;m= 1;2; : : :), then b_n;m= _nr+m+2¹ and a_n=nb_n;m=n

X1 k=n

1 k^r+m+2

X1 k=n

k k^r+m+2 =

X1 k=n

1

k^r+m+2 !0; when n! 1: So, using Bernstein’s inequality, the integral

Z

0

X1 k=n

b_k;mD^(r+1+m)_k (x) dx Z

0

X1 k=n

b_k;mD^(r+1+m)_k (x) dx X1

k=n

j b_k;mj Z

0

D_k^(r+1+m)(x) dx X1

k=n

k^r+1+m k^r+m+2

Z

0 jDk(x)jdx=O X1 k=1

logk k

!

;

is divergent, which means(a_n)62Ce_r+1;m. On the other side, the integral Z

0

X1 k=n

bk;mD_k^(r+m)(x) dx Z

0

X1 k=n

bk;mD^(r+m)_k (x) dx X1

k=n

j bk;mj Z

0

D^(r+m)_k (x) dx X1

k=n

k^r+m k^r+m+2

Z

0 jD_k(x)jdx=O X1 k=1

logk k²

!

; is convergent, which means(an)2Cer;m.

De…nition 5 A sequence (ak) tending to zero belongs to the classSer;m,(r = 0;1;2; : : :; m= 1;2; : : :), if there exists a non-increasing sequence(Bk)so that,j bk;mj Bk,8k2 f1;2; : : :g, andP₁

k=1k^r+mBk <1, wherebk= _k^am^k.

Remark 3 It is clear that Ser+1;m Ser;m,(r= 0;1;2; : : :; m= 1;2; : : :), however, the converse inclusion need not to be true as shown in the next example.

Example 2 De…neb_n;m=P₁

k=n 1

k^r+m+2,(r= 0;1;2; : : :;m= 1;2; : : :), then b_n;m= _nr+m+2¹ and a_n=nb_n;m=n

X1 k=n

1 k^r+m+2

X1 k=n

k k^r+m+2 =

X1 k=n

1

k^r+m+1 !0; when n! 1:

ChoosingBn =_nr+m+2¹ ,(r= 0;1;2; : : :;m= 1;2; : : :), thenBn#0andj bn;mj Bn. Now, the series X1

k=1

k^r+mBk= X1 k=1

k^r+m 1 k^r+m+2 =

X1 k=1

1 k² <1 is convergent, which means(an)2Ser;m. However, the series

X1 k=1

k^r+1+mB_k = X1 k=1

k^r+1+m 1 k^r+m+2 =

X1 k=1

1 k is divergent, which means(an)62Ser+1;m.

De…nition 6 A zero sequence (a_k)belongs to the class BVg_r;m,(r= 0;1;2; : : :;m= 1;2; : : :), if X1

k=1

k^r+mj bk;mj<1; wherebk;m= _k^am^k.

Remark 4 It is clear that BVg_r+1;m BVg_r;m, (r = 0;1;2; : : :; m = 1;2; : : :), however, the converse inclusion may not be true.

Remark 5 We note that Cer;m Cer,Ser;m Ser, andBVgr;m BVgr form= 1, and Cer;m C,e Ser;m S,e andBVgr;m BVg form= 1andr= 0.

The objective of this paper is to prove some theorems more general than Theorems 1–5 and Corollary 1 involving new classes Ce_r;m, Se_r;m, and BVg_r;m. To achieve this objective we need to recall some lemmas which have already proved elsewhere. Throughout this paper, for two positive quantities uand v, we write u=O(v), if there exists a positive constantC so thatu Cv.

2 Helpful Lemmas

Lemma 1 ([15]) Let m be a non-negative integer. Then for all 0 <jxj and all n 1 the estimate jDe^(m)n (x)j ⁴ⁿ_jx^rj holds true, whereDen^(m)(x)denotes m-th derivative of the conjugate Dirichlet kernel

De_k(x) = Xk j=1

sinjx= cos^x₂ cos k+¹₂ x 2 sin^x₂ :

Lemma 2 ([12]) Let m be a non-negative integer. Then for all0 < " x and all n 1 the estimate jD^(m)n (x)j ^Cnx^r holds true, whereC denotes a positive absolute constant.

Lemma 3 ([12]) kD^(m)n (x)kL¹ =O(n^mlogn),m2 f0;1;2; : : :g, holds true, whereD^(m)n (x) denotes m-th derivative of the Dirichlet kernel.

Lemma 4 ([7]) If Dn(x),Den(x), andFn(x)are the Dirichlet, the conjugate Dirichlet and the Fejér kernel respectively, thenDe_n⁰(x) = (n+ 1)Dn(x) (n+ 1)Fn(x).

Lemma 5 ([14]) Let the real numbers _i, i = 1;2; : : : ; k; satisfy conditions j ij 1. Then the following estimations hold true

Z

0

Xk i=0

i

sin i+¹₂ x

2 sin^x₂ dx C(k+ 1);

whereC is a positive constant.

3 Main Results

At …rst, pertaining to the BVgr;m class, r 2 f0;1;2; : : :g and m 2 f1;2; : : :g, we can raise the following natural question: What about inclusion of classesBVg_r;m with respect to m? The answer is given in next simple proposition.

Proposition 1 If

X1 k=1

(k+ 1)^rj a_kj<1;

then

BVg_r;m BVg_r;m+1; for allr2 f0;1;2; : : :g andm2 f1;2; : : :g.

Proof. We have X1 k=1

k^r+m+1j bk;m+1j

X1 k=1

k^r+m+1 ak

k^m+1

ak+1

k(k+ 1)^m +

X1 k=1

k^r+m+1 ak+1

k(k+ 1)^m

ak+1

(k+ 1)^m+1 X1

k=1

k^r+m a_k k^m

a_k+1 (k+ 1)^m +

X1 k=1

k^{r 1}ja_k+1j X1

k=1

k^r+mj b_k;mj+ X1 k=1

k^{r 1} X1 j=k+1

j a_jj

= X1 k=1

k^r+mj bk;mj+ X1 j=1

j ajj

j+1X

k=1

k^{r 1} X1

k=1

k^r+mj b_k;mj+ X1 j=1

(j+ 1)^rj a_jj;

which implies thatBVgr;m BVgr;m+1;for allr2 f0;1;2; : : :gandm2 f1;2; : : :g. The proof is completed.

Theorem 6 The following relation holds true Ser;m Cer;m\BVgr;m for each r 2 f0;1;2; : : :g and m 2 f1;2; : : :g.

Proof. Let(ak)2Ser;m,(r= 0;1;2; : : :;m= 1;2; : : :). Then there exists a non-increasing sequence(Bk)of numbers so that,j bk;mj Bk, 8k2 f1;2; : : :g, and P₁

k=1k^r+mBk <1. Whence, we clearly have X1

k=1

k^r+mj bk;mj X1 k=1

k^r+mBk<1; (1)

which means thatSer;m BVgr;m for eachr2 f0;1;2; : : :g and m2 f1;2; : : :g. So, it remains to prove the inclusionSer;m Cer;m for eachr2 f0;1;2; : : :g andm2 f1;2; : : :g. Let(ak)2Ser;m. Then applying Abel’s transformation we get

Z

0

X1 k=n

bk;mD^(r+m)_k (x) dx lim

s!1

"_{s 1} X

k=n

Bk

Z

0

Xk j=0

bj;m

Bj

D^(r+m)_j (x) dx

+B_s Z

0

Xs j=0

b_j;m Bj

D_j^(r+m)(x) dx

+B_n Z

0 nX1 j=0

b_j;m Bj

D_j^(r+m)(x) dx

# :

Applying, in last inequality, the well-known Bernstein’s inequality and Lemma5, we obtain Z

0

X1 k=n

bk;mD_k^(r+m)(x) dx lim

s!1

"_{s 1} X

k=n

k^r+m Bk

Z

0

Xk j=0

bj;m

B_j Dj(x) dx

+s^r+mBs

Z

0

Xs j=0

bj;m

Bj

Dj(x) dx

+(n 1)^r+mB_n Z

0 n 1

X

j=0

b_j;m Bj

D_j(x) dx

#

C lim

s!1

"_{s 1} X

k=n

(k+ 1)^r+m+1 B_k

+s^r+m+1Bs+n^r+m+1Bn

# : Since (B_k) is a non-increasing sequence and P₁

k=1k^r+mB_k <1, we see thatk^r+m+1B_k ! 0 as k ! 1, and thus

Z

0

X1 k=n

b_k;mD_k^(r+m)(x) dx C

" ₁ X

k=n

(k+ 1)^r+m+1 B_k+n^r+m+1B_n

#

C ( ₁

X

k=n

Bk (k+ 1)^r+m+1 k^r+m+1 +n^r+m+1Bn

)

C(r; m) ( ₁

X

k=n

k^r+mBk+n^r+m+1Bn

)

"

2; fornlarge enough, says n > n0.

Finally, using the fact that

D^(r+m)_k (x) = Xk j=1

j^(r+m)sin jx+(r+m)

2 k^r+m+1;

for any1 n s; we can write as follows Z

0

Xs k=n

b_k;mD^(r+m)_k (x) dx Z

0 n0

X

k=n

b_k;mD^(r+m)_k (x) dx

+ Z

0

Xs k=n0+1

b_k;mD_k^(r+m)(x) dx

n0

X

k=n

k^r+m+1j bk;mj

+ Z

0

X1 k=n0+1

b_k;mD_k^(r+m)(x) dx

"

2 +"

2 =";

for small enough. This means thatSer;m Cer;m for each r2 f0;1;2; : : :g andm2 f1;2; : : :g. The proof is completed.

Remark 6 Form= 1Theorem6reduces to Theorem 1.

Theorem 7 Let (a_n)2Ce_m\BVg_m for each m2 f1;2; : : :g, andlim_n!1a_nlogn= 0. Then

nlim!1kz_n;m^s gk= 0:

Proof. We have

z_n;m^s (x) = Xn k=1

2 4 ak+1

(k+ 1)^m+ Xn j=k

2 aj

j^m 3

5k^msinkx

= Xn k=1

aksinkx+ an+2

(n+ 2)^m

an+1

(n+ 1)^m Xn k=1

k^msinkx

= S_n^s(x) (bn+1;m) Xn k=1

k^msinkx (2)

After some transformation we have found that

S_n^s(x) = 8>

>>

<

>>

>: Pn

k=1b_k;m(coskx)^(m); ifm= 4p 3;

Pn

k=1bk;m(sinkx)^(m); ifm= 4p 2;

+Pn

k=1b_k;m(coskx)^(m); ifm= 4p 1;

+Pn

k=1bk;m(sinkx)^(m); ifm= 4p,

(3)

and

Xn k=1

k^msinkx= 8>

>>

><

>>

>>

:

D^(m)n (x); ifm= 4p 3;

De^(m)n (x); ifm= 4p 2;

+D^(m)n (x); ifm= 4p 1;

+De^(m)n (x); ifm= 4p,

(4)

where in all casesp2N. Combining (2) along with (3) and (4) we obtain

z^s_n;m(x) = 8>

>>

>>

>>

<

>>

>>

>>

>: Pn

k=1b_k;m(coskx)^(m)+ (b_n+1;m)Dn^(m)(x); ifm= 4p 3;

Pn

k=1b_k;m(sinkx)^(m)+ (b_n+1;m)De^(m)n (x); ifm= 4p 2;

+Pn

k=1b_k;m(coskx)^(m) (b_n+1;m)Dn^(m)(x); ifm= 4p 1;

+Pn

k=1b_k;m(sinkx)^(m) (b_n+1;m)De^(m)n (x); ifm= 4p,

(5)

for allp2N. The use of Abel’s transformation in (5), implies

z_n;m^s (x) = 8>

>>

>>

>>

>>

>>

>>

<

>>

>>

>>

>>

>>

>>

>: Pn

k=1 bk;mD_k^(m)(x)

bn;mDn^(m)(x) + (bn+1;m)Dn^(m)(x); ifm= 4p 3;

Pn

k=1 bk;mDe_k^(m)(x)

bn;mDen^(m)(x) + (bn+1;m)Den^(m)(x); ifm= 4p 2;

Pn

k=1 bk;mD_k^(m)(x)

+bn;mDn^(m)(x) (bn+1;m)Dn^(m)(x); ifm= 4p 1;

Pn

k=1 bk;mDe_k^(m)(x)

+bn;mDen^(m)(x) (bn+1;m)Den^(m)(x); ifm= 4p,

(6)

for allp2N. Applying Abel’s transformation in (3), we also get

S_n^s(x) = 8>

>>

><

>>

>>

: Pn

k=1 bk;mD^(m)_k (x) bn;mD^(m)n (x); ifm= 4p 3;

Pn

k=1 bk;mDe^(m)_k (x) bn;mDe^(m)n (x); ifm= 4p 2;

+Pn

k=1 bk;mD^(m)_k (x) +bn;mD^(m)n (x); ifm= 4p 1;

+Pn

k=1 bk;mDe^(m)_k (x) +bn;mDe^(m)n (x); ifm= 4p,

(7)

for allp2N. Using Lemmas1 and2, in (6) and (7), we have that jz_n;m^s (x)j O x ¹

Xn k=1

k^mj b_k;mj+ja_nj+ja_n+1j+ja_n+2j

!

; (8)

and

jS_n^s(x)j O x ¹ Xn k=1

k^mj bk;mj+janj

!

; (9)

for allm2N.

Whence, letting n! 1 in (8) and (9), and taking into account that (a_k)2BVg_r;m, m= 1;2; : : :), we conclude that seriesP₁

k=1 bk;mD_k^(m)(x)andP₁

k=1 bk;mDe_k^(m)(x)converge absolutely, and

nlim!1z^s_n;m(x) = lim

n!1S_n^s(x) =g(x) exists for allx2["; ], where" >0as small as. Now, we have

g(x) z^s_n;m(x) = 8>

>>

>>

>>

>>

>>

>>

<

>>

>>

>>

>>

>>

>>

>: P₁

k=n+1 b_k;mD^(m)_k (x)

+b_n;mD^(m)n (x) (b_n+1;m)Dn^(m)(x); ifm= 4p 3;

P₁

k=n+1 b_k;mDe^(m)_k (x)

+b_n;mDe^(m)n (x) (b_n+1;m)Den^(m)(x); ifm= 4p 2;

+P₁

k=n+1 b_k;mD^(m)_k (x)

b_n;mD^(m)n (x) + (b_n+1;m)Dn^(m)(x); ifm= 4p 1;

+P₁

k=n+1 b_k;mDe^(m)_k (x)

bn;mDe^(m)n (x) + (bn+1;m)Den^(m)(x); ifm= 4p,

(10)

for allp2N. Thus, based on (10), we have

kg z_n;m^s k 8>

>>

>>

>>

>>

>>

><

>>

>>

>>

>>

>>

>>

: R

0

P₁

k=n+1 b_k;mD^(m)_k (x) dx +jb_n;mjR

0 Dn^(m)(x) dx +j (b_n+1;m)jR

0 Dn^(m)(x) dx; ifm= 4p 3_m= 4p 1;

R

0

P₁

k=n+1 b_k;mDe^(m)_k (x) dx +jbn;mjR

0 Den^(m)(x) dx +j (bn+1;m)jR

0 Den^(m)(x) dx; ifm= 4p 2_m= 4p,

(11)

for allp2N. Let us estimate the terms in right hand side of (11). Namely, since(a_n)2Ce_m\BVg_mfor each m2 f1;2; : : :g, then for " >0 there exists >0, such that

Z

0

X1 k=n+1

bk;mD_k^(m)(x) dx "

2;

for alln 0. Consequently, form= 4p 3_m= 4p 1and Bernstein’s inequality we get Z

0

X1 k=n+1

b_k;mD^(m)_k (x) dx = Z

0

X1 k=n+1

b_k;mD^(m)_k (x) dx (12)

+

Z X1

k=n+1

bk;mD_k^(m)(x) dx

"

2 + X1 k=n+1

j b_k;mj Z

D_k^(m)(x) dx

"

2 + X1 k=n+1

k^{m 1}j bk;mj Z

jD_k⁰(x)jdx

"

2 +C X1 k=n+1

k^mj b_k;mj Z dx

x²

"

2 +C X¹

k=n+1

k^mj bk;mj

< "

2 +"

2 ="; (13)

and in a similar way, form= 4p 2_m= 4p, Z

0

X1 k=n+1

bk;mDe^(m)_k (x) dx < ": (14)

The other terms also tend to zero, since they can be estimated asanlogn (using Lemma3). Using these facts, (11), (12) and (14), we have proved that

nlim!1kz_n;m^s gk= 0:

The proof is completed.

Remark 7 Form= 1Theorem7reduces to Theorem 2.

Theorem 8 Let (an)2Cem\BVgm for each m2 f1;2; : : :g, andlimn!1anlogn= 0. Then

nlim!1kS^s_n gk= 0:

Proof. Using Theorem7, equalities (6), and equalities (7), we get kS_n^s gk = kS^s_n z_n;m^s +z_n;m^s gk

kz_n;m^s S_n^sk+kz_n;m^s gk (

j (bn+1;m)jR

0 D^(m)n (x)dx+o(1); ifm= 4p 3^m= 4p 1;

j (bn+1;m)jR

0 De^(m)n (x)dx+o(1); ifm= 4p 2^m= 4p;

for allp2N. Now applying Lemma3, Bernstein’s inequality, Lemma 4, and conditions of our theorem, we obtain

kS_n^s gk 8>

><

>>

:

C(n+ 1)^mj (bn+1;m)jlog(n+ 1) +o(1); ifm= 4p 3^m= 4p 1;

(n+ 1)j (bn+1;m)j R

0 D^(mn ¹⁾(x)dx+R

0 Fn^{(m 1)}(x)dx +o(1); ifm= 4p 2^m= 4p;

8>

<

>:

C(n+ 1)^mj (b_n+1;m)jR

0 D_n(x)dx+o(1); ifm= 4p 3^m= 4p 1;

(n+ 1)^mj (bn+1;m) j R

0 D_n(x)dx+R

0 F_n(x)dx +o(1); ifm= 4p 2^m= 4p;

= O(an+1log(n+ 1) +an+2log(n+ 2) +o(1)) =o(1) as n! 1: The proof is completed.

Remark 8 Form= 1Theorem8reduces to Theorem 3.

The following statements also hold true.

Theorem 9 Let (an)2Cer;m\BVgr;m,r2 f0;1; : : :g,m2 f1;2; : : :g, andlim_!1n^ranlogn= 0. Then

nlim!1k[z_n;m^s ]^(r) g^(r)k= 0:

Proof. The proof is similar to the proof of Theorem7. Therefore, we omit it.

Remark 9 Forr= 0, Theroem8reduces to Theorem4

Theorem 10 Let(an)2Cer;m\BVgr;m,r2 f0;1; : : :g,m2 f1;2; : : :g, andlimn!1n^ranlogn= 0. Then

nlim!1k[S_n^s]^(r) g^(r)k= 0:

Proof. The proof is similar to the proof of Theorem8. Therefore, we omit it.

Remark 10 Forr= 0, Theorem10 reduces to Theorem 5.

Using Theorems9 and10, we obtain next consequence.

Corollary 2 Let (a_n)2Se_r;m,r2 f0;1; : : :g,m2 f1;2; : : :g, andlim_n!1n^ra_nlogn= 0. Then (i) lim_n!1k[z_n;m^s ]^(r) g^(r)k= 0:

(ii) lim_n!1k[S_n^s]^(r) g^(r)k= 0:

Form= 1 we have:

Corollary 3 ([2]) Let (an)2Ser,r2 f0;1; : : :g, andlimn!1n^ranlogn= 0. Then (i) lim_n!1k[z_n^s]^(r) g^(r)k= 0:

(ii) limn!1k[S_n^s]^(r) g^(r)k= 0:

Acknowledgment. The author would like to express his sincere thanks to anonymous referee for her/his useful comments and suggestions.

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