C om mun.Fac.Sci.U niv.A nk.Series A 1 Volum e 67, N umb er 1, Pages 211–224 (2018) D O I: 10.1501/C om mua1_ 0000000843 ISSN 1303–5991
http://com munications.science.ankara.edu.tr/index.php?series= A 1
THE BINOMIAL ALMOST CONVERGENT AND NULL SEQUENCE SPACES
MUSTAFA CEM ·IL B·I¸SG·IN
Abstract. In this paper, we introduce the sequence spaces f (Br;s), f 0(Br;s) and f s(Br;s)which generalize the Kiri¸sçi’s work [16]. Moreover, we show that these spaces are BK-spaces and are linearly isomorphic to the sequence spaces f, f0 and f s, respectively. Furthermore, we mention the Schauder basis and give , -duals of these spaces. Finally, we determine some matrix classes related to these spaces.
1. Introduction
The family of all real(or complex) valued sequences is a vector space under usual coordinate-wise addition and scalar multiplication and is denoted by w. Every vector subspace of w is called a sequence space. The notations of `1, c0, c and `p
are used for the spaces of all bounded, null, convergent and absolutely p-summable sequences, respectively, where 1 p < 1.
A BK-space is a Banach sequence space provided each of the maps pi: X ! C,
pi(x) = xi is continuous for all i 2 N, where X is a sequence space. According to
this de…nition, the sequence spaces `1, c0and c are BK-spaces with their sup-norm
de…ned by kxk1= sup
n2Njxnj and `p
is a BK-space with its `p-norm de…ned by
kxk`p = 1 X k=0 jxkjp !1 p where 1 p < 1 [2].
Received by the editors: August 12, 2016, Accepted: March 16, 2017.
2010 Mathematics Subject Classi…cation. Primary 40C05, 40H05; Secondary 46B45.
Key words and phrases. Matrix domain, Schauder basis , and duals, Banach limits, almost convergence and matrix classes.
c 2 0 1 8 A n ka ra U n ive rsity C o m m u n ic a tio n s d e la Fa c u lté d e s S c ie n c e s d e l’U n ive rs ité d ’A n ka ra . S é rie s A 1 . M a th e m a t ic s a n d S t a tis tic s .
Let A = (ank) be an in…nite matrix of complex numbers. For any x = (xk) 2 w,
the A-transform of x is written by y = Ax and is de…ned by yn= (Ax)n=
1
X
k=0
ankxk (1.1)
for all n 2 N and each of these series being assumed convergent [3]. For simplicity in notation, we henceforth prefer that the summation without limits runs from 0 to 1.
Given two arbitrary sequence spaces X and Y , the class of all matrices A = (ank)
such that Ax 2 Y for all x 2 X is denoted by (X : Y ).
The domain of an in…nite matrix A = (ank) in a sequence space X is denoted
by XA de…ned by
XA= fx = (xk) : Ax 2 Xg (1.2)
which is also a sequence space. The domain of summation matrix S = (snk) in
sequence spaces c and `1are called the spaces of all convergent and bounded series and are denoted by cs and bs, respectively, where S = (snk) is de…ned by
snk= 10 ;; 0k > nk n
for all n; k 2 N.
A matrix is called a triangle if ank= 0 for k > n and ann 6= 0 for all n; k 2 N.
Also, a triangle matrix A = (ank) uniquely has an inverse A 1such that A 1 is a
triangle matrix.
As an application of the Hahn-Banach theorem to the sequence space `1, the notion of Banach Limits was …rst introduced by the Stefan Banach. Banach …rst recognized certain non-negative linear functionals on `1 which remain invariant under shift operators and which are extension of l, where l : c ! R, l(x) = lim
n!1xn
is de…ned and l is linear functional on c. Such functionals were later termed "Banach Limits" [1].
A functional L : `1 ! R is called a Banach Limit if the following conditions hold
(i) L(axn+ byn) = aL(xn) + bL(yn) a; b 2 R
(ii) L(xn) 0 if xn 0, n = 0; 1; 2; :::
(iii) L(Pj(xn)) = L(xn), Pj(xn) = xn+j, j = 1; 2; 3; :::
(iv) L(e) = 1 where e = (1; 1; :::)
Lorentz continued the study of Banach Limits and brought out a new concept called Almost Convergence. The bounded sequence x = (xn) is called almost
convergent and the number Limxn = is called its F -limit if L(xn) = holds for
every limit L [4].
The theory of matrix transformation has a great importance in the theory of summability which was obtained by Cesàro, Norlund, Borel, Riesz... . Therefore, many authors have constructed new sequence spaces by using matrix domain of
in…nite matrices. For instance, (`1)Nq and cNq in [5], Xpand X1 in [6], e
r 0 and erc
in [7], er
p and er1 in [8] and [9], er0( ); erc( ) and e1r ( ) in [10], er0( m); erc( m)
and er
1( m) in [11], er0(B(m)); erc(B(m)) and er1(B(m)) in [12], er0( ; p); erc( ; p)
and er
1( ; p) in [13], ^f0 and ^f in [14], f0(B) and f (B) in [15], f0(E) and f (E) in
[16].
In this paper, we introduce the sequence spaces f (Br;s), f
0(Br;s) and f s(Br;s)
which generalize the Kiri¸sçi’s work [16]. Moreover, we show that these spaces are BK-spaces and are linearly isomorphic to the sequence spaces f , f0 and f s,
respectively. Furthermore, we mention the Schauder basis and give , -duals of these spaces. Finally, we determine some matrix classes related to these spaces.
2. The Binomial Almost Convergent And Null Sequence Spaces In this part, we give some historical informations and de…ne the sequence spaces f0(Br;s), f (Br;s) and f s(Br;s). Furthermore, we show that these spaces are
BK-spaces and are linearly isomorphic to the sequence BK-spaces f0, f and f s, respectively.
Lorentz obtained the following characterization for almost convergent sequences. Theorem 1 (see [4]). In order that F -limit, Limxn = exists for the sequence
x = (xn), it is necessary and su¢ cient that
lim
k!1
xn+ xn+1+ ::: + xn+k
k + 1 =
holds uniformly in n.
By taking into account the notion of almost convergence and Theorem 1, the space of all almost convergent sequences, almost null sequences and almost conver-gent series are de…ned by
f = ( x = (xk) 2 w : 9 2 C 3 lim i!1 i X k=0 xn+k i + 1 = uniformly in n ) ; f0= ( x = (xk) 2 w : lim i!1 i X k=0 xn+k i + 1 = 0 uniformly in n ) ; and f s = 8 < :x = (xk) 2 w : 9 2 C 3 limi!1 i X k=0 n+kX j=0 xj i + 1 = uniformly in n 9 = ;; respectively.
By considering the notion of (1.2), the sequence space f s can be rearranged by means of the summation matrix S = (snk) as follows:
Theorem 2(see [17]). The inclusions c f `1 strictly hold.
Theorem 3(see [17]). The sequence spaces f and f0are BK-spaces with the norm
kxkf = sup i;n2N i P k=0 xn+k
i+1 and f s is a BK-space with the norm kxkf s= kSxkf.
In order to de…ne sequence spaces, the Euler matrix was …rst considered by Altay, Ba¸sar and Mursaleen in [7], [8] and [9]. They constructed the Euler sequence spaces er 0, erc, er1 and erp as follows: er0= ( x = (xk) 2 w : lim n!1 n X k=0 n k (1 r) n krkx k = 0 ) ; erc = ( x = (xk) 2 w : lim n!1 n X k=0 n k (1 r) n krkx k exists ) ; er1= ( x = (xk) 2 w : sup n2N n X k=0 n k (1 r) n krkx k < 1 ) and erp= ( x = (xk) 2 w : 1 X n=0 n X k=0 n k (1 r) n krkx k p < 1 ) : where 1 p < 1, and the Euler matrix Er= (er
nk) is de…ned by ernk= n k (1 r) n krk ; 0 k n 0 ; k > n
for all n; k 2 N, where 0 < r < 1.
Afterward, Kiri¸sçi used the Euler matrix in [16] for de…ning Euler almost null and Euler almost convergent sequence spaces. These spaces are de…ned by
f0(E) = 8 < :x = (xk) 2 w : limm!1 m X j=0 n+j X k=0 n+j k (1 r)n+j krkxk m + 1 = 0 uniformly in n 9 = ; and f (E) = 8 < :x = (xk) 2 w : 9 2 C 3 limm!1 m X j=0 n+jX k=0 n+j k (1 r) n+j krkx k m + 1 = uniformly in n 9 = ;: Recently, Bi¸sgin has de…ned the Binomial sequence spaces br;s0 , br;s
c , br;s1 and br;sp in [18] and [19] as follows: br;s0 = ( x = (xk) 2 w : lim n!1 1 (s + r)n n X k=0 n k s n krkx k= 0 ) ;
br;sc = ( x = (xk) 2 w : lim n!1 1 (s + r)n n X k=0 n k s n krkx k exists ) ; br;s1 = ( x = (xk) 2 w : sup n2N 1 (s + r)n n X k=0 n k s n krkx k < 1 ) and br;sp = ( x = (xk) 2 w : X n 1 (s + r)n n X k=0 n k s n krkx k p < 1 )
where 1 p < 1 and the Binomial matrix Br;s= (br;snk) is de…ned by br;snk= 1 (s+r)n n k sn krk ; 0 k n 0 ; k > n
for all k; n 2 N, r; s 2 R and rs > 0. Here, we would like to touch on a point, if we take r + s = 1, we obtain the Euler sequence spaces er
0, erc, er1 and erp. Therefore
Bi¸sgin has generalized the Altay, Ba¸sar and Mursaleen’s works.
Now, we de…ne the sequence spaces f0(Br;s), f (Br;s) and f s(Br;s) by
f0(Br;s) = 8 < :x = (xk) 2 w : limi!1 i X j=0 n+j X k=0 n+j k sn+j krkxk (i + 1)(r + s)n+j = 0 uniformly in n 9 = ;; f (Br;s) = 8 < :x = (xk) 2 w : 9 2 C 3 limi!1 i X j=0 n+j X k=0 n+j k s n+j krkx k (i + 1)(r + s)n+j = uniformly in n 9 = ; and f s(Br;s) = 8 < :x = (xk) 2 w : 9 2 C 3 limi!1 i X j=0 n+j X =0 X k=0 k s krkx k (i + 1)(r + s) = uniformly in n 9 = ;; respectively. By taking into account the notation (1.2), the sequence spaces f0(Br;s),
f (Br;s) and f s(Br;s) can be rede…ned by means of the domain of the Binomial ma-trix Br;s= (br;s
nk) as follows:
f0(Br;s) = (f0)Br;s; f (Br;s) = fBr;s and f s(Br;s) = f sBr;s (2.2)
In addition, given an arbitrary sequence x = (xk) 2 w, the Br;s-transform of
x = (xk) is de…ned by yk = (Br;sx)k= 1 (s + r)k k X j=0 k j s k jrjx j (2.3) for all k 2 N.
Theorem 4. The sequence spaces f0(Br;s), f (Br;s) and f s(Br;s) endowed with the norms kxkf (Br;s)= kxkf 0(Br;s)= kB r;s xkf and kxkf s(Br;s)= kBr;sxkf s
are BK-spaces, respectively.
Proof. We know that f , f0 and f s are BK-spaces. Also, Br;s= (br;snk) is a triangle
matrix and the condition (2.2) holds. By combining these three facts and Theorem 4.3.12 of Wilansky[3], we deduce that f (Br;s), f0(Br;s) and f s(Br;s) are BK-spaces.
This completes the proof.
Theorem 5. The sequence spaces f0(Br;s), f (Br;s) and f s(Br;s) are linearly
iso-morphic to the sequence spaces f0, f and f s, respectively.
Proof. Since the relations f0(Br;s) = f0 and f s(Br;s) = f s can be shown by using
a similar way, we give the proof of theorem for only the sequence space f (Br;s).
For this, we should show the existence of a linear bijection between the sequence spaces f (Br;s) and f .
Let us consider the transformation L : f (Br;s) ! f such that L(x) = Br;sx.
Then it is obvious that for every x = (xk) 2 f(Br;s), L(x) = Br;sx 2 f. Moreover,
it is clear that L is a linear transformation and x = 0 whenever L(x) = 0. Because of this, L is injective.
Now, we de…ne a sequence x = (xk) by means of the sequence y = (yk) 2 f by
xk= 1 rk k X j=0 k j ( s) k j(s + r)jy j
for all k 2 N. Then, we have (Br;sx)k = 1 (s + r)k k X j=0 k j s k jrjx j = 1 (s + r)k k X j=0 k j s k j j X i=0 j i ( s) j i(s + r)iy i = yk
for all k 2 N. This shows us that lim i!1 i X j=0 n+j X k=0 n+j k s n+j krkx k (i + 1)(r + s)n+j = limi!1 i X j=0 yn+j i + 1 = F limyn
namely, x = (xk) 2 f(Br;s) and L(x) = y. Therefore L is surjective. Moreover, for
all x = (xk) 2 f(Br;s), we know that
So, L is norm preserving. As a results of these, L is a linear bijection which says us that the sequence space f (Br;s) is linearly isomorphic to the sequence space f ,
that is f (Br;s) = f . This completes the proof.
Theorem 6. The inclusion c f (Br;s) is strict.
Proof. It is obvious that the inclusion c f (Br;s) holds. Now, we consider the
sequence x = (xk) de…ned by xk = ( 1)k for all k 2 N. Then, x = (xk) =2 c but
Br;sx = s r s+r
k
2 f, namely x 2 f(Br;s). So, the inclusion c f (Br;s) strictly
holds. This completes the proof.
3. The Schauder Basis And , -Duals
In this part, we speak of the Schauder basis and give , -duals of the spaces f (Br;s) and f s(Br;s).
Let us start with the de…nition of the Schauder basis. For a given normed space (X; k:kX), a sequence b = (bk) of elements of X is called a Schauder basis for X, if
and only if, for all x 2 X, there exists a unique sequence = ( k) of scalar such that x =P
k k
bk; i.e. such that
x n X k=0 kbk X ! 0 as n ! 1.
Corollary 1(see [14]). Almost convergent sequence space f has no Schauder basis. Remark 1. For an arbitrary sequence space X and a triangle matrix A = (ank),
it is known that XA has a basis if and only if X has a basis [20].
By combining this fact and Corollary 1, we can give the next result.
Corollary 2. The sequence spaces f (Br;s) and f s(Br;s) have no Schauder basis. The multiplier space of two arbitrary sequence spaces X and Y is de…ned by
M (X; Y ) =na = (ak) 2 w : xa = (xkak) 2 Y for all x = (xk) 2 X
o By using this de…nition and sequence spaces cs and bs, the - and -duals of a sequence space X are de…ned by
X = M (X; cs) and X = M (X; bs) respectively.
Now, we give some statements which are used in the next lemma. Let A = (ank)
be an in…nite matrix of complex numbers. sup
n2N
X
k
lim
n!1ank= k for each …xed k 2 N (3.2)
lim n!1 X k ank= (3.3) lim n!1 X k j (ank k)j = 0 (3.4) sup n2N X k j ankj < 1 (3.5) lim
k!1ank= 0 for each …xed n 2 N (3.6)
lim n!1 X k j 2ankj = (3.7) where ank= ank an;k+1and 2ank= ( ank).
Lemma 1. For an in…nite matrix A = (ank), the following statements hold:
(i) A = (ank) 2 (f : `1) , (3.1) holds (see [21])
(ii) A = (ank) 2 (f : c) , (3.1), (3.2), (3.3) and (3.4) hold (see [21])
(iii) A = (ank) 2 (fs : `1) , (3.5) and (3.6) hold (see [14])
(iv) A = (ank) 2 (fs : c) , (3.2), (3.5), (3.6) and (3.7) hold (see [22])
Theorem 7. Given the sets tr;s1 , tr;s2 , tr;s3 , tr;s4 , tr;s5 , tr;s6 and tr;s7 as follows:
tr;s1 = 8 < :a = (ak) 2 w : supn2N n X k=0 n X j=k j k ( s) j k(r + s)kr ja j < 1 9 = ; tr;s2 = 8 < :a = (ak) 2 w : limn!1 n X j=k j k ( s) j k(r + s)kr ja
j exists for each k 2 N
9 = ; tr;s3 = 8 < :a = (ak) 2 w : limn!1 n X k=0 2 4 k X j=0 k j ( s) k j(r + s)jr k 3 5 ak exists 9 = ; tr;s4 = 8 < :a = (ak) 2 w : limn!1 X k 2 4 n X j=k j k ( s) j k(r + s)kr ja j k 3 5 = 0 9 = ; tr;s5 = 8 < :a = (ak) 2 w : supn2N X k 2 4 n X j=k j k ( s) j k(r + s)kr ja j 3 5 < 1 9 = ; tr;s6 = 8 < :a = (ak) 2 w : limk!1 n X j=k j k ( s) j k(r + s)kr ja j= 0 for each n 2 N 9 = ;
and tr;s7 = 8 < :a = (ak) 2 w : limn!1 X k 2 2 4 n X j=k j k ( s) j k(r + s)kr ja j 3 5 exists 9 = ; where lim n!1 n P j=k j k ( s) j k(r + s)kr ja j= k for all k 2 N.
Then, the following statements hold. (i) ff(Br;s)g = tr;s1 \ t r;s 2 \ t r;s 3 \ t r;s 4 (ii) ff(Br;s)g = tr;s1 (iii) ffs(Br;s)g = tr;s 2 \ t r;s 5 \ t r;s 6 \ t r;s 7 (iv) ffs(Br;s)g = tr;s 5 \ t r;s 6
Proof. To avoid the repetition of similar statements, the proof of theorem is given for only part (i). For any a = (ak) 2 w, we consider the sequence x = (xk) de…ned
by xk= 1 rk k X j=0 k j ( s) k j(r + s)jy j
for all k 2 N. Then, we get
n X k=0 akxk = n X k=0 2 4 1 rk k X j=0 k j ( s) k j(r + s)jy j 3 5 ak = n X k=0 2 4 n X j=k j k ( s) j kr j(r + s)ka j 3 5 yk = Dr;sy n for all n 2 N, where the matrix Dr;s= (dr;s
nk) is de…ned by dr;snk= 8 < : n P j=k j k ( s)j kr j(r + s)kaj ; 0 k n 0 ; k > n
for all k; n 2 N. So, ax = (akxk) 2 cs whenever x = (xk) 2 f(Br;s) if and only
if Dr;sy 2 c whenever y = (yk) 2 f. This gives us that a = (ak) 2 ff(Br;s)g if
and only if Dr;s 2 (f : c). By combining this and Lemma 1 (ii), we obtain that
a = (ak) 2 ff(Br;s)g if and only if sup n2N X k jdr;snkj < 1; lim n!1d r;s
lim n!1 X k dr;snk = and lim n!1 X k j (dr;snk k)j = 0: As a consequence ff(Br;s)g = tr;s 1 \ t r;s 2 \ t r;s 3 \ t r;s
4 . This completes the proof.
4. Matrix Classes
In this part, we determine some matrix classes related to the sequence spaces f (Br;s) and f s(Br;s).
For simplicity of notation, from now on, we use the following connections. gnkr;s= 1 X j=k j k ( s) j kr j(r + s)ka nj (4.1) hr;snk= 1 (s + r)n n X j=0 n j s n jrja jk (4.2)
for all n; k 2 N, respectively.
Theorem 8. For a given sequence space X, assume that the in…nite matrices A = (ank), Gr;s = (gnkr;s) and Hr;s = (h
r;s
nk) are connected with the relations (4.1)
and (4.2). Then, the following statements hold. (i) A 2 (f(Br;s) : X) , Gr;s 2 (f : X) and fa
nkgk2N 2 ff(Br;s)g for all
n 2 N,
(ii) A 2 (X : f(Br;s)) , Hr;s2 (X : f).
Proof. (i) We suppose that A 2 (f(Br;s) : X). By considering the fact that f (Br;s)
and f are linearly isomorphic, we take an arbitrary sequence y = (yk) 2 f, where
y = Br;sx. Then, Gr;sBr;s exists and fa
nkgk2N 2 ff(Br;s)g for all n 2 N. This
gives us that fgr;snkgk2N2 `1 for each n 2 N. Thus, Gr;sy exists and
X k gnkr;syk = X k ankxk
for all n 2 N, namely Gr;sy = Ax. So, Gr;s2 (f : X).
Conversely, we suppose that Gr;s 2 (f : X) and fa
nkgk2N 2 ff(Br;s)g for all
Ax exists. Also, we have X k=0 ankxk = X k=0 2 4 1 rk k X j=0 k j ( s) k j(r + s)jy j 3 5 ank = X k=0 2 4X j=k j k ( s) j kr j(r + s)ka nj 3 5 yk
for all n 2 N. By passing to limit as ! 1, we deduce that Ax = Gr;sy. This
leads us A 2 (f(Br;s) : X).
(ii) For any x = (xk) 2 X, we consider the following equality:
fBr;s(Ax)gn = 1 (r + s)n n X k=0 n k s n krk(Ax) k = X k 1 (r + s)n n X j=0 n j s n jrja jkxk = fHr;sxgn
for all n 2 N. By going to the generalized limit, we obtain that Ax 2 f(Br;s) if and
only if Hr;sx 2 f. This completes the proof.
Now, we list some properties in order to give next lemma. Let A = (ank) be an
in…nite matrix of complex numbers.
F lim
n!1ank= k for all …xed k 2 N (4.3)
F lim n!1 X k ank= (4.4) F lim n!1 n X j=0
ajk= k for all …xed k 2 N (4.5)
sup n2N X k n X j=0 ajk < 1 (4.6) sup n2N X k n X j=0 ajk < 1 (4.7) X n
ank= k for all …xed k 2 N (4.8)
X n X k ank= (4.9) lim n!1 X k hXn j=0 ajk k i = 0 (4.10)
lim #!1 X k 1 # + 1 # X j=0 an+j;k k = 0 uniformly in n (4.11) lim #!1 X k h 1 # + 1 # X j=0 an+j;k k i = 0 uniformly in n (4.12) lim #!1 X k 1 # + 1 # X i=0 hn+iX j=0 ajk k i = 0 uniformly in n (4.13) lim #!1 X k 1 # + 1 # X i=0 2h n+i X j=0 ajk k i = 0 uniformly in n (4.14) Lemma 2. Let A = (ank) be an in…nite matrix of complex numbers. Then, the
followings hold:
(i) A = (ank) 2 (c : f) , (3.1), (4.3) and (4.4) hold (see [23])
(ii) A = (ank) 2 (`1: f ) , (3.1), (4.3) and (4.11) hold (see [24])
(iii) A = (ank) 2 (f : f) , (3.1), (4.3), (4.4) and (4.12) hold (see [24])
(iv) A = (ank) 2 (f : cs) , (4.7), (4.8), (4.9) and (4.10) hold (see [26])
(v) A = (ank) 2 (cs : f) , (3.5) and (4.3) hold (see [25])
(vi) A = (ank) 2 (cs : fs) , (4.5) and (4.6) hold (see [25])
(vii) A = (ank) 2 (bs : f) , (3.5), (3.6), (4.3) and (4.13) hold (see [27])
(viii) A = (ank) 2 (bs : fs) , (3.6), (4.5), (4.6) and (4.13) hold (see [27])
(ix) A = (ank) 2 (fs : f) , (3.6), (4.3), (4.12) and (4.13) hold (see [28])
(x) A = (ank) 2 (fs : fs) , (4.5), (4.6), (4.13) and (4.14) hold (see [28])
By combining Lemma 1, relations (4.1), (4.2), Theorem 8 and Lemma 2, the following results can be given.
Corollary 3. Let us replace the entries of the matrix A = (ank) by those of the
matrix Gr;s= (gr;s
nk) in (3.1)-(3.7) and (4.3)-(4.14), then the followings hold:
(i) A = (ank) 2 (f(Br;s) : c) if and only if fankgk2N 2 ff(Br;s)g for all
n 2 N and (3.1), (3.2), (3.3) and (3.7) hold.
(ii) A = (ank) 2 (f(Br;s) : `1) if and only if fankgk2N 2 ff(Br;s)g for all
n 2 N and (3.1) holds.
(iii) A = (ank) 2 (f(Br;s) : cs) if and only if fankgk2N 2 ff(Br;s)g for all
n 2 N and (4.7), (4.8), (4.9) and (4.10) hold.
(iv) A = (ank) 2 (f(Br;s) : bs) if and only if fankgk2N 2 ff(Br;s)g for all
n 2 N and (4.8) holds.
Corollary 4. Let us replace the entries of the matrix A = (ank) by those of the
matrix Hr;s= (hr;snk) in (3.1)-(3.7) and (4.3)-(4.14), then the followings hold: (i) A = (ank) 2 (c : f(Br;s)) , (3.1), (4.3) and (4.4) hold,
(iii) A = (ank) 2 (f : f(Br;s)) , (3.1), (4.3), (4.4) and (4.12) hold,
(iv) A = (ank) 2 (cs : f(Br;s)) , (3.5) and (4.3) hold,
(v) A = (ank) 2 (bs : f(Br;s)) , (3.5), (3.6), (4.3) and (4.13) hold,
(vi) A = (ank) 2 (fs : f(Br;s)) , (3.6), (4.3), (4.12) and (4.13) hold,
(vii) A = (ank) 2 (cs : fs(Br;s)) , (4.5) and (4.6) hold,
(viii) A = (ank) 2 (bs : fs(Br;s)) , (3.6), (4.5), (4.6) and (4.13) hold,
(ix) A = (ank) 2 (fs : fs(Br;s)) , (4.5), (4.6), (4.13) and (4.14) hold.
5. Conclusion
By taking into account the de…nition of the Binomial matrix Br;s = (br;snk), we deduce that Br;s = (br;snk) reduces in the case r + s = 1 to the Er = (ernk) which is called the method of Euler means of order r. So, our results obtained from the matrix domain of the Binomial matrix Br;s = (br;snk) are more general and more extensive than the results on the matrix domain of the Euler means of order r. Moreover, the Binomial matrix Br;s = (br;s
nk) is not a special case of the weighed
mean matrices. So, the paper …lls up a gap in the existent literature. References
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