JOURNAL OF COMPUTER AND MATHEMATICAL SCIENCES An International Open Free Access, Peer Reviewed Research Journal www.compmath-journal.org
ISSN 0976-5727 (Print) ISSN 2319-8133 (Online) Abbr:J.Comp.&Math.Sci. 2014, Vol.5(2): Pg.155-161
A Note on Matrix Summability of an Infinite Series Chitaranjan Khadanga1, Kiran Mishra2 and Lituranjan Khadanga3 1
Professor, Department of Mathematics RCET, Bhilai, C.G., INDIA. 2 Assistant Professor, Department of Mathematics, G.D. Rungta College of Sc. & Tech. Dept. of Mathematics, Bhilai, C.G., INDIA. 3 Assistant Professor, Department of Mathematics, CEC Bhilai, C.G., INDIA. (Received on: March 1, 2014) ABSTRACT In the present Paper we establish an analogue theorem for
A, δ k , k ≥ 1 summability. Keywords: Matrix Transformation , Infinite Series, Summability Theory.
Let T = (a nk ) and σ = {s n } . Then
INTRODUCTION Let
∑u
n
be an infinite series with
sequence of partial sums
( )
Let T = a n , k
{sn }.
be an infinite matrix with
real and complex elements, then the transform {t n } of {s n } is given by n
tn =
∑a
n,k
sk .
Tσ exists for all bounded sequence, if Tσ exists for all sequences convergent to 0. A necessary and sufficient condition for Tσ to exists for all sequence of either class is that ∞
∑a
n ,k
converges for all m .
n =1
The Matrix T transforms all bounded sequences into sequences if and
k =0
Journal of Computer and Mathematical Sciences Vol. 5, Issue 2, 30 April, 2014 Pages (123 - 257)
156
Chitaranjan Khadanga, et al., J. Comp. & Math. Sci. Vol.5 (2), 155-161 (2014)
only if it transforms all sequences which converge to 0 into bounded sequences. If T = a n ,k , a necessary and sufficient
( )
condition for T to transform all sequences of either class into bounded sequences is that there exists a constant C such that ∞
∑a
≤ C , for all n.
n ,k
k =1
If for k > n , then the method is called triangular matrix method. If lim t n = s , then the sequence {s n } or the n →∞
( ) simply T-summable to s. A series ∑ u
series or
∑u
n
, is said to be summable a n ,k
n
is said to be absolutely summable by Tmethod or simply T -summable, if
∑
t n − t n −1 < ∞
∑a
n
be an infinite series with
sequence of partial sums {s n } such that n
An =
∑
a nv s v .
∑a
Then the series
∞
k
n
is said to be
< ∞.
n =1
triangular matrices A and Aˆ as follows: n
r =v
and
nr
Theorem-1: Let A be a lower triangular matrix with non-negative entries satisfying a n0 = 1, n = 0,1,... (i) (ii)
a n −1,v ≥ a nv for n ≥ v + 1
(iii)
n ann = O(1) If the sequence {s n }is bounded and the sequence {λn } is such that (iv)
∑ ∆λ
n
n =1 m
(v)
∑a
nn
λn
= O(1) , k
= O (1) .
n =1
Then the series
∑a
n
λn
is summable
KNOWN RESULT Theorem-2
k
Associate with A we define two lower
∑a
summability.
k ≥ 1, , if
∑ n k −1 An − An−1
a nv =
Euler-Totient function φ (k ) as the number of positive integers not exceeding k and relatively prime to k . In 2007, E.Savas and B.E.Rhodes proved the following theorem on A k -
A k , k ≥ 1.
v =0
summable A
For any positive integer k ≥1 we define
m
Let A = (a mn ) be a lower-triangular matrix and
aˆ nv = a n ,v − a n −1 , n = 1,2,3,... and v = 0,1,2,...
, n = 0,1,2,... and v = 0,1,2,...
Let A be a lower-triangular matrix with non-negative entries satisfying
a n , 0 = 1, n = 0,1,2,... a n −1,v ≥ a n ,v for n ≥ v + 1
nann = O(1) Journal of Computer and Mathematical Sciences Vol. 5, Issue 2, 30 April, 2014 Pages (123 - 257)
Chitaranjan Khadanga, et al., J. Comp. & Math. Sci. Vol.5 (2), 155-161 (2014)
We need the following lemmas for the proof of our theorem. Lemma-1
and for 1 n +1
tn =
( O( w
k
(v + 1)2k + 2
)
1 n +1
n
∑a
k
log k
n −1
k =1
∑∆
,
Lemma-2 For
(v + 1)2 k + 2
n > 1, λn = ∑ φ (k ) =
m +1
∑∆
v
aˆ n ,v = O a vv
k ≤n
1 n ∑ t v aˆ n,v = O k −1 v =1 (v + 1) a vv ∞ ∞ tv wv = O ( 1 ) ; = O(1) ∑ ∑ v =1 a vv v =1 a vv k
1 tn n
k
Then
k −1
1 = O 2k + 2 (n + 1)
∑a
n
3 2 n + O(n log n) π2
Lemma-3
n = v +1
n −1
aˆ nv = a n,n
v
v =0
1
a vv =
v
wn =
1
a vv = k
,
k =1
)
k
O tv
n
∑k a
1 ; n wn
k
m
∑ aˆ
n ,v
=1
n = v +1
Proof of Main Theorem: We have
1 = O 2k + 2 (n + 1)
λ n is summable A k , k ≥ 1 .
n n i Tn = ∑ a ni si = ∑ a ni ∑ λv a v i =0 i =0 v =0 n
n
v =0 n
i =v
In the present paper, we establish an analogue theorem for A, δ k , k ≥ 1
= ∑ λv a v ∑ a ni
summability. We prove:
= ∑ λv a v a nv = ∑ λv a v a nv
MAIN THEOREM
Then
Let A be a triangular matrix with non-negative entries satisfying for
Tn − Tn −1 = ∑ λr a v a nv − ∑ λv a v a n −1,v
n
v =0
v =1 n −1
n
tn =
1 n +1
n
∑k a , k
wn =
k =1
1 n +1
n
∑a
k
log k
∑n
δk
v =1
∆ v aˆ n ,v = O a vv
δk ∑ (ν + 1) v =1
Then
∑a
v =1 n
v =1 n
v =1 n
∞ tv δk wv = O(1) ; ∑ (ν + 1) = O(1) a vv avv v =1
≤
λ n is summable A , δ k , k ≥ 1 .
=
n
n
= ∑ (a nv − a n −1,v )λv a v = ∑ aˆ nv λv a v
n = v +1 ∞
v =1
n
= ∑ λv a v a nv − ∑ λv a v a n−1,v
k =1
and m +1
157
∑ aˆ
v =1
n ,v
(
)
a v v 2 + v log v (using Lemma -2)
v =1 n
n
v =1
v =1
∑ aˆ nv av v 2 + ∑ aˆ nv av v log v
Journal of Computer and Mathematical Sciences Vol. 5, Issue 2, 30 April, 2014 Pages (123 - 257)
158
Chitaranjan Khadanga, et al., J. Comp. & Math. Sci. Vol.5 (2), 155-161 (2014) n
n
∑ (v aˆ nv )(v av ) + ∑ (v aˆ nv )(av
=
v =1
n −1
log v )
=
n −1 v n = ∑ ∆ v (v aˆ nv )⋅ ∑ (r a r ) + (n aˆ nn ) ∑ vav v =1 r =1 v =1 n −1
v
∑ ∆v (v aˆ )⋅ ∑ a
+
nv
v =1
n ,v
v
v
nn
n −1 v =1
log r + (n aˆ nn ) ∑ (a v log v )
v =1
v =1
(By using Abel’s Lemma.) n −1
=
∑ {(v + 1) ⋅ ∆ (aˆ ) + (aˆ ) (−1) } (v + 1) t + (naˆ ) (n + 1) t v
n ,v
n ,v
v
nn
n
v =1 n −1
+ ∑ {(v + 1) ⋅ ∆ v (aˆ n ,v ) + (aˆ n ,v ) ( −1) } (v + 1) ⋅ wv + (naˆ n 0 ) ( n + 1) wn = +
v =1 n −1
n −1
v =1 n −1
v =1 n −1
2 ∑ (v + 1) ⋅ t v ⋅ ∆ v (aˆ n,v ) − ∑ (aˆ n,v ) (v + 1) t v + (n aˆ nn ) (n + 1) t n
∑ (v + 1)
2
⋅ wv ⋅ ∆ v (aˆ n ,v ) − ∑ (aˆ n ,v ) (v + 1) wv + (n aˆ nn ) ( n + 1) wn
v =1
v =1
= T1 − T2 + T3 + T4 − T5 + T6 . In order to establish the theorem it is sufficient to prove that ∞
∑n
δk + k −1
Tn, r
k
< ∞, r = 1,2,3,4,5,6 .
n =1
Now m +1
∑n
δk + k −1
m +1
Tn ,1
n =1
≤
=
k
=
∑n
δk + k −1
n =1
n −1
k
∑ (v + 1)
2
t v ⋅ ∆ v (aˆ n,v )
v =1 1 k
n−1 1 k k −1 n−1 2 δk + k −1 k k ˆ ˆ n ( v + 1 ) ⋅ t ⋅ ( ∆ a ) ( ∆ a ) ∑ ∑ ∑ v v n , v v n , v v =1 n =1 v =1 m +1
m +1
∑n
δk + k −1
n =1 m +1
n −1 2k ∑ (v + 1) t v v =1
≤ O (1)∑ n δk + k −1 n =1
n −1
n
+ ∑ ∆ v (v aˆ n ,v ) (v + 1) wv + (n aˆ nn )(n + 1) wn
n
r
∑ ∆ (vaˆ )(v + 1) ⋅ t + (n aˆ )(n + 1) t v =1
v =1
∑ (v + 1)
2k
k
∆ v aˆ n,v k
n −1 ∑ ∆ v aˆ nv v =1
t v ∆ v aˆ n ,v ⋅ a nn
k k −1
k −1 k k
k −1
k −1
v =1
Journal of Computer and Mathematical Sciences Vol. 5, Issue 2, 30 April, 2014 Pages (123 - 257)
Chitaranjan Khadanga, et al., J. Comp. & Math. Sci. Vol.5 (2), 155-161 (2014) m +1
= O (1)∑ n
δk
n =1
n −1
∑ (v + 1)
2k
tv
k
∆ v aˆ n ,v
δk
∆ v aˆ n ,v
v =1
m
2k
= O(1)∑ (v + 1) v =1 m
m +1
k
tv
∑n n = v +1
≤ O (1)∑ (v + 1) v =1 m
2k
2k
= O(1)∑ (v + 1) v =1
k
t v ⋅ a vv
⋅
1 = (v + 1) 2 k + 2
m
∑ v =1
1 = O(1), as m → ∞ . (v + 1) 2
Next m +1
∑n
δk + k +1
m +1
Tn , 2 =
n =1
n =1
m +1
=
∑n
δk + k −1
n =1
m +1
= ∑n
=
δk + k −1
n =1
=
∑ aˆ
n ,v
(v + 1) t v
v =1
n −1
k −1
1
k
∑ (v + 1)⋅ (t v aˆ n,v )k (t v a n,v ) k 1 k n −1 ∑ (v + 1) (t v ⋅ aˆ n ,v )k v =1
δk + k −1
∑n m +1
k
n −1
v =1
n =1
m +1
∑n
δk + k −1
n −1
∑ (v + 1) (t k
v
n −1
⋅ aˆ n ,v )
∑t
m +1
∑ n δk −1
k
v
v =1 n −1
k −1 k k
k −1
aˆ n ,v
v
⋅ aˆ n,v ) n
k
k −1
n −1
∑t
v
aˆ n,v
v =1
1 aˆ n,v ) ⋅ O k −1 n =1 v =1 ( r + 1) a vv m +1 n −1 1 ≤ ∑ n δk −1 ∑ (v + 1) t v aˆ n,v ⋅ avv n =1 v =1 m m +1 t = ∑ (v + 1) v ⋅ ∑ n δk −1 aˆ n,v avv n= v +1 v =1 m m +1 t 1 ≤ ∑ (v + 1) v ⋅ ⋅ aˆ n,v ∑ avv (v + 1)1−δk n=v +1 v =1 =
k k −1
v =1
∑ n ∑ (v + 1) (t n =1
n −1 k −1 ˆ ( t a ) ∑ v n ,v k v =1
n −1
v =1 δk −1
1 k
∑ (v + 1) (t k
v
Journal of Computer and Mathematical Sciences Vol. 5, Issue 2, 30 April, 2014 Pages (123 - 257)
159
160
Chitaranjan Khadanga, et al., J. Comp. & Math. Sci. Vol.5 (2), 155-161 (2014) m
=
tv
∑a v =1
(v + 1) δk ⋅ 1 (using Lemma-3)
vv
= O(1) as m → ∞ ) Next m +1
∑n
k δk + k −1
k
m +1
∑n
Tn ,3 =
n =1
δk + k −1
n aˆ nn (n + 1) t n
n =1 k
m +1
k
≤ ∑ n k −1 (n + 1) t n n =1 m +1
=
∑n
δk
.n .(n + 1) k
k
n =1 m +1
1 tn n
k
k
1 ≤ ∑ n (n + 1) tn n n =1 m +1 1 = ∑ (n + 1)δk (n + 1)2 n =1 m +1 1 = O(1) as m → ∞ = ∑ 2 −δk n =1 (n + 1) = O(1) as m → ∞ , since 2 − δk >1 . 2k
δk
Further m +1
∑ n δk + k −1 Tn,4
m +1
k
=
n =1
m +1
≤ = ≤
∑n
δk + k −1
∑ (v + 1)
n =1
v =1
m +1
n −1
2
∑ n δk +k −1
∑ (v + 1)2 k
n =1
v =1
m +1
n
∑n
δk + k −1
n =1
m +1
=
n −1
∑ (v + 1)2 wv
n =1
v =1
wv (∆ v (aˆ n,v )) wv
k
1 k
(∆ (aˆ )) v
n ,v
k −1 k
k
k −1
n −1
∆ v aˆ n ,v
∆ v (aˆ n ,v )
∑ ∆ v aˆ n,v v =1
∑ (v + 1)
2k
wv
k
∆ v aˆ n ,v aˆ n ,n
k −1
v =1
m +1
∑ n ∑ (v + 1) n =1
k
n −1
∑ n δk + k −1
δk
v =1
2k
wv
k
∆ v aˆ n ,v
m
=
2k ∑ (v + 1) v =1
wv
k
m +1
∑n
δk
∆ v aˆ n ,v
n = v +1
Journal of Computer and Mathematical Sciences Vol. 5, Issue 2, 30 April, 2014 Pages (123 - 257)
Chitaranjan Khadanga, et al., J. Comp. & Math. Sci. Vol.5 (2), 155-161 (2014) m
=
2k ∑ (v + 1) v =1 m
=
k
wv
=
a vv
v =1
≤ 2k + 2
=
∑n
δk + k −1
∑ aˆ (v + 1) w
Reihe
v
v =1
(proceeding in the lines of m +1
∑n
δk + k −1
Tn , 2
k
)
2.
n =1
and m +1
∑n
3. δk + k −1
Tn ,6
k
n =1 m +1
=
∑n n =1 m +1
≤
∑n n =1
k δ k + k −1
δk
n aˆ n , n w n
.n .(n + 1) k
.
1. Abel, N. H. Üntersuchungen über die
k
n −1
n ,v
n =1
= O(1) as m → ∞,
REFERENCES
n =1
m +1
This completes the proof of Theorem.
k
Tn ,5
1
(n + 1)2−δk
1 O 2 (n + 1 )
sin ce 2 − δk >1.
Finally δk + k −1
∑ n =1
1 =∑ = O(1) as m → ∞ 2 v =1 (v + 1) m +1
δk
n =1
m
∑n
∑ (n + 1) m +1
1
2k
∑ (v + 1) (v + 1)
m +1
161
2k
4. 1 wn n
(proceeding in the line of
k
m +1
∑n n =1
5. δ k + k −1
T n ,3
k
)
1 + mx +
m(m − 1) 2 x +L , 2
Journal fur reine and angewants mathematik (crelle) I, 311-339 (1826). Hardy, G.H. Divergent Series, Clarendow Press, Oxford, (1949). Misra, M., Misra, M., Rauto, K. Absolute Banach Summabilty of Fourier Series, International Journal of Mathematical Sciences, Vol. 1, No.1,39 – 45 June (2006). Paikaray, S.K. Ph.D. thesis submitted to Berhampur University, (2010) Petersen, M. Regular matrix transformations, McGraw-Hill Publishing Company Limited (1996).
Journal of Computer and Mathematical Sciences Vol. 5, Issue 2, 30 April, 2014 Pages (123 - 257)