J. Comp. & Math. Sci. Vol.3 (3), 320-328 (2012)
A Note on
α
N , pn ; δ
– Summability of A Series
k
S.K. PAIKRAY1, R. K. JATI2, U. K. MISRA3 and N. C. SAHOO4 1
Dept. of Mathematics, Ravenshaw University, Cuttack, Odisha, INDIA. 2 Dept. of Mathematics, DRIEMS, Cuttack. Odisha, INDIA 3 Dept. of Mathematics, Berhampur University, Odisha, INDIA. 4 S. B Women’s College (Auto), Cuttack, Odisha, INDIA (Received on: May 22, 2012) ABSTRACT In this paper a theorem on
α
N , pn ; δ – summability has been k
proved. Keywords: Summability, Bounded, partial sum. AMS Classification No.: 40 G05. α
Where An =
1. INTRODUCTION
{sn } {p n }
Let ∑ an be an infinite series and be the sequence of its partial sums.
be a sequence of non-negative numbers with
Let
α
(1.1)
γ =0
Let us define α
pn =
n
∑ γ =0
n +α
α ≥1
α
n
α
γ =0
α
α
p− i = p− i = 0 , i≥ 1 Let
Tn
α
1 = α Pn
α −1
pγ
of the sequence (1.2)
(1.4)
n
pγ ∑ γ =0
α
sγ
{ }is (N , p ), mean
Then the sequence Tn
An −γ
(1.3)
Pn = ∑ pγ , with
n
Pn = ∑ pγ , p0 ≠ 0
( ),
α
{sn }
α
n
generated by the
{ }. α
sequence of coefficients pn
Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)
321
S. K. Paikray, et al., J. Comp. & Math. Sci. Vol.3 (3), 320-328 (2012)
The series α
N , pn
k
n
is said to be summable
α
Pn n =1 n
α k n −1
α
Tn − T
Where Xn =
<∞
∑a
n
is said to be
summable N , pn ; δ , k ≥ 1, δ ≥ 0
n =1
k
If
Pn . npn
α k
Tn − Tn−1
n
n
n
an λn X n is summable N , pn k .
<∞
Let {sn} be a bounded
Theorem B:
α
α
If α = 0 , δ= 0, k = 1, then N , pn , δ
k
summability is same as
sequence and the sequence {λn } and { pn } satisfy the following conditions : α
n
2. KNOWN THEOREMS Dealing with N , pn
(ii) k
summability (iii)
Let k ≥ 1 and let the sequence {pn } and {λn } be such that
1 n
(i) ∆ X n = O
n =1
k −1
pn −v p − nα−v−1 α Pn Pn−1
∑ nδ
| λn |k + | λn +1 |k <∞ n
(2.1)
∞
(iv)
k + 2k
∑ nδ
k +2k
n =1
= O(1)
α
( pn ) k | λn |k < ∞ and α
( pn )
k
| ∆λ n | k < ∞
Then the series ,
∑ λ Pα a n n
(2.2)
k k −1
α
n =1
Theorem – A
n
α
∞
Bor proved the following theorem :
∞
∑ γ =0
1
∑X
α
(i) Pn = O(n, pn )
N , pn – summability.
(ii)
(2.3)
Subsequently, Misra, sahoo and Paikray2 prove the following theorem
δk + k −1 α
∞
∑a λ X
α
Pnα α ∑ n =1 pn
)
+ 1 | ∆λn |< ∞
If {sn} is bounded, then the series
Again, the series
∞
k n
n =1
k −1
∑ p α
∑ (X ∞
(iii)
,k ≥1
∞
If
∑a
α
n
is | N , pn , δ | k summability
where k ≥ 1 and α > – 1
Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)
322
S. K. Paikray, et al., J. Comp. & Math. Sci. Vol.3 (3), 320-328 (2012)
In this paper we have extended α
N , pn ; δ
theorem –B for of the series
∑a λ X n
n
– summability
(3.3)
| ∆λ n | <∞ n n =1
(3.4)
∞
k
∑
n
n−1 ∑ ∆X γ γ =1
Theorem: Let k ≥ 1, α ≥1, δ k < 1 and
Xn =
| λn | k <∞ ∑ n n =1 ∞
3. MAIN RESULT :
α
Pn α npn
k −1
(3.5)
1 = O γ +1
(3.6)
1 | X n |k = O n
(3.7)
| ∆X γ | | Xγ | γ k
If the sequence {sn} is bounded and α
α
the sequences { pn } and {λn } are such that α
α
Pn−1 = O (npn ) α
Pnα α p n
k +1
δk −1
(3.1)
Then the series,
(3.2)
∑a λ X
∞
α
| X γ |k γ δk ( pγ + Pγ ) = O (1)
1 = O k γ
n =1
n
n
α
is summable N , pn , δ .
n
k
4. PROF OF THE THEOREM α
∞
α
Tn be the {N , pn } mean of the series n
∑a λ X n =1
n
Tn =
1 α Pn
pγ ∑ γ
=
1 α Pn
υ α p a z λz X z ( x0 = 0) ∑ ∑ γ γ =0 z =0
α
1 α Pn 1 = α Pn
=
=0
α
n
n
sγ
n
1 n α α α p ( a λ X ) + p a λ X + ... + p a z λz X z ∑ ∑ 1 z z z n 0 0 0 0 z =0 z =0
[p
α 0
α
(a0λ0 X 0 ) + p1 (a0λ0 X 0 + a1λ1 X 1 )
+ ... + Pn (a0 λ0 X 0 + a1λ1 X 1 + ... + an λn X n )] α
Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)
323
S. K. Paikray, et al., J. Comp. & Math. Sci. Vol.3 (3), 320-328 (2012)
[
1 α α α ( p0 + p1 ..... + pn )a0λ0 X 0 α Pn
=
α
α
+ ( p1 + ..... + pn )a1λ1 X 1 + α
+ pn .an λn X n ]
[
1 α α α Pn (a0λ0 X 0 ) + ( Pn − P0 ) a1λ1 X 1 α Pn
=
α
α
+… + ( Pn − Pn−1) an λn X n ]
1 n α α α ( Pn − Pγ −1 ) (aγ λγ X γ ) , where P−1 = 0 α ∑ Pn γ =0
= For n ≥ 1, α
1 n α α ( Pn − Pγ −1 ) aγ λγ X γ α ∑ Pn γ =0 α − pγ −1 ) aγ λγ X γ n 1 1 α α = α ∑ ( Pn − Pγ −1 ) aγ λγ X γ − α Pn γ =1 Pn−1 α
Tn − Tn−1 = −
1 α
pn−1
n−1 α ∑ ( pn−1 γ =0
=
−
1 α Pn
n
α
aγ λγ X γ −
P ∑ γ =1
n
1
n
α
∑ pn−1 aγ λγ X γ +
α
pn−1
γ =1
α
p = α nα Pn Pn−1
α
p = α nα Pn Pn−1
n
α Pγ ∑ γ =1
α
Pγ ∑ γ =1
1 α
Pn−1
−1
n
=1
n −1
α
− Pγ −1 ) aγ λγ X γ
aγ λγ X γ α
Pγ ∑ γ =1
α
(P ∑ γ
−1
aγ λγ X γ
n α ∑ Pr −1 aγ λγ X γ γ =1
1 1 + α α Pn−1 Pn
= −
=
1 α pn
n
n
−1
aγ λγ X γ
α n−1 pn α α ∑ sγ ∆ ( Pγ −1 λγ X γ ) + sn ( Pn−1 λn X n ) by Abel’s lemma α α Pn Pn−1 γ =1 n−1 α α ∑ − sγ {λγ X γ pγ + pγ X γ (∆λγ ) γ =1
Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)
S. K. Paikray, et al., J. Comp. & Math. Sci. Vol.3 (3), 320-328 (2012) α
]
α
+ Pγ λγ +1 ( ∆X γ )} + sn ( Pn−1 λn X n )] α
n −1
pn α α Pn Pn−1
= −
α
α
∑ sγ pγ λγ X γ + γ =1
α
pn α α Pn Pn−1
n −1
Pα X γ (∆λγ ) sγ ∑ γ =1
α
n −1
pn p α α Pγ λγ +1 (∆X γ ) sγ + α n α sn λn X n Pn−1 ∑ α Pn Pn−1 γ =1 Pn Pn−1 = Tn ,1 + Tn , 2 + Tn ,3 + Tn , 4 +
α
The theorem will be proved if we can prove that
Pnα α ∑ n =1 pn
∞
(A)
δk + k −1
Pnα α ∑ n =1 p n ∞
k
Tn ,r < ∞, r = 1, 2, 3, 4
Pnα = ∑ α n =1 pn ∞
δk + k −1
Tn,1 δk + k −1
k
α
− pn α α Pn Pn−1
k
n −1
α γ
p ∑ γ =1
λγ X γ sγ
Let us consider δk + k −1
pnα α ∑ n = 2 pn
m +1
pα = ∑ nα n = 2 pn
pα = ∑ nα n = 2 pn
m+1
m +1
δk −1
δk −1
pα = 0 (1) ∑ nα n=2 pn m
= 0(1)
sγ ∑ γ =1
k
k
k
n −1
α γ
p ∑ γ =1
λγ X γ sγ
k −1
k
k −1 1 n −1 1 1 α k α k s λ X ( p ) ( p ) ∑γ γ γ γ γ α α pn−1 pn +1 γ =1 k −1 1 n−1 k k k α 1 n−1 α ∑ sγ λγ X γ pγ α ∑ pγ α p pn−1 γ =1 n−1 γ =1
m +1
α
− pn α α pn pn−1
δk −1
λγ X γ
1 α
pn−1 k
α γ
p
h −1
k
sγ ∑ γ =1
λγ k X γ k pγα
pnα α ∑ h =γ +1 p n m +1
δk −1
1 α
pn−1
Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)
324
325
S. K. Paikray, et al., J. Comp. & Math. Sci. Vol.3 (3), 320-328 (2012) m
k
sγ ∑ γ
=
k
λγ X γ
=1
m
≤
k
sγ ∑ γ
k
λγ X γ
=1
m
≤
k
sγ ∑ γ
k
λγ X γ
=1
m
≤
sγ ∑ γ
k
k
λγ X γ
=1
m
≤
sγ ∑ γ =1
≤
∑ γ m
≤
∑ γ
λγ X γ
k
k
k
p
α γ
p
pγ
pγ pγ
α
(p ∑ (p (p ∑ (p (p ∑ (p n =γ +1
) ) ) ) ) )
α 1−δk
m +1
n
α 1−δk
n
m +1
2−δk
α
n −1 n
n =γ +1
2 −δk
α
n −1
m +1
1 ∑ n =γ +1 n
α
(
α
O γ δk −1
(1 – δk>0)
2−δk
α
m +1
pn −1
α 1−δk
n
n =γ +1
1 α
2 −δk
(Using 3.1)
)
k
α
X γ Pγ γ δk
λγ k
=1
k
k
γ
=1
k
α γ
k
λγ
m
k
γ
O(1)
using 3.2
< ∞ as m → ∞ using 3.3
Pnα α ∑ n =1 pn ∞
(B)
δk + k −1
Tn, 2
pα = ∑ nα n=1 pn
∞
1
Now,
α
Pn−1 ≤
1 α
Pn−1
n −1
δk + k −1
α
Pγ ∑ γ =1
α
Pn−1
k
α
pn α α Pn Pn−1
n −1
k
α γ
P ∑ γ =1
X γ (∆X γ ) sγ
∆λγ
n −1
∆λγ ∑ γ =1`
= O(1) Let us consider Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)
326
S. K. Paikray, et al., J. Comp. & Math. Sci. Vol.3 (3), 320-328 (2012)
Pnα α ∑ n = 2 pn
m+1
Pα ≤ ∑ nα n = 2 pn
m +1
α
δk −1
pα ≤ ∑ nα n =2 p n
Pnα ≤ ∑ α n = 2 pn
m +1
δk −1
k
n −1
α γ
P ∑ γ =1
X P (P ) ∑ k
γ
k −1
α
(∆λγ )( sγ )
(
n−1 α ∑ X γ sγ Pγ (∆λγ γ =1
α
k
α
Pn−1
α γ
k
k
(
k
pα ≤ O(1)∑ | X γ | | Pγ | | (∆λγ ) | ∑ nα γ =1 n=γ +1 pn
≤
)
α γ
m+1
α
k
γ
n−1 | X | | s | P ( ∆ λ ) ∑ γ γ γ ∑ | ∆λγ γ =1 γ =1 n−1
1
m
) (P 1 k
1 | X γ | | sγ | ( P (∆λγ )) α ∑ γ =1 Pn−1 n−1
1 pn−1
δk −1
γ
γ =1
1 1 α α Pn−1 Pn−1
δk −1
X γ (∆X γ ) sγ k
n −1
1
n−1
m +1
α
pn α α Pn Pn−1
α
P = ∑ nα n=2 pn m +1
δk + k −1
m
α | X γ | | Pγ | | ∆λγ | O(γ δ ∑ γ k
k −1
δk −1
α
(∆λγ )
)
k −1 k
k
P (∆λγ ) ∑ γ =1 n−1
|
k −1
α γ
k −1
1 α n−1
P
)
=1
m
=
∑ γ
| ∆λγ |
α
| X γ | k | Pγ | γ δk
γ | ∆λγ | =∑ O(1) γ γ =1 =1
m
using 3.2
< ∞ as m → ∞
Pn α α ∑ n=1 pn ∞
(C)
Pn α = ∑ α n=1 pn ∞
using 3.4 δk + k −1
Tn ,3
δk + k −1
k
α
pn α α Pn Pn−1
n −1
k
α γ
P ∑ γ =1
λγ +1 (∆X γ ) sγ
Let us consider,
Pnα α ∑ n = 2 pn m +1
δk + k −1
α
pn α α Pn Pn−1
n−1
k
α γ
P ∑ γ =1
λγ +1 (∆X γ ) sγ
Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)
327
S. K. Paikray, et al., J. Comp. & Math. Sci. Vol.3 (3), 320-328 (2012) δk −1
Pnα = ∑ α n = 2 pn
Pα = ∑ nα n = 2 pn
Pnα = ∑ α n = 2 pn
Pα = ∑ n α n=2 pn
δk −1
m +1
m +1
m +1
m+1
δk −1
δk −1
1 α P n−1
k
n−1
λγ ∑ γ =1
s ( Pγ (∆X γ ) ( Pγ (∆X γ ))
+1 γ
n−1
1
| λγ ∑ γ =1
1 α |k Pγ (∆X γ )O k γ
+1
α m+1 Pn ∑ α n=γ +1 pn m 1 α ≤ ∑ | λγ +1 |k Pγ | ∆X γ | O k O γ δk −1 γ =1 γ m 1 α = ∑ | λγ +1 |k k +1 Pγ | ∆X γ | γ δk
1 = ∑ | λγ +1 | P | ∆X γ | O k γ =1 γ α γ
(
m
| λγ ∑ γ =1
|k
+1
| λγ ∑ γ =1
1 | ∆X γ | γ k +1 | X γ |k
1 |k O γ +1
m
=
k −1
α
k −1
using 3.5 1 α
Pn−1
)
γ
γ =1
=
δk −1
k −1 k k
Pγ ∆X γ ∑ γ =1 n−1
n−1 1 n−1 α k ∑ ∆X γ | λ | ( P ∆ X ) ∑ γ γ γ + 1 α Pn−1 γ =1 γ =1
P
k
α
1 1 n−1 ∑ | λγ +1 |k ( Pγ α ∆X γ ) α α P Pn−1 γ =1 n−1
α n −1
m
1 k
α
+1
< ∞ as m → ∞
Pα = ∑ nα n=1 pn
∞
using 3.6
using 3.4
δk + k −1
Pα (D) ∑ n α n =1 pn ∞
using 3.2
Tn, 4 δk + k −1
k
k
α
pn s n λn X n α Pn
Let us consider
Pnα α ∑ n =1 pn m
δk + k −1
α
k
pn s n λn X n α Pn
Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)
328
S. K. Paikray, et al., J. Comp. & Math. Sci. Vol.3 (3), 320-328 (2012)
Pnα = ∑ α n=1 pn m | λ |k = ∑ n n n =1 m
δk −1
| λn | k | X n |k using 3.7
< ∞ as m → ∞. This completed the proof of the theorem. REFERENCES 1. H. Bor : On the local property of summability of Factored N , pn k
Fourier Series, Journal of Mathematical Analysis and Applications. 163, 220-226
(1992). 2. U. K. Mishra, N.C. Sahoo, S. K. Paikray:
A
note
on
α
N , pn ; δ
k
summability, Journal of the Indian Academy of Mathematics Vol. 30 No.2 p.p. 481-487 (2008).
Journal of Computer and Mathematical Sciences Vol. 3, Issue 3, 30 June, 2012 Pages (248-421)