J. Comp. & Math. Sci. Vol.2 (1), 123-128 (2011)
N , pn , α n ; δ
k
Summability of An Infinite Series
U. K. MISRA1, M. MISRA2 and P. SAMANTA3 1
Department of Mathematics, Berhampur University, Berhampur-760 007, Orissa, India. 2 Principal, Government Science College, Malkangiri, Orissa, India 3 Department of Mathematics, Gopalpur College, Gopalpur on sea, Orissa, India. ABSTRACT
N , pn , α n ; δ
In this paper we establish a theorem on
k
summability factors of an infinite series. Key words:
N , pn , α n ; δ k summability.
AMS Classification: 40G05
∑a
1. INTRODUCTION
∑ an be an infinite series with
Let
sequence of partial sums {s n } . Let {p n } be a sequence of non-negative numbers such that
Pn =
( P−i
n
∑ pv → ∞, as n → ∞,
v =0
n
is
Pn ∑ n =1 p n ∞
(1.2)
tn =
1 Pn
Tn =
(1.3)
n
∑ p s , (P v =0
v v
n
(
)
mean of the sequence {s n } generated by the
{p n }.
summable
k −1
t n − t n −1 < ∞ . k
1 Pn
n
∑p v =0
n −v
s v , (Pn ≠ 0)
defines the sequence {Tn }of the ( N, p n )
≠ 0)
defines the sequence {t n } of the N , p n
sequence of coefficients
be
Similarly, the sequence-to-sequence transformation
The sequence-to-sequence transformation (1.1)
to
N , p n k , k ≥ 1 if (Bor7)
.
= p−i = 0; i ≥1)
said
The series
mean of the sequence {s n } generated by the sequence of coefficents
∑a
n
is
said
to
{p n }. The series be
N , p n k , k ≥1 if
Journal of Computer and Mathematical Sciences Vol. 2, Issue 1, 28 February, 2011 Pages (1-169)
summable
U. K. Misra, et al., J. Comp. & Math. Sci. Vol.2 (1), 123-128 (2011)
124
Pn ∑ n =1 p n ∞
(1.4)
Let
{α n }
k −1
Tn − Tn−1
be any sequence of positive
numbers. The series
is said to be
n
N , p n , α n k , k ≥ 1 , if
summable ∞
∑α
(1.5)
∑a
n =1
k −1 n
Tn − Tn −1 < ∞ .
k + k −1
n =1
n
then the series
Qnk Qn −1
αn ∑ βn
k −1
∞
n =1
∈n
k
Tn
α k −1 q k −1 0 ν k v , Qv
= k
k
Pn qn pn Qn , k
< ∞
k −1
∈n
k
Tn
k
<
∞,
and
= N , pn
k
αn ∑ n =1 β n
k
k
is said to be
n
k
N , qn , α n k , k ≥1 .
Tn − Tn −1 < ∞ ,
∑a
k −1
Pn−1 k k ∆ ∈n Tn < ∞ , pn then the series ∑ a n ∈n is summable ∞
and
Further, if
∑α δ
n = v +1
αn ∑ n =1 β n
N , pn , 1 1 = N , pn . ∞
α nk −1 q nk
∞
k
P Clearly N , p n , n pn
(1.6)
∞
∑
< ∞.
k
N , p n , α n ; δ k , k ≥1, δ > 0 summable. Clearly for
δ = 0, N , p n , α n ; δ k = N , p n , α n k . We consider {α n }, {β n } and {q n } as sequences of positive numbers such that n
Q n = ∑ qv =→ ∞ on n → ∞ .
Recently, Misra, Misra and Jena42 have proved an analogue theorem. They proved: Theorem-B. Let t n be the n-th N , p n mean
1−
Tn = β n ∞
∑
n = v +1
v =0
On dealing with N , q n , α n , k ≥1
1 k
the
q n −v Qn
∑a
series
n
)
and
let
∆t n −1 . If
α nk −1
αn ∑ n =1 β n ∞
2. KNOWN THEOREM
of
(
k −1
=
qn Qn
α k −1 q v 0 v Qv
,
k
∈n
k
Tn
<
k
∞,
and
k
summability, Sulaiman52 proved the following theorem. Theorem-A. Let {t n } denote the N , q n -
(
mean of the series
Tn = β
1 1− k n
∆Tn −1 . If
∑a
n
)
. Let us write
αn ∑ n =1 β n ∞
k −1
k
k
Pn −1 ∆ ∈n Tn pn
then the series
∑a
n
∈n
k
<
∞,
is summable
N , q n , α n k , k ≥1 .
Journal of Computer and Mathematical Sciences Vol. 2, Issue 1, 28 February, 2011 Pages (1-169)
U. K. Misra, et al., J. Comp. & Math. Sci. Vol.2 (1), 123-128 (2011)
In this Theorem-B
paper,
we
generalize for N , q n , α n ; δ k , k ≥1, δ ≥ 0 , summability
methods.
Theorem:
of the series
n
Tn = β
1 1− −δ k n
∞
(3.1)
∑α
n = v +1
n
δ k +k −1
α (3.2) ∑ n n=1 βn ∞
(3.3)
∑a
n
and let
∆t n −1 . If
δk + k −1
∞
t n be the n-th
Let
α ∑ βn n =1 n
α vδk + k −1 q v = 0 Qv
k
,
k
P q k k ⋅ n ⋅ n ∈n Tn < ∞ , p Q n n qn ∈n Qn
δ k + k −1
(3.4)
α ∑ βn n =1 n
then the series
Tn
<
∞
k
k
Pn −1 ∆ ∈n Tn pn
∑a
n
∈n
k
<∞,
is summable
4. PROOF OF THE THEOREM Let τ n be the n-th ( N , q n ) -mean of
∑ an ∈n .
Then τ n − τ n −1 = n
∑ ( Qn qn − v v =1
qn − v − Qn − v qn ) Pv−−11 ∈v
n
qn − v − Qn − v qn ) Pv−−11 ∈v
}
}
n + ∑ Pr −1 a r Pn−−11 (Qn q 0 − Q0 q n ) ∈n , r =1
1 Qn Qn −1 − Qn − v qn ) av ∈v
=
1 n −1 P P −1−δ 1 ∑ v v −1 β vk Tv Qn Qn −1 v =1 pv
{( Qn −1 qn −ν
− Qn qn −ν −1 ) Pv−−11 ∈v
+ ( Qn qn − v −1 − Qn −v −1 qn ) pv
k
N , q n , α n ; δ k , k ≥1, δ > 0 .
the series
v =1
1 n −1 v ∑ ∑ Pr −1ar ⋅ ∆ QnQn −1 v =1 r =1
{( Q
k
and ∞
n
∑ ( Pv =1 av )
by Abel’s partial summation formula
q n−v Qn
δ k + k −1
n
1 Qn Qn −1
{(Q =
3. MAIN RESULT
(N , p ) -mean
=
125
Pv Pv −1
∈v + ( Qn qn − v −1 − Qn − v −1 qn ) Pv−1 ∆ ∈ν
}
1 Pn Pn −1 k −1−δ + βn Tn Pn−−11 q0 Qn −1 ∈n . pn
=
n −1 Pv 1 −1−δ 1 ∑ β vk Qn Qn −1 v =1 pv
( Qn −1 qn −v − Qn qn−ν −1 ) ∈v Tv + ( Qn qn − v −1 − Qn − v −1 qn )
1 −1−δ ∈v β vk
Tv
P + v −1 ( Qn qn − v −1 − Qn −v −1 qn ) × pv 1
× β vk
−1−δ
∆ ∈v Tv +
= Tn ,1 + Tn , 2
1 −1−δ Pn q 0 ∈n β nk Tn p n Qn +T n ,3+Tn , 4 + Tn ,5 + Tn ,6 + Tn , 7 ,
say. To prove the theorem, by Minkowski’s inequality it is sufficient to show that
Journal of Computer and Mathematical Sciences Vol. 2, Issue 1, 28 February, 2011 Pages (1-169)
U. K. Misra, et al., J. Comp. & Math. Sci. Vol.2 (1), 123-128 (2011)
126 ∞
∑α n =1
δk + k −1 n
Tn ,r
k
P = 0(1) ∑ v v =1 p v using (3.1) m
< ∞, r = 1,2,3,4,5,6,7 .
Now m +1
∑α n=2
δk + k −1
k
Tn ,1
n
=
m +1
∑ αn
n=2
1 Qn
m +1
∑α n=2
≤
n −1
∑ qn−v v =1
kδ + k −1 n
Tn , 2
k
=
k
m +1
β v1− k −δ k Tv
kδ + k −1 n
n=2
m +1
δk + k −1
n −1
1
⋅ Q nk−1
P
∑ α nkδ + k −1 ⋅ Qn −1 ∑ pvv 1
qv Qv
t v
k
k Tv ,
q n−v −1 ∑ v =1 n −1
1 −1−δ Pv q n − v −1 ∈v β vk Tv v =1 v
n −1
q n − v −1 ∈v
k
β v1− k − kδ Tv
k
×
k −1
, as above
k −1 k
m P = 0(1) ∑ v ∈v v =1 p v
m P = 0(1) ∑ v v =1 p v
using Holder’s inequality. k
m P = 0 (1) ∑ v ∈v v =1 pv
k
β v1− k − kδ Tv
k
m +1
qn − v Qn
∑ α nδ k + k −1
n = v +1
k
k
β v1− k − kδ Tv
αv βv
m +1
∑ αδ
k
δk + k
n = v +1
qv Qv
k + k −1
n
∈v
q n −v −1 Qn −1 k
Tv
k
= 0(1), as m → ∞ , using (3.2).
Next, m +1
∑α n=2
k −1 n
Tn,3
k
=
m +1
∑α n=2
≤
δk + k −1 n
m +1
∑α n=2
1 Qnk−1
m
k
v =1
α = 0 (1) ∑ v v =1 β v m
∑q v =1
n − v −1
∈v β
1 −1− k k v
k
Tv
1 n −1 k k 1 ∑ q n −v −1 ∈v β v1− k − kδ Tv Qn −1 v =1 Qn −1
kδ + k −1 n
= 0(1) ∑ ∈v
n −1
β v1−k −kδ Tv k −1
qv ∈v Qv
k
m +1
∑α δ
n = v +1
k
Tv
k + k −1 n
k
∑ p
k
v =1
1 × Qn
k
, using (3.2).
∑α
n=2
n −1 P k ∑ v qn − v ∈v v =1 pv
αv βv
Next,
1 Qn
δ k + k −1
k
= 0 (1) , as m → ∞
k
1 −1−δ Pv q n −v ∈v β vk Tv ∑ v =1 p v
≤
k + k −1
n
n=2
n −1
m +1
1 Qnk
∑α δ
q n −v −1 ∑ v =1 n −1
q n−v −1 Qn −1
k
= 0(1), as m → ∞ , using (3.3). Journal of Computer and Mathematical Sciences Vol. 2, Issue 1, 28 February, 2011 Pages (1-169)
k −1
U. K. Misra, et al., J. Comp. & Math. Sci. Vol.2 (1), 123-128 (2011)
127
Further, m +1
∑α n=2
≤
kδ + k −1 n
k
Tn , 4
m +1
δk + k −1
∑α
=
n
n=2
m +1
∑ α nkδ +k −1 Q k
n=2
n
qnk Qn−1
q nk Qnk Qnk−1
Qn −v −1 1k −1−δ β v q n −v ∈v Tv ∑ v =1 q n−v n −1
k n−1 Qn−v −1 ∑ qn−v ∈v v =1 qn−v
m
= 0(1) ∑ ∈v
k
v =1 m
= 0(1) ∑ ∈v
k
v =1
β v1− k −kδ
k 1 Tv Qn−1
n −1
∑ qn−v v =1
k
q n q n −v Qn −v −1 β Tv ∑ α n = v +1 Qn Qn −1 q n−v m +1 q k β v1− k − kδ Tv ∑ α nkδ + k −1 n − v n = v +1 Q n −1 1− k v
k
m +1
kδ + k −1
α k q = 0(1) ∑ v ∈v v v =1 β v Qv = 0(1), as m → ∞ , using (3.3). m
k
k
k −1
k
kδ + k −1 n
Tv
k
Next, m +1
∑α n=2
δk + k −1 n
≤
Tn,5 m +1
∑α n=2
k
=
m +1
∑α n=2
δk + k −1 n
δk + k −1 n
n −1
1 −1−δ Pv −1 k q ∆ ∈ β Tv ∑ n − v −1 v v v =1 p v
k 1 n −1 Pv −1 ∑ qn−v−1 ∆ ∈v Qn −1 v =1 p v k
P = 0(1) ∑ v −1 ∆ ∈v v =1 p v m
1 Qnk−1
k
β v1−k −kδ Tv
δk + k −1
k
β
k
m +1
∑α δ
n = v +1
1− k − kδ v
k + k −1
n
Tv
k
k
1 Qn −1
q n −v −1 ∑ v =1 n −1
q n −v −1 Qn − v
k
m α Pv −1 ∆ ∈v = 0(1) ∑ v p v =1 β v v = 0(1), as m → ∞ , using (3.4).
k
k
Tv ,
Further,
Journal of Computer and Mathematical Sciences Vol. 2, Issue 1, 28 February, 2011 Pages (1-169)
k −1
,
U. K. Misra, et al., J. Comp. & Math. Sci. Vol.2 (1), 123-128 (2011)
128 m +1
∑α n=2
δk + k −1
Tn , 6
n
k
=
m +1
∑α n=2
≤
m +1
δk + k −1 n
δk + k −1
∑α n n=2
q nk Q0k Qnk−1
1 −1−δ Pv −1 Qn − v −1 ( ∆ ∈v ) β vk q n −v Tv ∑ v =1 p v q n− v n −1
P q n −v v −1 ∑ v =1 pv n −1
q nk Qnk Qn −1
k
k
Qn− v −1 ∆ ∈v q n −v
1 × Qn −1 k
P = 0(1) ∑ v −1 ∆ ∈v v =1 p v m
β
k
1− k − kδ v
k −1
Tv
k
m +1
∑α
n = v +1
δk + k −1 n
qn Qn
k
Qn −v −1 q n−v
k
k
k
k β v1−k −kδ Tv ×
q n−v ∑ v =1 n −1
k −1
q n −v Qn −1
k
m α P k = 0(1) ∑ v v −1 ∆t v Tv v =1 β v pv = 0(1), as m → ∞ , using (3.4).
k
Finally, m
∑α n =1
k
δk + k −1 n
Tn , 7
=
m
∑α
n= f
δk + k −1 n
Pn pn
k
q0 Qn
δk + k −1
k
∈n k
k
β n1−k −kδ Tn k
α Pn q n = 0(1) ∑ n β n =1 n p n Qn = 0(1), as m → ∞ , using (3.2). m
This completes the proof of the theorem. 5. REFERENCES 1. Bor, H., On Two Summability Methods, Math. Proc. Cambridge. Philos., 97, pp.147-149 (1985). 2. Jena, K., On N , pn ,α n k summability
∈n
k
Tn
k
method, Ph.D. thesis, Berhampur Thesis, (2009). 3. Sulaiman, W.T., On a New Absolute Summability Method, Internat. J. Math and Math. Sci., Vol. 21, No.3, pp. 603606 (1998).
Journal of Computer and Mathematical Sciences Vol. 2, Issue 1, 28 February, 2011 Pages (1-169)