Generalization of some absolute summability methods

PII. S016117120100309X

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GENERALIZATION OF SOME ABSOLUTE

SUMMABILITY METHODS

W. T. SULAIMAN

(Received 6 January 1999)

Abstract.Mazhar (1971) gave the characterization for the seriesannto be summable |N,pn|wheneveranis summable|C,α|k, α≥0, k≥1. Here we prove two theorems, the first concerns the sufficient conditions and the second the necessary conditions satisfied bynin order to haveannsummable|N,p¯ n|kwheneveranis summable|C,α|k, k≥1.

2000 Mathematics Subject Classification. Primary 40G.

Letanbe a given infinite series with partial sumssn.Letσnδandrnδdenote thenth Cesàro mean of orderδ (δ >−1)of the sequences{sn}and{nan}, respectively. The seriesanis said to be summable|C,δ|k, k≥1 if

∞

n=1

nk−1_σδ

n−σn−δ 1k <∞, (1)

or equivalently

∞

n=1 n−1_rδ

nk<∞. (2)

LetPnbe a sequence of positive real constants such that

Pn=p0+p1+···+pn, Pn → ∞asn → ∞,P−1=p−1=0. (3)

A seriesanis said to be summable|N,p¯ n|k, k≥1, (see [1]), if

∞

n=1

_P n

pn k−1

tn−tn−1k<∞, (4)

where

tn= ₁

Pn ∞

ν=0

Forpn=1/(n+1),the summability|N,p¯ n|kreduces to the well-known summability |R,logn,1|k.In general, the two summabilities|N,p¯ n|kand|C,α|kare independent of each other, but forpn=1 andα=1 they are equivalent. For any realαand integer

n≥0,we define

∆α n=

∞

ν=nA −α−1

ν−n ν, (6)

whenever the series is convergent.

TheoremA(see [2]). The necessary and sufficient conditions for the seriesann to be summable|N,p¯ n|wheneveranis summable|C,α|k,α≥0,k≥1are

(i) nα+1−(1/k₎

∆α_(n/n)∈k

,1/k+1/k_=1; (ii) n−1/k

n∈k,0≤α≤1; (iii) nα−(1/k₎

(pn/Pn)n∈k,α >1;

where (a) pn = O(pn+1), (b) (n+1)pn = O(Pn), and (c) Pn =O(nαpn) (α >1). Theorem A includes as special cases k=1and pn=1,the result of Mohapatra[4] and Mehdi[3], respectively.

We need the following lemmas. Lemma1(see [5]). Ifσ > δ >0,then

∞

n=ν+1

(n−ν)δ−1

nσ =O

νδ−σ_{. (7)}

Lemma2. Ifσ > δ≥0, then

∞

n=ν+1 Aδ−1

n−ν

Aσ n =O

νδ−σ_{. (8)}

Proof. Ifδ >0,then byLemma 1,

∞

n=ν+1 Aδ−1

n−ν

Aσ

n =O(1) ∞

n=ν+1

(n−ν)δ−1

nσ =O

νδ−σ_{. (9)}

Ifδ=0,then byLemma 1,

∞

n=ν+1 A−1

n−ν

Aσ

n =O(1) 1 Aσ

n ∞

n=ν+1

_A−1

Lemma3(see [6]). Letk 1,Pn=p0+p1+ ··· +pn→ ∞asn→ ∞, pn>0. Then for allν≥1,

1 kPk

ν−1

∞

n=ν

pn

PnPn−k 1

1

Pk ν−1

. (11)

We prove the following theorems.

Theorem4. Let{n}be monotonic nonincreasing sequence of constants. Suppose that{Pnn/n}is nonincreasing, and

npn=OPn. (12)

Then the following are sufficient conditions forannto be summable|N,p¯ n|k when-everanis summable|C,α|k,0< α≤αk≤1:

(i) n=O(npn/Pn), (ii) ∆n=O(pn/Pn).

Proof ofTheorem4. Define

Tn=

pn/Pn1/k

Pn−1

n

ν=1

Pν−1aνν, tnα=_A1α n

n

ν=1 Aα−1

n−νναν,

Tn=

pn/Pn1/k

Pn−1

n

ν=1 Pν−1ν

ν

n

r=1 A−α−1

ν−r Aαrtrα

=

pn/Pn1/k

Pn−1

n

r=1 Aα

rtrα n

ν=r

Pν−1ν

ν A−α−ν−r1

=

pn/Pn1/k

Pn−1

n

r=1 Aα

rtrα n−r

ν=0

Pν−r−1ν+r

ν+r A−α−ν 1

=

pn/Pn1/k

Pn−1

n

r=1 Aα

rtrα _n−r−1

ν=0

 ν

µ=0 A−α−1

µ 

_∆νPν+r−1ν+r

ν+r

+  n−r

µ=0 A−α−1

µ  Pn−1n

n

=

pn/Pn1/k

Pn−1

n

r=1 Aα

rtrα _n−r−1

ν=0 A−α

ν

Pν+r−1ν+r

(ν+r )(ν+r+1)−

pν+rν+r

ν+r+1

+Pν+r_ν+∆_rν₊ν+r₁

+A−α

n−rPn−_n1n

,

=Tn,1+Tn,2+Tn,3+Tn,4. (13)

In order to prove sufficiency, by Minkowski’s inequality, it is sufficient to show that

m

r=1

m

n=1

_Tn,1k₌ m n=1

pn

PnPn−k 1

n r=1 Aα

rtrα n−r−1

ν=0 A−α

ν _(ν+Pν+r−_{r )(ν}1₊ν+r_r+1) k ≤ m n=1 pn

PnPn−k 1

n r=1 Aα

rtrαPr−_r1r n−r−1

ν=0 A−α

ν

ν+r+1 k

as Pn_nn being nonincreasing

= m

n=1 pn

PnPn−k 1

n r=1 Aα

rtrαPr−_r1r n

ν=r

(ν−r )−α

ν

k

=O(1)

m

n=1 pn

PnPn−k 1

n r=1 Aα

rtαrPr−_r1rr−α k

≤O(1)m

n=1 pn

PnPn−k 1

n r=1

prtαr_{r p}Pr rr

k

≤O(1)m

n=1 pn

PnPn−1

n

r=1

prtαrk _P

r

r pr k

_rkn r=1

pr

Pn−1

k−1

=O(1)

m

r=1

prtrαk _P

r

r pr k

_rkm n=r

pn

PnPn−1

=O(1)

m r=1 pr Pr _tα

rk _P

r

r pr k

rk

=O(1)

m

r=1 1 rtrαk.

(15)

m

n=1

_Tn,2k₌ m n=1

pn

PnPn−k 1

n r=1 Aα

rtrα n−r−1

ν=0 A−α

ν −_νp₊ν+r_r+ν+r₁ k ≤ m n=1 pn

PnPn−k 1

n r=1 Aα

rtrαprr n−r−1

ν=0 A−α

ν

ν+r+1 k

≤ m n=1

pn

PnPn−k 1

n r=1 tα

rprr k

≤O(1)m

n=1 pn

PnPn−1

n

r=1

_tα

rkprrk _n

r=1 pr

Pn−1

k−1

=O(1)

n

r=1

_tα

rkprrk m

n=r

pn

PnPn−1

=O(1)

m r=1 pr Pr _tα

rkrk

=O(1)

m

r=1 1

rtαrkrk

=O(1)

m

r=1 1

rtαrk asr=O(1)being nonincreasing.

m

n=1

_Tn,3k₌ m n=1

pn

PnPn−k 1

n r=1 Aα

rtrα n−r−1

ν=0 A−α

ν Pν+r_ν+∆_rν₊ν+r₁ k = m n=1 pn

PnPn−k 1

n r=1 Aα

rtrαPr∆r n−r−1

ν=0 A−α

ν

ν+r+1 k

≤O(1)

m

n=1 pn

PnPn−k 1

n r=1 tα

r pr∆r k

=O(1)

m

n=1 pn

PnPn−1

n

r=1

prtαrk _P

r

pr k

_∆rkn r=1

pr

Pn−1

k−1

=O(1)

m

r=1

prtrαk _P

r

pr k

_∆rkm n=r

pn

PnPn−1

=O(1)

m r=1 pr Pr _tα

rk _P

r

pr k

_∆rk

=O(1)

m

r=1 1 rtrαk.

(17)

m

n=1

_Tn,4k₌ m n=1

pn

PnPn−k 1

n r=1 Aα

rtrαA−αn−rPn−_n1n k = m n=1 pn|n|k

Pnnk n r=1 Aα r

k_tα rk

A−α n−r k _n r=1 1 k−1 = m n=1 pn|n|k

Pnn n r=1 Aα r

k_tα rk

A−α n−r k = m r=1 Aα rktαrk

m

n=r

pn|n|k

nPn

A−α n−rk

≤O(1)

m

r=1

Aα

rktrαkrk m n=r A−αk n−r n2

=O(1)m

r=1

Aα

rktrαkrkr−1−αk

=O(1)

m

r=1 1 rtrα

k_.

(18)

Theorem5. The necessary conditions forannto be summable|N,p¯ n|kwhenever

anis summable|C,α|k, k >1, are (i) n=O{(Pn/pn)1/kν1−α−(1/k)}, (ii) ∆n=O(ν1−α−(1/k)).

Proof ofTheorem5. Fork 1,we define

A=ai:

aiis summable|C,α|k

, B=ai:

aiiis summableN,p¯ nk

These are BK-spaces, if normed by

a1=

_∞

n=1 1 ntn

k1/k_{, a}

2=

_∞

n=1

_Tnk1/k_{, (20)}

respectively. Since(1/n)|tn|k_{<∞ ⇒|Tn|}k_{<∞by the hypothesis ofTheorem 5,} then

a1<∞ ⇒ a2<∞. (21) Consider the inclusion mapi:A→Bdefined byi(x)=x,iis continuous which follows asAandBare BK-spaces. Therefore, there exists a constantMsuch that

a2≤Ma1. (22) Applying the values oftn andTn, stated inTheorem 5, toa=eν−eν+1(whereeν is theνth coordinate vector), we have

tn=                 

0, ifn < ν,

ν Aα

ν, ifn=ν, ∆νAα−n−ν1

Aα

n , ifn > ν,

(23)

Tn=                   

0, ifn < ν,

_p ν

Pν

1/k

, ifn=ν,

pn/Pn1/k

Pn−1 ∆

Pν−1ν, ifn > ν.

(24)

Equalities (20) give

a1=

1 ν

_ν

Aα ν

k +

∞

n=ν+1 1 ν

∆νAα−n−ν1ν

Aα n

k1/k,

a2=

pν

Pν

_νk₊ ∞ n=ν+1

pn

PnPn−k 1

_∆

pν−1νk 1/k

.

(25)

By inequality (22), pν

Pν

_νk₊ ∞ n=ν+1

pn

PnPn−k 1

_∆

Pν−1νk≤Mk

1 ν

_ν

Aα ν

k +

∞

n=ν+1 1 ν

∆ννAα−n−ν1

Aα n

k.

(26) Since∆νAα−1

n−ν

= −Aα−1

n−ν+ν∆νAα−n−ν1, then by Minkowski’s inequality

Mk

1 ν

_ν

Aα ν

k +

∞

n=ν+1 1 ν

∆ννAα−n−ν1

Aα n

k=Ovk−αk−1_+O

1 ν

∞

n=ν+1

(n−ν)αk−k

Aαk n

+O

νk

ν

∞

n=ν+1

(n−ν)αk−2k

nαk

=Oνk−αk−1_+Oν−k_+Oν−k_{,byLemma 1} =Oνk−αk−1_.

As each part of the left-hand side of (26) isO(νk−αk−1_{), we have}

ν=O _P

v

pv

1/k

ν1−α−(1/k)_,

_∆

Pν−1νk ∞

n=ν+1 pn

PnPn−k 1

=Oνk−αk−1_, (28)

which, byLemma 3, implies _∆

Pν−1νk=OPν−k 1νk−αk−1. (29)

Now, as∆(Pν−1ν= −pνν+Pν∆ν,

∆ν=p_Pν νν+

1 Pν∆

Pν−1ν

=p_Pν νO

_P ν

pν

1/k

ν1−α−(1/k)

+_P1 νO

Pν−1ν1−α−(1/k)

=O

_p ν

Pν

1−(1/k)

ν1−α−(1/k)

+_P1 νO

Pν−1ν1−α−(1/k)

=Ov1−α−(1/k)_.

(30)

This completes the proof of the theorem.

References

[1] H. Bor,On two summability methods, Math. Proc. Cambridge Philos. Soc.97(1985), no. 1, 147–149.MR 86d:40004. Zbl 554.40008.

[2] S. M. Mazhar,On the absolute summability factors of infinite series, Tôhoku Math. J. (2)23 (1971), 433–451.MR 46#7758. Zbl 227.40005.

[3] M. R. Mehdi,Summability factors for generalized absolute summability. I, Proc. London Math. Soc. (3)10(1960), 180–200.MR 22#9760. Zbl 093.26203.

[4] R. N. Mohapatra,On absolute Riesz summability factors, J. Indian Math. Soc. (N.S.)32(1968), 113–129.MR 43#7811. Zbl 197.34404.

[5] W. T. Sulaiman,On absolute Cesàro summability of infinite series, Pure Appl. Math. Sci.31 (1990), no. 1-2, 25–30.MR 91j:40005. Zbl 713.40002.

[6] ,A study on absolute weighted mean summability method, Math. Sci. Res. Hot-Line 1(1997), no. 11, 25–29.MR 98m:40004.