Inclusion And Equivalence Relations Between Absolute Nörlund And Absolute

abstract: In this paper, a set of conditions under which the absolute Nörlund summability method include in the absolute weighted mean method have been established. Three non-trivial examples to show that this inclusion holds have been given, and other three examples to show that even if both (N, r) and (N, q) are regular, the inclusion fails to holds have been constructed. The paper give two non-trivial examples to show that the equivalence of these two methods may holds. Finally, we give two examples to show that inclusion may holds in only one way without the other.


Introduction
Let A be a sequence-to sequence transformation The sequence {S n } is said to be summable (A) to s if t n −→ s as n −→ ∞, and if in addition {t n }, is of bounded variation, then {S n } is said to be absolutely summable (A) or summable |A|.

Amjed Zraiqat
We define the sequence of constants {C n } formally by means of the identity and will write c(z) for then we shall write r n ∈ µ.Let (N, r) denote the Nörlund method in which the sequence {S n } is transformed into the sequence {t r n }, where The special case in which r n = 1 (n ≥ 0), then (N, r) reduces to a simple arithmetic mean of (C, 1).
Each sequence {q n } for which Q n = q 0 + q 1 + • • • , q n = 0 (all n ≥ 0) for each n defines the weighted mean method (N , q) of the sequence {S n }, where A method of summability is called regular, if it sums every convergent series to its ordinary sum.It follows from Toeplitz's Theorem ( [6]; Theorem 2) that (N, r) is regular if, and only if, and And (N , q) is regular if, and only if, and Inclusion And Equivalence Relations

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Let A be a sequence-to sequence transformation given by (1.1).If whenever {S n } has a bounded variation it follows that {t n } has a bounded variation, and if the limits are preserved, we say that A is absolutely regular.
(A) ⊆ (B) means that any series summable by (A) to sum S is necessary summable (B) to the same sum.(A) and (B) are equivalent if (A) ⊆ (B) and (B) ⊆ (A).For any sequence {u n } we shall write (1.11)

Object Of The Paper
The author ( [2], Theorem 6.1) obtained necessary and sufficient conditions for which (N , q) ⊆ |(N, r)|.The object of this paper is to obtain a set of conditions for the other way round, and to give some non-trivial special cases to show that this inclusion may holds, and we will give some other special cases to show that this inclusion fails to hold even if both (N, r) and (N , q) are regular.Finally, we will give two examples involving the equivalence of these two methods, and another two examples to show that the inclusion may holds in only one way without the other.These results will be concluded in sections 5, 6 and 7.

Results Required
This section is devoted to results that are necessary for our purposes: Theorem 4.1.[8] The sequence-to-sequence transformation given by (1.1) is absolutely regular if, and only if, and ], Theorem 6.1) Suppose that (N , q) and (N, r) are regular, q n = 0 (all n ≥ 0 ), then (N , q) ⊆ |(N, r)| if, and only if, where  2), then: 3.

Main Result
In this section we shall state and prove our main result: Theorem 5.1.Let (N , q) and (N, r) are regular; then |(N, r)| ⊆ (N , q) if, and only if, where 2) is alone is necessary and sufficient condition for Proof: Let {t r n } and {t q n } be respectively the (N, r) and (N , q) transforms of {S n } , then and To prove the result, we need to find t q n in terms of t r n .Observe that r 0 = 0 , ∞ n=0 r n z n = r(z), say, is non-zero in some neighborhood of the origin, we have 1 r(z) = c(z), say, is regular in some neighborhood of the origin, and so has a power series expansion

which by
Inclusion And Equivalence Relations so that Comparing the coefficient of in (5.8), we have (5.9) Using (5.9), it follows from (5.6) that where B n.v is given by (5.3) and (5.4)The special case in which S n = 1 , (n ≥ 0), then (5.5), (5.6) and (5.10) imply that (5.11) This implies (4.2).Using (5.4), it follows from (5.11) that the left hand sides of (4.3) and (5.2) are equivalent, and Mears Theorem (5.1) implies the result.Next, if r n ∈ µ, then (N, r) is regular.Using Kaluza Theorem (5.3), it follows that{c n } is bounded and c c → 0 as n → ∞.Using this and the regularity of (N , q), (5.1) holds and the proof is completed.✷ Remark 5.2.We remark that the condition B n,n = O(1) is necessary (but not sufficient) for (5.2) to be satisfied.It follows from Theorems 4.2 and 5.1 the following Lemma: , and only if (4.4), (5.1) and (5.2) are satisfied.

Equivalence Relations
With the aid of Lemma (5.3), it is natural to give some examples to show that the equivalence may holds in some trivial and non-trivial cases.In this section we will construct two examples to show that |(N, r)| ∼ N , q) , and two other examples to show that the inclusion may valid in only one way without the other.
Proof: The result that |(N, r)| ⊆ N , q) follows from example (6.2).Next, observe that the first term of the left hand side of (4.4) is equivalent to necessary condition for (4.4) to be satisfied.Using (6.12), (6.13) and (6.15), we have Therefore, (4.4) is not satisfied, and so N , q) ⊆ |(N, r)|.✷ Example 7.4.Let the assumptions on {r n } and {q n } be given as in (6.40) and (6.41), then N , q) ⊆ |(N, r)| but the converse is not valid.