A Tauberian theorem for the generalized Nörlund summability method

Remark 1.1.

The (N,p,q) summability method reduces to the Cesàro summability method when pn=1 and qn=1 for all n∈ℕ, to the weighted mean method (N¯,p) when qn=1 for all n∈ℕ, to the method (C,α,β) (see [1]) when pn=(n+ββ), qn=(n+α-1α) and to the Nörlund method (N,q) when pn=1 for all n∈ℕ (see [6]).

Theorem 2.1.

Let (pn) and (qn) be any two non-negative real sequences such that

where λn:=[λ⁢n] denotes the integral part of λ⁢n for every n∈N, and let ∑k=0∞ak be a series of real numbers which is (N,p,q) summable to a finite number L. Then ∑k=0∞ak is convergent to the same number L if and only if the following two conditions hold:

Remark 2.2.

Following Schmidt [11], we say that a real sequence (sn) is slowly decreasing if

or, equivalently,

Conditions (2.2) and (2.3) are satisfied if (sn) is slowly decreasing.

Remark 2.3.

The classical one-sided Tauberian condition of Landau (see [9])

is sufficient for (2.4) to hold.

Remark 2.4.

If we take qn=1 for all n∈ℕ as a special case of Theorem 2.1, we obtain the Móricz and Rhoades conditions (see [10])

and

where Pn:=(p⋆1)n=∑k=0npk.

Theorem 2.5.

Let condition (2.1) be satisfied and let ∑k=0∞ak be a series of complex numbers which is (N,p,q) summable to a finite number L. Then ∑k=0∞ak is convergent to the same number L if and only if one of the following two conditions holds:

or

Remark 2.6.

Following Schmidt [11], we say that a real sequence (sn) is said to be slowly oscillating if

or, equivalently,

Conditions (2.5) and (2.6) are satisfied if (sn) is slowly oscillating.

Remark 2.7.

The classical two-sided Tauberian condition

is sufficient for (2.5) and (2.6) to hold.

Remark 2.8.

If we take qn=1 for all n∈ℕ as a special case of Theorem 2.1, we obtain the Móricz and Rhoades conditions (see [10])

and

where Pn:=(p⋆1)n=∑k=0npk.

Lemma 3.1.

The condition given by relation (2.1) is equivalent to the condition

Proof.

Suppose that relation (2.1) is valid, 0<λ<1 and m=λn=[λ⁢n], n∈ℕ. Then it follows that

From the above relation and the definition of the positive real sequences (pn) and (qn), we obtain

Conversely, suppose that relation (3.1) is valid. Let λ>1 be a given number and let λ1 be chosen so that 1<λ1<λ. Set m=λn=[λ⁢n]. From 0<1λ<1λ1<1, it follows that

provided that λ1≤λ-1n, which is the case when n is large enough. Under these conditions, we have

Lemma 3.2.

Let condition (2.1) be satisfied and let ∑k=0∞ak be a series of complex numbers which is generalized Nörlund summable to L. Then

and

Proof.

Consider the case where λ>1. Then we obtain

provided that Rλn>Rn.

Since, by (2.1),

(3.2) follows from (3.4), (3.5) and the definition of the sequence (qn).

Consider the case where 0<λ<1. Then we obtain

provided that Rn>Rλn.

Since, by (2.1),

(3.3) follows from (3.6), (3.7) and the definition of the sequence (qn). ∎

Proof of Theorem 2.1.

Necessity. Suppose that limn→∞⁡sn=L, and (1.1) holds. From Lemma 3.2, we have

for every λ>1 . In the case where 0<λ<1, we find that

Sufficiency. Assume that conditions (2.2) and (2.3) are satisfied. In what follows, we will prove that limn→∞⁡sn=L. Given any ϵ>0, by relation (2.2), we can choose λ1>0 so that

where λn1=[λ1⁢n]. By the assumed summability (N,p,q) of (xn) and Lemma 3.2, we have

for any λ>1. Taking (4.1) and (4.2) into account, we obtain

On the other hand, if 0<λ<1, then, for every ϵ>0, we can choose 0<λ2<1 so that

where λn2=[λ2⁢n]. By the assumed summability (N,p,q) of (xn) and Lemma 3.2, we have

for any 0<λ<1.

Taking (4.4) and (4.5) into account, we obtain

Since ϵ>0 is arbitrary, combining relations (4.3) and (4.6), we obtain

Proof of Theorem 2.5.

Necessity. If both (1.1) and (1.2) hold, then Lemma 3.2 yields (2.5) for every λ>1 and (2.6) for every 0<λ<1.

Sufficiency. Suppose that (1.1), (2.1) and one of conditions (2.5) or (2.6) is satisfied. For any given ϵ>0, there exists some λ1>1 such that

where λn1=[λ1⁢n]. Taking into account the fact that ∑k=0∞ak is (N,p,q) summable to L, we get the following estimation:

For any given ϵ>0, there exists some 0<λ2<1 such that

where λn2=[λ2⁢n]. Taking into account the fact that ∑k=0∞ak is (N,p,q) summable to L, we get the following estimation:

Since ϵ>0 is arbitrary, in either case, we get limn→∞⁡sn=L. ∎