Abstract
Let $(x_k)$, for $k\in \mathbb{N}\cup \{0\}$ be a sequence of real or complex numbers and set $(EC)_{n}^{1}=\frac{1}{2^n}\sum_{j=0}^{n}{\binom{n}{j}\frac{1}{j+1}\sum_{v=0}^{j}{x_v}},$ $n\in \mathbb{N}\cup \{0\}.$ We present necessary and sufficient conditions, under which $st-\lim_{}{x_k}= L$ follows from $st-\lim_{}{(EC)_{n}^{1}} = L,$ where L is a finite number. If $(x_k)$ is a sequence of real numbers, then these are one-sided Tauberian conditions. If $(x_k)$ is a sequence of complex numbers, then these are two-sided Tauberian conditions.
Highlights
We present necessary and sufficient conditions, under which st − lim xk = L follows from st − lim (EC)1n = L, where L is a finite number
We will prove under which conditions statistical convergence st − lim xn, follows from (EC)1n-statistically convergence
Let us suppose that st − limk xk = L; is (EC)1n− statistically convergent and relation (2.1) satisfies, for every t > 1, is valid the following relation: st
Summary
Let us suppose that sequence (xn)-statistically convergent to L, and |xn −L| ≤ M for every n ∈ N. Proof: From fact that (xn) converges statistically to L, we get lim |{k ≤ n : |xk − L| ≥ ǫ}| = 0. Tauberian) under which the converse implication holds, in Theorem 1.3, for defined convergence. Where tn, denotes the integral parts of the [tn] for every n ∈ N, and let (xk) be a sequence of real numbers which converges to L, (EC)1n− statistically.
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