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I am interested in conditions in terms of standard scales of summation methods that guarantee the existence of an averaged limit for all almost convergent sequences. For the Cesaro summation method $(C, 1)$ this fails; is this true, e.g., for the Cesaro methods $(C, \alpha)$ with $\alpha<1$?

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The paper G.G. Lorentz: A contribution to the theory of divergent sequences; Acta mathematica, Volume 80, Number 1, 1960, 167-190; DOI: 10.1007/BF02393648, contains several interesting results related to your questions on Banach limits.

A characterization of matrix methods that sum all almost convergent sequences is given (Theorem 7). In particular, each $C_\alpha$ sums all almost convergent sequences.

However, it is shown that almost convergence cannot be represented by a regular matrix method. Also the following stronger result about a class of matrix methods is shown.

Theorem 11. For every sequence $\{A_k\}$ of methods of the class $\mathfrak A$ there is a bounded sequence $x = \{x_n\}$ which is not almost convergent but is summable to the value zero by every one of the methods $A_k$.

The class $\mathfrak A$ in Lorentz's paper is the class of matrices fulfilling $$\lim\limits_{m\to\infty} \max\limits_n |a_{mn}|=0.$$ I think it's not that hard to show that each matrix $C_n$ belongs to $\mathfrak A$.

Some further references for almost convergence are mentioned e.g in the book Boos: Classical and Modern Methods in Summability.

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  • $\begingroup$ @Martin: I have already got the paper, thank you. Also I realized that what I meant is not what I asked, but you have already answered my real question :) $\endgroup$
    – kap44
    Apr 10, 2012 at 18:08
  • $\begingroup$ I saw in Lorentz' paper that $(C, \alpha)$-convergence for any $\alpha>0$ is equivalent to $(C, 1)$ convergence (apparently, for bounded sequences); thus my question is equivalent to the previous one. Is there any reference to the proof? $\endgroup$
    – kap44
    Apr 10, 2012 at 18:09
  • $\begingroup$ @vanja Sorry, I do not know a reference for that result. $\endgroup$ Apr 11, 2012 at 11:22

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