Given a sequence of bounded real numbers $a_n\in [0,1], n\in\mathbb{N}$, are there convenient conditions on the $a_n$ sufficient for convergence of the running means $A_n=(\sum_{i=1}^na_i)/n$? If the $a_n$ themselves converge, then $A_n$ does, but I am hoping for a weaker condition that would cover, e.g., $a_n=(-1)^n$. It seems for $A_n$ to diverge $a_n$ will oscillate with decreasing frequency, enough to overcome the increasing stability of $A_n$ as $n$ grows. (Sorry for the imprecise language.) But I am not sure how to formulate a condition excluding this behavior.

There is a mathoverflow question here where, in a comment, an answerer says he isn't aware of any convenient conditions: https://mathoverflow.net/questions/84772/bounded-sequences-with-divergent-ces%C3%A0ro-mean There is also a question on here that directly asks this question, but the op accidentally phrased it in terms of cesaro summability rather than convergence of the cesaro means, got answers to the wrongly asked questions, and when he corrected his question in the comments, didn't get any answer. Sufficient conditions to guarantee Cesaro summarble . But I thought it was worth seeing if this question could get answers as a separate question.

  • $\begingroup$ Have you read this comment on the MathOverflow question you linked? (About Tauberian conditions) $\endgroup$ – Clement C. Mar 7 '17 at 18:14
  • $\begingroup$ @ClementC. I read that comment to be about hypotheses on the cesaro means $A_n$ (plus some extra hypothesis) that guarantee convergence of the original sequence $a_n$, no? I'm looking for hypotheses on the sequence to get convergence of the cesaro means. $\endgroup$ – Hasse1987 Mar 7 '17 at 19:51
  • $\begingroup$ My bad -- I read that way too quickly. $\endgroup$ – Clement C. Mar 7 '17 at 19:53

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