It is well known that there are bounded sequences with divergent Cesàro mean, i.e., a bounded $a\_n$ for which given $$c\_N := \frac{1}{N}\sum\_{n=1}^N a\_n,$$ the sequence $(c\_N)\_{N\geq1}$ has no limit. A simple example is $a\_n = (-1)^{\lfloor \log\_2 n\rfloor}$, for which the $(2^{n+1}-1)$-th term of the sequence of means is $-1/3$ for odd $n$, and the $(2^{n+1}-2)$-th is $1/3$ for even $n$.


But I don't know of any example which is substantially different.


My question is, then, can the Cesàro mean of a bounded sequence diverge in some other way or this slow but stubborn oscillation is essentially the only example?




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EDIT:
Is there a simple characterization of the bounded sequences for which the Cesàro means converges?



Bounded sequences with divergent Cesàro mean
Choose any bounded infinite sequence $\{ b\_m \}\_{m \geq 0}$ of integers that is not eventually stationary, and let $a\_n = b\_{\lfloor \log (\log (n+2)) \rfloor}$. When $m$ is a large integer, and $N+2$ is almost $e^{e^m}$, the Cesàro mean $c\_N$ will be very close to $b\_{m-1}$.


If you want arbitrary iterated Cesàro means to diverge, you can replace the double log with a slower-growing function, like the inverse Ackerman function $\alpha$.


Regarding your question, it is not hard to show that if your sequence is bounded, that the Cesàro means $c\_N$ can't move very quickly for large $N$. In other words, you can avoid oscillation in the literal sense, but whatever you have will be slow.


