It is well known that there are bounded sequences with divergent Cesàro mean, i.e., a bounded $a_n$ for which $$\lim_{N\to\infty} c_N = \frac{1}{N}\sum_{n=1}^N a_n$$ does not exist. 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?

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

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    $\begingroup$ $a_n=\{\log\log n\}$ is substantially different (here $\{\cdot\}$ means the fractional part). With this sequence of $a_n$'s, $c_N$ is very close to $a_N$ (unless $a_N$ happens to be very close to 0). $\endgroup$ – Anthony Quas Jan 2 '12 at 23:25
  • $\begingroup$ Hmm, I don't have the exact description at hand - but if I recall it right then Cesaro-summation can be seen as matrix-multiplication of a vector A (containing the terms of the sequence) and a lower triangular matrix C containing the coefficients for "implementing" the Cesaro-summation such that $\small C \cdot A = S $ and S contains a sequence approaching the convergent. Then it seems simple to invent any non-converging vector S (say, having periodic entries) and premultiply with the inverse of C: $\small C^{-1} \cdot S = B $ which produces then in B an interesting(?) sequence. $\endgroup$ – Gottfried Helms Jan 3 '12 at 0:15
  • $\begingroup$ A huge amount of information is to be found in G.H.Hardy's last book Divergent Series published in 1948 (which I haven't got, unfortunately). This old-fashioned book, by one of the last surviving champions of "hard analysis", is unfortunately not very well-known nowadays, probably because "soft analysis", e.g. functional analysis was much more fashionable then - but renewed versions of "hard analysis" are making comebacks! $\endgroup$ – Zen Harper Jan 3 '12 at 3:02
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    $\begingroup$ There are many Tauberian theorems which say that if the Cesaro means (or other means) converge and the sequence $(a_n)$ satisfies an extra condition (called a Tauberian condition) then the sequence itself converges (necessarily to the same limit). Famous Tauberian conditions include Littlewood's $na_n > -K$ condition, and stuff about slowly oscillating sequences, and many, many other conditions also; as well as stuff used to prove the Prime Number Theorem. As well as Hardy's book, I recommend Tauberian theory, a century of developments, by Jacob Korevaar. $\endgroup$ – Zen Harper Jan 3 '12 at 3:15
  • $\begingroup$ @Zen: "by one of the last surviving champions" - is this a new definition of "surviving", or do you know something necromantic I don't? Also, not sure I agree with your version of history... $\endgroup$ – Yemon Choi Jan 3 '12 at 3:16

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.

  • $\begingroup$ That's an excellent answer, thanks! But it seems that my question was not very good, I'll edit it a bit. If you can answer the new version, I'd be glad. $\endgroup$ – Mateus Araújo Jan 3 '12 at 20:00
  • $\begingroup$ Sorry, I do not know of any simple characterization of bounded sequences that have convergent Cesàro means. $\endgroup$ – S. Carnahan Jan 5 '12 at 8:11

Here is a construction: Choose real numbers a < c < d < b. Construct a sequence a, b, a, a, ..., a, b, b, ..., b, a, a, ..., a, b, b, ..., b, a, ...... consisting of blocks of a's and b's as follows: Start with the first two terms a, b. For the first block of a's put enough terms equal to a so that the Cesaro average so far is less that c. Then start inserting terms equal to b, enough of them so that the average so far exceeds d. Then start inserting terms equal to a until the average is less than c, and continue with inserting terms equal to b until the average exceeds d, and so on.


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