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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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$a_n=\{\log\log n\}$ 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). –  Anthony Quas Jan 2 '12 at 23:25
    
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. –  Gottfried Helms Jan 3 '12 at 0:15
    
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! –  Zen Harper Jan 3 '12 at 3:02
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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. –  Zen Harper Jan 3 '12 at 3:15
    
@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... –  Yemon Choi Jan 3 '12 at 3:16

1 Answer 1

up vote 3 down vote accepted

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.

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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. –  Mateus Araújo Jan 3 '12 at 20:00
    
Sorry, I do not know of any simple characterization of bounded sequences that have convergent Cesàro means. –  S. Carnahan Jan 5 '12 at 8:11

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