Is it possible to construct a bounded positive sequence $a_i$, ($0 < a_i < K < \infty$) such that the limit of its Cesaro mean does not exist but the limit of the Cesaro mean of its Cesaro means does exist?

That is a sequence $a_i$ s.t. : $$ \nexists \lim_{n \to \infty} \frac{\sum_{i=1}^n a_i}{n} $$ but $$ \exists \lim_{N \to \infty} \frac{\sum_{n=1}^N\frac{\sum_{i=1}^n a_i}{n}}{N} $$

  • $\begingroup$ Crostul gave a nice answer but ignored the conditions of positivity and boundedness. Does anyone know if it possible to construct a sequence keeping the conditions? $\endgroup$ – benny Apr 4 '16 at 17:04

Of course. Such a sequence can be constructed forcing $$(-1)^n= \frac{1}{n}\sum_{k=1}^n a_k$$ so that you have $a_1=-1$, and $a_n = n(-1)^n - (a_1 + \dots +a_{n-1})$: this gives you a way to compute $\{ a_n \}_{n=1}^{\infty}$ recursively.

Obviously, if $\{ i_n \}_{n=1}^{\infty}$ is any sequence having no limit as $n \to \infty$, you can construct another sequence $\{ x_n \}_{n=1}^{\infty}$forcing $$i_n= \frac{1}{n}\sum_{k=1}^n x_k$$ In particular, you can use your previously defined $\{ a_n \}_n$ to define another $\{ a_n' \}_n$, and then $\{ a_n'' \}_n$, and so on. This allows to construct sequences such that the Cesaro mean of the Cesaro mean of the Cesaro mean of ... etcetera does not exist.

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    $\begingroup$ how does the above answer relate to the requirement that $0 < a_i < K < \infty$? $\endgroup$ – Mirko Mar 31 '16 at 23:41
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    $\begingroup$ @Mirko You are right! I totally forgot to consider that condition. I'll try to fix my answer. $\endgroup$ – Crostul Apr 1 '16 at 8:15
  • $\begingroup$ @Crostul: Thank you for the answer but I am looking for bounded positive sequences. Do you think it is possible to construct such a sequence? $\endgroup$ – benny Apr 2 '16 at 16:28
  • $\begingroup$ @benny I have no idea. $\endgroup$ – Crostul Apr 2 '16 at 16:31

Consider the sequence $a:=(a_{n})_{n\in\mathbb{N}_{\geq 1}}$ given by

$$a_{n} = \begin{cases} \;\;1 &\text{ for } 2^{k}\leq n < 2^{k+1}, k \text{ even} \\ -1 &\text{ for } 2^{k}\leq n < 2^{k+1}, k \text{ odd.} \\ \end{cases}$$

(Hanul Jeon gave this as an example in: Bounded sequence with divergent Cesaro means) Define the Cesàro mean $C(a):=(c_{n})_{n\in\mathbb{N}_{\geq 1}}$:

$$c_{n}:= \frac{1}{n}\sum_{k=1}^{n} a_{k} $$

Then via the geometric series we have

$$c_{2^{n}-1}=\frac{1}{2^{n}-1}\sum_{k=1}^{2^{n}-1} a_{k} = \frac{1}{2^{n}-1}\sum_{k=0}^{n-1} (-2)^{k}=\frac{(-2)^{n}-1}{(-3)\cdot(2^{n}-1)}.$$

As noted in the previous post, this is a divergent subsequence of C(a) and thus, the Cesàro mean diverges also. I would suspect that the Cesàro mean of the Cesàro mean $C^{2}(a)$ converges to zero (since $C(a)$ "oscillates smoothly" between $\frac{1}{3}$ and $-\frac{1}{3}$), but I might be wrong and I don't have time for a proof right now. Maybe you can try yourself. I'd be interested!

EDIT: The requirement that the sequence be positive is unnecessary because if it's bounded by $K$ you can always just add $K$ to every member of the sequence and the Cesàro mean of $a$ will be $C(a)+K$. But still, it remains to be proved that for the above sequence $C^{2}(a)\to 0$.


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