3
$\begingroup$

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 $$ \exists \lim_{N \to \infty} \frac{\sum_{n=1}^N\frac{\sum_{i=1}^n a_i}{n}}{N} $$

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ how does the above answer relate to the requirement that $0 < a_i < K < \infty$? $\endgroup$ – Mirko Mar 31 '16 at 23:41
  • 1
    $\begingroup$ @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
0
$\begingroup$

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 $C^{2}(a)\to 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.