While studying the convergence of a certain iterative algorithm, I have come across the following generalization of the Cesàro mean: given a sequence $\{a_k\}$ and an integer $m\geq 0$, define

$c_k^{(m)} = \frac{(k-1)!(m+1)}{(k+m)!} \sum_{i=1}^k \frac{(m+i-1)!}{(i-1)!} a_i$

For $m=0$ we have the usual Cesàro mean:

$c_k^{(0)} = \frac{1}{k} \sum_{i=1}^k a_i$

It is known that there exist bounded sequences $\{a_k\}$ for which the sequence of Cesàro means $\{c_k^{(0)}\}$ diverges (see the discussion here). My question is: given a bounded sequence $\{a_k\}$, is it always possible to find an integer $m$ such that $\{c_k^{(m)}\}$ converges?

Note that $c_k^{(m)}$ can be written in the more intuitive form

$c_k^{(m)} = \left(\frac{(m+1)k}{k+m} \right)\left[\frac{1}{k}\sum_{i=1}^k \frac{i}{k} \frac{i+1}{k+1} \cdots \frac{i+m-1}{k+m-1} a_i \right]$

The first factor in the right-hand side goes to $m+1$ as $k\rightarrow\infty$ and is not a problem. The factor in brackets looks like a weighted Cesàro mean in which each term $a_i$ is multiplied by a weight given by the product of $m$ positive factors, all smaller than 1 except for the last one corresponding to $i=k$, which is exactly 1. Thus it looks like these weights could "bring the means down to convergence" if $m$ is chosen large enough.

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    $\begingroup$ No. No matter what countable family of faithful (giving you the true limit of a sequence when it exists) methods you use, there exists a bounded sequence for which neither of them works. The idea is actually very simple. You are saying 0 or 1 every second and you have countably many experts who must tell you some number every time you say 0 or 1 so that if you say only x after a certain time, their numbers should converge to x as well. Now start saying 0 until you force the first expert to blurt out a number <1/3. Then say 1 until the first two experts produce numbers >2/3, and so on. $\endgroup$ – fedja Nov 14 '12 at 19:09

Observe that

$$ c_k^{(m)}= \frac{(k-1)! (m+1)!}{(m+k)!} \sum_{i=1}^k \binom{m+i-1}{m} a_i $$

$$= \frac{1}{\binom{m+k}{k-1}}\sum_{i=1}^k \binom{m+i-1}{m} a_i. $$

Observe that if we set

$$p_i =\binom{m+i-1}{m}, \;\; P_k=p_1+\cdots + p_k, $$


$$ P_k= \binom{m+k}{k-1} $$ and we can write

$$ c_k^{(m)}=\sum_{i=1}^k \frac{p_i}{P_k} a_i $$

so indeed this is an averaging technique. This type of averaging method is investigated in the book Divergent Series by G.H. Hardy. Check Thm. 15 in Section 3.8 of the book. I think it might help.

Edit To elaborate fedja's excellent idea. Choose a sequence of integers

$$0=N_0 < N_1 < N_2< \cdots $$

with the following properties

$$\frac{P_{N_{2k+1}}-P_{N_{2k}}}{P_{N_{2k+1}}} \to 1,\;\; \frac{P_{N_{2k-1}}}{P_{N_{2k}}}\to 0\;\;\mbox{as $k\to \infty$}. $$

Then define $a_i$ so that $a_i=1$ if $N_{2k}< i\leq N_{2k+1}$ for some $k\geq 0$ and $a_i =0$ if $N_{2k+1}< i\leq N_{2k+2}$ for some $k\geq 0$.

This sequence thus constructed will not converge in the above generalized Cesaro sense.

  • $\begingroup$ I was afraid so. Thank you very much - I'll check Hardy's book, if I can get ahold of it. $\endgroup$ – Roberto López-Valcarce Nov 14 '12 at 23:42

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