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](https://mathoverflow.net/questions/84772/bounded-sequences-with-divergent-cesaro-mean)). 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.



Generalized Cesàro means of a bounded sequence
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


$$p\_i =\binom{m+i-1}{m}, \;\; P\_k=p\_1+\cdots + p\_k, $$


then 


$$ P\_k= \binom{m+k}{k-1} $$ 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.


