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.