Let $V$ be a normed vector space, let $l \in V$, and let $(a_n)$ be a sequence in $V$. We say that $a_n$ is Cesàro-convergent to $l$ if $\frac{1}{n}\sum_{i=1}^n a_i \to l$ as $n\to\infty$.

Now I will say that $a_n$ is $(\ast)$-convergent to $l$ if for any unbounded increasing sequences of positive integers $(m_n)$ and $(k_n)$, we have $$ \frac{1}{m_n} \sum_{i=k_n+1}^{k_n+m_n} a_i \,\to\, l \textrm{ as $n\to\infty$}. $$

This is equivalent to saying that for all $\varepsilon>0$ there exists $M>0$ such that for all $m,k>M$, $$ \left|l - \frac{1}{m} \sum_{i=k+1}^{k+m} a_i \right| < \varepsilon. $$ This is also equivalent to saying that the Cesàro-convergence of $(a_{k+n})_{n\geq 1}$ to $l$ is uniform across $k\geq 0$. (To see that this is implied by the previous formulation, fix $\varepsilon$ and choose $M$ in the previous formulation corresponding to $\frac{\varepsilon}{2}$, and then choose $N>M$ sufficiently large that $\frac{1}{N}\sum_{i=1}^M |a_i - l|<\frac{\varepsilon}{2}$. Then for our given $\varepsilon$ this $N$ should work uniformly across all $k\geq 0$.)

Is there a name for what I have called $(\ast)$-convergence? Have its relationships to other notions of "average"/"statistical" convergence been studied before?

Remark. As above, $(\ast)$-convergence implies Cesàro-convergence. But the converse is not true; e.g. with $V=\mathbb{R}$, take $$ 1, \ 0, \ \textrm{two $1$s}, \ \textrm{four $0$s}, \ \textrm{three $1$s}, \ \textrm{nine $0$s}, \ \textrm{four $1$s}, \ \textrm{sixteen $0$s}, \ \text{etc.} $$ This sequence has Cesàro-convergence to $0$, but not $(\ast)$-convergence to $0$.