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$bullet$ A real sequence $x=(x_n)_n$ is called convergent to $alpha$ in usual sense if for any $epsilon>0$ the set $x_n-alpha$ is finite.

$bullet$ A real sequence $x=(x_n)_n$ is called statistically convergent to $alpha$ if for any $epsilon>0$ the set $x_n-alpha$ has natural density $0$. The natural density $d$ of $Asubsetmathbb N$ is defined by $d(A)=limlimits_ntoinftyfracn$, (provided the limit exists) where $|A|$ denotes the cardinality of $A$.

$bullet$ A bounded real sequences $x=(x_n)_n$ is said to be almost convergent to $alpha$ if all the Banach limit functionals give an unique value for the sequence $x$.

A family $mathcal I$ of subsets of $mathbb N$ is said to be an ideal in $mathbb N$ if

(i) $A,Bin mathcal I$ $implies $ $Acup Bin mathcal I$

(ii) $Ain mathcal I$ and $Bsubset A$ $implies$ $Bin mathcal I$

$bullet$ A real sequence $x=(x_n)_n$ is called $mathcal I$-convergent to $alpha$ in usual sense if for any $epsilon>0$ the set $x_n-alphain mathcal I$.

$$dotsdotsdotsdotsdotsdotsdotsdotsdotsdotsdotsdotsdotsdotsdotsdotsdotsdotsdotsdotsdotsdots$$

$mathcal I_f=Asubsetmathbb N: A text is finite$ and $mathcal I_d=Asubsetmathbb N: d(A)=0$ become ideals in $mathbb N$. Moreover, $mathcal I_f$-convergence and $mathcal I_d$-convergence coincide with usual convergence and statistical convergence respectively. But, what is the ideal in the case of almost convergence?

My Question : Find the ideal $mathcal I$ for which $mathcal I$-convergence coincides with the almost convergence. Is it available in literature?

Such an ideal does not exist.

Indeed, suppose the contrary, and let $I$ be such an ideal. The sequence $(x_n)=(1,0,1,0,dots)$ is almost convergent, and therefore $I$-convergent, to $1/2$. So,
$$mathbb N=ninmathbb Ncolonin I,
$$
and hence $I$ is the powerset of $mathbb N$. So, every sequence is $I$-convergent, and therefore almost convergent, to every real limit, which is of course absurd.

This consideration also shows that the Cesàro convergence -- which is implied by the almost-convergence -- is also not the $I$-convergence, for any ideal $I$.