$\bullet$  A real sequence $x=(x_n)_n$ is called convergent to $\alpha$ in usual sense if for any $\epsilon>0$ the set $\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}$ is finite.

$\bullet$ A real sequence $x=(x_n)_n$ is called statistically convergent to $\alpha$ if for any $\epsilon>0$ the set $\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}$ has natural density $0$. The natural density $d$ of $A\subset\mathbb N$ is defined by $d(A)=\lim\limits_{n\to\infty}\frac{|A\cap\{1,2,\dots,n\}|}{n}$, (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][1] 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) $A,B\in \mathcal I$ $\implies $ $A\cup B\in \mathcal I$

(ii) $A\in \mathcal I$ and $B\subset A$ $\implies$ $B\in \mathcal I$

$\bullet$ A real sequence $x=(x_n)_n$ is called $\mathcal I$-convergent to $\alpha$ in usual sense if for any $\epsilon>0$ the set $\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}\in \mathcal I$.   

$$\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots$$


$\mathcal I_f=\{A\subset\mathbb N: A \text{ is finite}\}$ and $\mathcal I_d=\{A\subset\mathbb N: d(A)=0\}$ 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?

  


  [1]: https://en.wikipedia.org/wiki/Banach_limit