Characterisation of convergence of bounded sequences via ultra-filters Let $\{a_n\}_{n\in\mathbb N}$ be a bounded sequence of real or complex numbers and $\mathscr F\subset\mathscr P(\mathbb N)$ be a non-principal ultra-filter. Then $a=\lim_{\mathscr F}a_n$ is well-defined and corresponds to a subsequential limit of  $\{a_n\}_{n\in\mathbb N}$, i.e., there exists an infinite set
$$
A=\{i_1<i_2<\cdots<i_n<\cdots\}\subset\mathbb N,
$$
such that $a_{i_n}\to a$. 
Clearly,  $\lim_{\mathscr G}a_n=a$, for every non-principal ultrafilter containing $A$.
I was wondering whether the inverse is also true. By inverse I mean the following:
If $a=\lim_{\mathscr G}a_n$, then $\mathscr G$ contains the index set of a subsequence of $\{a_n\}_{n\in\mathbb N}$ converging to $a$, i.e., there exists an $A=\{i_n\}_{n\in\mathbb N}\in\mathscr G$, such that $a_{i_n}\to a$. 

If I understood your question correctly, the answer cannot be given in ZFC.  It depends on whether the ultrafilter is a P-point or not, by a result of Saharon Shelah; see here. My favorite paper dealing with the issue is
Cutland, Nigel; Kessler, Christoph; Kopp, Ekkehard; Ross, David. On Cauchy's notion of infinitesimal. British J. Philos. Sci. 39 (1988), no. 3, 375–378.
See here.
The question is easily paraphrasable in the following terms: is a hyperreal infinitesimal representable by a sequence tending to zero? The answer in general is negative, as Cutland et al. showed.
