Suppose we had a sequence say $K=\langle1,8,5,1,8,5,1,8,5,\ldots\rangle$ periodic on these $3$ numbers and our ultrafilter contained the odd numbers.
Then am I right in thinking that K could be $\langle1\rangle$ or $\langle8\rangle$ or the periodic sequence $\langle1,8,1,8,1,8,\ldots\rangle$?
Also since hyperreals are sets of equivalence classes of sequences, does this mean that the sequence $\langle1\rangle$, $\langle8\rangle$, $\langle1,8,1,8,\ldots\rangle$ all belong to the same equivalence class and are therefore all equal?
Thanks.
You are not distinguishing sufficiently in your notation between the sequence on the one hand and the hyperreal number generated by it (as an equivalence class), on the other, but if I understood your question correctly then yes, $K$ could be either of those things, provided there are infinitely many occurrences of "$1$" in the first case, for example. This depends on the ultrafilter.
To introduce better notation you could exploit the following convention. A sequence $u$ is $\langle u_n : n\in\mathbb{N}\rangle$ whereas its equivalence class could be denoted $[u]$ (or alternatively $[u_n]$).