Does ultralimit of sequence change after shift? Let


*

*$(a_n)$ be a bounded sequence of numbers,

*$\omega$ be an non-principal ultrafilter on $\mathbb N$,


then one can assign a limit along ultrafilter $(\omega-)\lim a_n$ to it as is said here. This limit remains the same, if one changes a finite number of the elements of sequence.
But will it be shift-invariant, i.e. equal for sequences $(a_n)$ and $(a_{n+1})$? If not, can one choose an ultrafilter with this property at least for sequences with $a_{n+1}-a_n \to 0$?
 A: To answer the question at the end of your post, any ultrafilter $U$ will be shift invariant (at least by one place) on sequences $(a_n)$ with the property $\lim_U (a_{n+1}-a_n) = 0$. 
Indeed, $\lim_U (a_{n+1}) = \lim_U(a_n + (a_{n+1} - a_n)) = \lim_U(a_n) + \lim_U(a_{n+1}-a_n) = \lim_U(a_n)$, just using linearity of $\lim_U$.
Above I've taken your suggested condition $a_{n+1}-a_n\to 0$ to mean convergence with respect to $U$. Of course, if the sequence of differences actually converges, this will agree with convergence according to $U$ as long as $U$ is nonprincipal. There's one subtlety with using $\lim_U$ here, though. The limit assigned to the sequence $(a_{n+1}-a_n)$ by $U$ might not be shift invariant! So if you want to shift $k$ places to the right, you need that $\lim_U(a_{n+k}-a_n) = 0$. This issue goes away if you use the stronger assumption that $\lim_{n\to\infty}(a_{n+1}-a_n) = 0$, since this notion of limit is shift-invariant.
A: No free ultrafilter on $\Bbb N$ has this property. Let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$. Let $E=\{2n:n\in\Bbb N\}$, and define a sequence
$$a:\Bbb N\to\Bbb R:n\mapsto\begin{cases}
0,&\text{if }n\in E\\
1,&\text{if }n\notin E\;.
\end{cases}$$
If $E\in\mathscr{U}$, then $\mathscr{U}$-$\lim a=0$, and if not, then $\mathscr{U}$-$\lim a=1$. Now let $b:\Bbb N\to\Bbb R:n\mapsto a_{n+1}$; clearly $\mathscr{U}$-$\lim b=1-\mathscr{U}$-$\lim a$.
Added: This is of course only an answer to the first question. Alex Kruckman has dealt nicely with the second question.
