Is a specific choice of ultrafilter necessary in order to get "concrete results" in nonstandard analysis? Suppose we have sequences of real numbers which are indexed by the natural numbers. We can then define an ultrafilter $\mathcal{U} \subset 2^{\mathbb{N}}$ (where $2^{\mathbb{N}}$ is the powerset of natural numbers), which we can use to do calculus in a nonstandard fashion. There are many choices for an ultrafilter. 
I am trying to get a handle on what impact the choice of ultrafilter has. Is it similar to choosing a set of basis vectors in order to do concrete calculations in linear algebra? 
For example, suppose I wanted to determine the (generalized) limit of the sequence $1, 0, 1, 0, 1, 0, \dots$. Whether $1, 0, 1, 0, 1, 0 \dots \approx 0, 0, 0, 0, \dots$ relies whether my choice of ultrafilter contains the set $0, 2, 4, 6, 8, \dots$ or $1, 3, 5, 7, 9...$. This seems to suggest to me that a choice of ultrafilter is necessary in order to concretely answer a question like "what is the limit of a sequence $x_i$"?   
 A: Let me unpack your $0,1,0,1,...$ example, since I think that will clarify the picture.
The point is that this sequence $S$ "lives in the nonstandard world" in two different ways:


*

*It names an element of the ultrapower, namely $[S]_\mathcal{U}$. As you say, this element is either $0$ or $1$, depending on the choice of $\mathcal{U}$.

*It also has a nonstandard version ${}^*S$. This nonstandard version is, intuitively, the continuation of $S$ through the nonstandard natural numbers. The ultrapower $^*\mathcal{R}$ thinks that $^*S$ does not converge, just as $\mathcal{R}$ thinks that $S$ does not converge.
So the issue is that $S$-as-a-name is a different thing than $S$-as-a-sequence. Questions about the behavior of specific names do indeed depend on the choice of ultrafilter; questions about the nonstandard versions of concrete objects do not, by the transfer principle. 
Roughly speaking, any question about the reals you can ask without using the language of model theory is going to have the same answer across all ultraproducts.
A: It might be, and that really depends on what you're trying to do.
For example, take the ultraproduct of all finite prime fields (i.e., $\Bbb Z/p\Bbb Z$ where $p$ is a prime), the result is a field of characteristics $0$, say $F$.
Question. Does $F\models\exists x(x\cdot x=1+1)$?
Well, that depends on the choice of ultrafilter! If the ultrafilter concentrates on those primes for which $2$ is not a quadratic residue, the answer is negative; otherwise it is positive. And you can then start asking more and more complicated questions, whose answers all depend on the choice of ultrafilter.
Another example, take an enumeration of $\Bbb Q\cap[0,1]=\{q_n\mid n\in\Bbb N\}$. Then for every $x\in[0,1]$ there is an ultrafilter $\mathcal U_x$, such that $\lim_{\mathcal U_x}q_n=x$.

Most "daily uses" of ultraproducts are such that by ensuring that the ultrafilter is free, we already guarantee that the statements we "care about" will be true. E.g., the characteristics of $F$ in the first example is $0$.
A: You include the tag (nonstandard-analysis), so I reply for that.  Other uses of ultraproducts may differ, as Asaf says.  
The "transfer principle" of nonstandard analysis insures that for propositions $P$ of an appropriate form, they are true in the nonstandard model if and only if they are true in the standard model.
In particular, "The sequence $(-1)^n$ converges" is false in the nonstandard models of "nonstandard analysis" as well as in the standard model.
