Cardinality of a Hypernatural Sequence Supposing we have a sequence of numbers $a_n$ that ranges from 1 to a particular hypernatural $\delta$. This is an infinite series in the sense that it "goes past" the natural numbers and the natural numbers are countably infinite. However, this sequence does have an end, in this case $\delta$. 
The cardinality of the natural numbers is $\aleph_0$ and the cardinality of the hypernaturals is the same as the real numbers, $2^{\aleph_0}$.
My question is what is the cardinality of $a_n$? Is it $2^{\aleph_0}$, $\aleph_0$, or something in-between the two?
Guessing, I would say that it is in-between the two as it goes past the cardinal number $\aleph_0$ but I am not sure if this reasoning even works.
 A: Cardinality is something you want to assign to sets, but a sequence is a function from $\mathbb N$ (or in this case $^*\mathbb N$) onto another set. So your question seems to not make sense. 
You could of course ask what the cardinality of $\{1,\ldots,\delta\}$ is. (by the way why do you chose $\delta$ as a hypernatural, this symbol is usually reserved for infinitesimal entities)
To which the answer would be it's either $\aleph_0$, if $\delta$ is finite, or $\aleph_1$ if $\delta$ is unlimited (i.e. $\delta\in{}^*\mathbb N\setminus\mathbb N$), if we assume that the continuum hypothesis holds true.
Edit: Actually we can prove that in the latter case it is $\aleph_1$ regardless of the continuum hypothesis. This is due to the structure of the hypernaturals (I'm assuming we use the standard ultrafilter construction of ${}^*\mathbb N$)
Claim: $\forall \delta\in {}^*\mathbb N\setminus\mathbb N\;\forall r\in\mathbb R_+\exists q\in {}^*\mathbb Q_+ : q\delta \in \text{hal}(r\delta)\cap{}^*\mathbb N$
Proof: Let $\mathcal F$ be our ultra-filter. Then $\delta$ is associated with a sequence $d_{n} \in \mathbb R^{\mathbb N}$ such that $\{n\in\mathbb N : d_n \in\mathbb N\}\in \mathcal F$. Since $\delta$ is assumed to be unlimited, this sequence must go to infinity. Now we simply take a sequence $b_n\in\mathbb N^{\mathbb N}$ such that $\lim\limits_{n\to\infty}|r-\frac{b_n}{d_n}| = 0$. Consequently let $q_n = \frac{b_n}{d_n}$ and $q\in{}^*\mathbb Q$ be the hyperrational number associated with that sequence. Then obviously $q\in\text{hal}(r)$, hence also $^{(1)}$ $q\delta\in\text{hal}(r\delta)$. But on the other hand $q\delta$ corresponds to the sequence $\frac{b_n}{d_n}d_n = b_n$, hence $q\in{}^*\mathbb N$.
Remark: So in fact if you have any unlimited hypernatural $\delta$ and take any positive real number $r$ then there is another hypernatural $\delta'$ such that $|r\delta-\delta'|$ is infinitesimal.
(1): Here we have to be a bit more careful, e.g. if $q = r +\epsilon$, $\epsilon$ infinitesimal, then $q\in\text{hal}(r)$, but $\frac{1}{\epsilon}q \notin \text{hal}(\frac{1}{\epsilon}r)$. We can fix this issue in our proof by taking $q$ close enough to $r$, i.e. such that $\delta|q-r|$ is still infinitesimal.
A: The cardinality of $\{1,2,3,\ldots,H\}$ where $H$ (your delta) is an infinite hypernatural is necessarily the continuum.  This is because there is a surjective map from this set to the real interval $[0,1]$ given by sending $x$ to the standard part of $\frac{x}{H}$.
A: The (finite/standard) Natural numbers are also Cardinal numbers, so typically we can use them interchangeably.  So we can talk about Natural numbers as having "countable cardinality" or "cardinality less than any uncountable Cardinal number", and we all know what we mean.
But the unlimited Hypernaturals do not overlap with Cardinal (or even Ordinal) numbers.  We notice this by the fact that each positive Hypernatural number has an immediate predecessor, but this is not true with Cardinal or Ordinal numbers (specifically: Limit Cardinals/Ordinals have no immediate predecessor).  Also, while there is a smallest infinite Cardinal number, there is no smallest unlimited Hypernatural number.  So, technically speaking, we can't really talk about Hypernaturals in terms of having "cardinality", per se.
However, we can easily can get around this issue, by modifying the question slightly:

*

*If H is an arbitrary unlimited Hypernatural number, what is the cardinality of the set { 1, ..., H } ?

Being a set, it most certainly has cardinality.  I believe that the answer to this question is that it has the cardinality of the Continuum.  This is true for any unlimited H.  It is certainly true that the entirety of the set of Hypernaturals has the cardinality of the Continuum.  And this is so of the Hyperreals, as well.
NOTE: I am assuming the traditional construction of the Hypernaturals using a countable sequence of Natural numbers with a defined equivalence relation using a non-principal ultrafilter (as described in Robert Goldblatt's excellent book "Lectures on the Hyperreals").  It could be that more esoteric or set-theoretic constructions of the Hypernaturals could yield larger cardinalities.  But when people speak of the Hypernaturals (or Hyperreals) without qualification, I presume they mean this more common construction.
