It would be possible to define an uniform distribution on $\Bbb N$ using infinitesimals? In standard analysis it is clear that it is impossible to define an uniform probability distribution on $\Bbb N$ because there is no constant $c\in\Bbb R$ such that $\sum_{k=1}^\infty c=1$.
Using Robinson's hypernaturals it doesn't seem feasible either because $\Bbb N$ is an external set and so there is no way that $\Bbb N$ appear as a set in an internal formula.
Another way to do this could be to try to define an external function of the kind $f:\{1,\ldots, N\}\to\Bbb N$ here $N$ is an unlimited hypernatural and $f$ is surjective and linear. However again this doesn't seem feasible because the structure of the hyperfinite set $\{1,\ldots, N\}$ seems very different that the structure of $\Bbb N$.
Someone knows some attempt like this, maybe using a different kind of infinitesimals different than those of Robinson's theory?

Another way to say the same: someone knows some attempt to extend the field of reals such that there is some constant $c\neq 0$ such that $\sum_{k=1}^\infty c\in\Bbb R\setminus\{0\}$?
 A: Of course we need to specify what properties the  extension should have.
The answer is no  under fairly  weak  assumptions:


If $F$ is a field with a topology such  that addition is separately continuous and infinite sums  are defined as limits of partial sums then $\sum c$ diverges for every non-zero $c\in F$.


Say $s=\sum  c$ and let $s_n$ be the $n$th partial sum. Then $s_{n+1}=s_n+c$, so continuity of addition shows that $s=s+c$, hence $c=0$.
A: I don't disagree with the answer, but I thought I would pose a different semi-answer - an answer to a slightly different question, but that someone might be thinking of when they look at this question, as other similar questions have arisen.  Instead of requiring that it be specifically $\mathbb{N}$, let's say that we just want to define a uniform distribution on the set of numbers from $1$ to some hyperinteger $\omega$.
As the questioner implies, for any probability distribution on the set $\Omega$, $\sum\limits_{n=1}^{\omega}\text{Pr}[x=n] = 1$.  If the probability is uniform, then $\text{Pr[x = n]}$ is a constant ($c$) for the whole distribution.  So what is this $c$?  Well, for any $k$, $\sum\limits_{n=1}^{k} c = c\cdot k$.  Therefore, for a hyperfinite $\omega$ this will also be true.  This means that, for a uniform distribution:
$$
\sum\limits_{n=1}^{\omega}\text{Pr}[x=n] = c\cdot \omega = 1 \\
c = \frac{1}{\omega}
$$
Likewise, we can also figure out other features such as the expected value of this distribution.  The expected value is given by:
$$
\mathbb{E}[x] = \sum_{n=1}^\omega Pr[x=n]\cdot n
$$
Since the probability is the constant $c$, it can replace the probability and be moved outside of the summation:
$$
\mathbb{E}[x] = c \sum_{n=1}^\omega n
$$
$\sum\limits_{n=1}^\omega n$ is just an arithmetic series.  Therefore, we can determine its value using the standard arithmetic series formula as $\frac{\omega}{2}(1 + \omega) = \frac{\omega^2 + \omega}{2}$.  Since $c = \frac{1}{\omega}$, this becomes:
$$
\mathbb{E}[x] = \frac{1}{\omega} \frac{\omega^2 + \omega}{2} = \frac{\omega}{2} + \frac{1}{2}
$$
So, using hyperreals, although we may not be able to mimic $\mathbb{N}$ precisely, there are similar questions that can be asked and answered using the system. This answers the question as you posed it at the end of your post, but not necessarily as you posed it in the beginning.
