Where $\frac{1}{\infty}$ and $\frac{\infty}{\infty}$ are both undefined terms that generally lead to nonsense, I'm wondering if we can assert that...:
$$\frac{1+1+1+\cdots}{1+1+1+\cdots} = 1$$
...or even
$$\frac n {1+1+1+\cdots} = 0$$
...for any natural number $n$. In terms of infinite probabilities, it might make sense to recognize something like this because then we can differentiate:
$$\frac{1+1+1+\cdots}{2+2+2+\cdots} = \frac 1 2\left(\frac{1+1+1+\cdots}{1+1+1+\cdots}\right) = \frac 1 2.$$
I ask because I independently developed the following proof for selecting a natural number using a method where all natural numbers have an equal, however undefined, probability of being selected.
Introduction:
It is helpful to consider a simple observation pertaining to finite sets in order to get started. Let $A = \{1, 2\}$ and $B = \{3, 4, 5, 6\}$. Let $f(x) = 1$ if $x$ is odd and $f(x) = 2$ if $x$ is even. Then, function $f$ is a surjection from $B$ onto $A$ that is “uniform” in the sense that selecting an element $x \in B$ uniformly at random will result in the selection of $f(x) \in A$ uniformly at random as well. Note that in order to do this, we have effectively partitioned $B$ into subsets the same size as $A$ so that we could biject those subsets with $A$. Likewise, this work shows how we can partition $[ \,0, 1) \,$ into countable sets that are then bijected with the natural numbers. Consideration is then given to whether selecting an element of $[ \,0, 1) \,$ uniformly at random allows for the selection of a natural number uniformly at random as well.
Definitions:
Let $V^{( \,0.5, 1) \,}$ be a set containing one and only one element from each Vitali equivalence class on the interval $( \, 0.5, 1 ) \,$ (Vitali equivalence classes are equivalence classes of the real numbers that partition $\mathbb{R}$ under the relation $x \equiv y \iff ( \, \exists q \in \mathbb{Q} ) \, ( \, x - y = q ) \,$). The axiom of choice allows for such a selection.
For any real number $r$, let $d(r)$ equal the one and only one element $v \in V^{( \,0.5, 1) \,}$ such that $r - v \in \mathbb{Q}$.
Let $h : \mathbb{N} \longmapsto \mathbb{Q} [ \, 0, 1 ) \,$ (here $\longmapsto$ denotes a bijection).
Let $x$ be an element of $[ \,0, 1) \,$ selected uniformly at random.
Let $k(x) = \begin{cases} h^{-1}(x - d(x) + 0.5) && x \geq d(x) - 0.5 \\ h^{-1}((x + 1) - d(x) + 0.5) && x < d(x) - 0.5 \end{cases}$.
For each natural number $1, 2, 3, …$, let $V^1, V^2, V^3, \ldots$, respectively, be the sets such that $( \, \forall x \in V^{n} ) \, ( \,k(x) = n ) \,$. We then have $V^{( \,0.5, 1) \,} = V^{h^{-1}(0.5)}$, for example. Each $V^{n}$ will be a Vitali set by definition with the collection $\{ V^{n} : n \in \mathbb{N} \}$ forming a partition of $[ \,0, 1) \,$.
Comments on Uniformity:
A uniform distribution is a concept of translation invariance. For example, if $S$ is a measurable set, we may want the probability of $S$ to be the same as the probability of $\{y : y = z + n, z \in S \}$ for each natural number $n$. In the case of function $k$ over the domain $[ \,0, 1) \,$, however, we end up mapping each element of each non-measurable Vitali set $V^n$ to a distinct natural number $n$. Where $a, b \in \mathbb{N}$, it is easy to see that the probability of $x$ falling within $V^a$ is equal to the probability of $x$ falling in $V^b$, but we cannot rely on a Lebesgue measure as a means of establishing probability or creating any sort of cumulative distribution function on $\mathbb{N}$. The probability of selecting any given natural remains undefined.
