Can one calculate the following operator? Summary
I recently defined some numbers which obey multiplication but not addition. To my surprise after some heuristic manipulations (ignoring convergence), it seems I can express the creation and annihilation operators from quantum mechanics in terms of these numbers! I am now interested in these numbers in their own right.
Question
Using the below arguments how does one calculate the $( \sum_{0 < R \leq 1} \hat R)^\dagger | \phi \rangle$  where one can define:
$$ I  - (I + \underbrace{(\sum_{0 < R < 1}\hat R)^\dagger}_{\hat O} )^{-1} = A^\dagger$$
where $I$ is the identity.
Hat Numbers
We define the following numbers 
$$\hat 1 = |1 \rangle \langle 1 | + |2 \rangle \langle 2 | + |3 \rangle \langle 3 | + \dots $$  
$$\hat 2 = |1 \rangle \langle 2 | + |2 \rangle \langle 4 | + |3 \rangle \langle 6 | + \dots  $$
$$\vdots$$
In general,
$$ \hat n = |1 \rangle \langle n | + |2 \rangle \langle 2n | + |1 \rangle \langle 3n | + \dots $$
We notice the following:
$$ \hat x \hat y = \hat y \hat x = \hat (xy)$$
For example:
$$ \hat 2 \cdot \hat 2 = \hat 4$$
Now we also their can create fractions by taking hermitian conjugate of the hat numbers:
For example,
$$ \hat 2^\dagger =  |2 \rangle \langle 1 | +  |4 \rangle \langle 2 | + |6 \rangle \langle 3 | + \dots$$ 
Then we can define rational numbers now as (again for example): 
$$ \hat {\frac{3}{2}} = \hat 3 \hat 2^\dagger $$
Creation and Annihilation operators
One can define a creation operator:
$$ A^\dagger | n \rangle = | n+1 \rangle$$
In fact, 
$$ A^\dagger = |1  \rangle \langle 2 | + |2  \rangle \langle 3 | + |3  \rangle \langle 4 | +  \dots$$
Now, $\hat 1$ is the identity element and if one allows for the geometric series:
$$(\hat 1 - A^\dagger)^{-1} = \hat 1 + \hat A^{\dagger} + \hat A^{\dagger 2} + \hat A^{\dagger 3} + \dots  $$
One also can express the above series using the number operators:
$$ (\hat 1 - A^\dagger)^{-1} = ( \sum_{0 < R \leq 1} \hat R)^\dagger $$
where $( \sum_{0 < R \leq 1} \hat R)$ represents the sum of all rational hat numbers less than $1$. I don't have a rigorous proof of the above equation but am very confident it is correct (in the sense the LHS and RHS have the bra-ket notation).
 A: I disagree with the previous comments. As Christian says, the sums all converge strongly if we interpret $|x\rangle \langle y|$ as a rank 1 operator. In particular, $\hat{n}$ is a coisometry which takes the orthonormal set $\{|n\rangle, |2n\rangle, |3n\rangle, \ldots\}$ to the standard basis $\{|1\rangle, |2\rangle, |3\rangle, \ldots\}$. The "rational operator" $\frac{\hat{m}}{n}$ takes the basis vector $|km\rangle$ to $|kn\rangle$ when $kn$ is an integer. This is math.
I am not sure exactly what the question is --- maybe whether one can make rigorous the series involving the sum over $0 < R \leq 1$? I don't have time to check it now, but at first blush it looks to me like the series converges strongly to the given answer. Maybe it isn't deep, but I think it's interesting and it looks like research math to me.
Edit: I had a chance to take another look, and the sums do not all converge strongly. The operator $(\hat{1} - A^\dagger)^{-1}$ is unbounded, but the series for it clearly makes sense and is correct on finite linear combinations of basis vectors. So this gets even more interesting.
A: comment
So far it doesn't look like math to me.  Let me guess some math meaning.
$$
|1 \rangle,\qquad |2 \rangle, \qquad |3 \rangle, \qquad \cdots
$$
are an orthonormal basis for a Hilbert space $H$.  And
$$
\langle1 |,\qquad \langle2 |,\qquad \langle3 |,\qquad 
$$
are the biorthogonal basis in the dual space $H^\dagger$.  Then products like
$$
\langle 2 |\;\ |5 \rangle := \langle 2 |5 \rangle
$$
are inner products, so thay are scalars, but things like
$$
|5 \rangle\langle 2 |
$$
are tensors; belonging to a tensor product $H \otimes H^\dagger$.  I guess the set of all $|m \rangle\langle n |$ is an orthonormal basis for $H \otimes H^\dagger$.  
If this is right, then of course
$$
\hat 1 = |1 \rangle \langle 1 | + |2 \rangle \langle 2 | + |3 \rangle \langle 3 | + \dots
$$
does not converge in the Hilbert space sense.  
But as an alternative, maybe we can think of things like
$$
|5 \rangle\langle 2 |
$$
as operators.  Or infinite matrices (with respect to the orthonormal bases chosen).  That is, $|m \rangle\langle n |$ is the matrix where the entry in row $m$ column $n$ is $1$, and all other entries are $0$.  Or maybe the adjoint of that?  Then the other things mentioned may be computed as infinite matrices also.
