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.

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.