One-mode CCR/CAR and a vector outside the number-operator domain

For bosons, the ladder formulas [F1] give $a(e)a^{†}(e)\mathrm{\mid }n⟩=(n+1)\mathrm{\mid }n⟩$ and $a^{†}(e)a(e)\mathrm{\mid }n⟩=n\mathrm{\mid }n⟩$ for every $n\ge 0$, where the second identity at $n=0$ uses $a(e)\mathrm{\Omega }=0$. Their difference is $\mathrm{\mid }n⟩$; linear extension proves $[a(e),a^{†}(e)]=I$ on the finite-particle core. For fermions, $c(e)\mathrm{\Omega }=0$, $c^{†}(e)\mathrm{\Omega }=\mathrm{\mid }1⟩$, $c(e)\mathrm{\mid }1⟩=\mathrm{\Omega }$, and $c^{†}(e)\mathrm{\mid }1⟩=0$. Hence $\{c(e),c^{†}(e)\}$ fixes both basis vectors, while $c(e{)}^2=(c^{†}(e){)}^2=0$; these identities hold on the whole two-dimensional fermionic Fock space.

The squared norm of $\mathrm{\Psi }$ is ${\sum }_{n\ge 0}(n+1{)}^{-3}$, which is finite: on each block $2^k\le n+1<2^{k+1}$ there are $2^k$ terms, each at most $2^{-3k}$, so the block sum is at most $2^{-2k}$ and the geometric series of these bounds converges. By [F3], $\mathrm{\Psi }\in D(N)$ would require ${\sum }_{n\ge 1}{n}^2\mathrm{/}(n+1{)}^3<\mathrm{\infty }$. Since $n+1\le 2n$ for $n\ge 1$, $n^2\mathrm{/}(n+1{)}^3\ge 1\mathrm{/}(8n)$. The harmonic series diverges because each block $2^k\le n<2^{k+1}$ has $2^k$ terms at least $2^{-(k+1)}$, so every block sum is at least $1\mathrm{/}2$. Therefore $\mathrm{\Psi }\mathrm{\notin }D(N)$.

By [F2], the annihilation-domain sum for $\mathrm{\Psi }$ is ${\sum }_{n\ge 1}n\mathrm{/}(n+1{)}^3\le {\sum }_{n\ge 1}{n}^{-2}$, and the creation-domain sum is ${\sum }_{n\ge 0}(n+1{)}^{-2}$. Each is finite: the $k$th dyadic block of ${\sum }_{n\ge 1}{n}^{-2}$ has at most $2^k$ terms each at most $2^{-2k}$, so its block sum is at most $2^{-k}$; the shifted series has the same convergence. Thus $\mathrm{\Psi }\in D(a(e))\cap D(a^{†}(e))$ while step 1.2 proves $\mathrm{\Psi }\mathrm{\notin }D(N)$. This is a domain distinction for the stated operators, not a physical or empirical claim.