= Solution
The <Holstein–Primakoff transformation> maps the finite <spin> basis $|S,m\rangle$ to bosonic <Fock states> $|n\rangle$ with $n=S-m$. Thus $n=0$ is the fully polarized reference state, and creating a <boson> lowers $S^z$ by one. The physical subspace has the <Holstein–Primakoff occupation constraint> $0\le n\le2S$, not the whole infinite oscillator <Hilbert space>.
The specified ordering of the square root gives
$$
S^+|n\rangle=\sqrt{n(2S-n+1)}\,|n-1\rangle,\qquad S^-|n\rangle=\sqrt{(n+1)(2S-n)}\,|n+1\rangle.
$$
These are the exact <spin ladder operator> matrix elements; in particular, the lowering amplitude vanishes at $n=2S$. The underlying oscillator obeys its usual <canonical commutation relation> on the enlarged space, while the <spin operators> preserve the constrained physical subspace. Projecting the oscillator operators themselves to a finite space would change their commutator at its upper endpoint.
Expanding $\sqrt{2S-n}=\sqrt{2S}(1-n/(4S)+\cdots)$ converts an ordered <spin> problem to approximately free <magnons>, with interactions in higher orders of $1/S$. Two complications are the nonpolynomial, order-sensitive square roots, which generate interacting <bosons>, and loss of the occupation bound when a truncated expansion uses the full bosonic <Fock space>. Large depletion, high <temperature>, or soft modes near a <phase transition> can invalidate the assumption $\langle n\rangle\ll S$ even when $S$ is large.
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