= Solution
For occupation much smaller than $2S$, expand the square root in the <Holstein–Primakoff transformation>. To leading order $S_n^+=\sqrt{2S}\,a_n+O(S^{-1/2})$, $S_n^-=\sqrt{2S}\,a_n^\dagger+O(S^{-1/2})$ and $S_n^z=S-a_n^\dagger a_n$. Consequently
$$
\mathbf S_n\cdot\mathbf S_{n+1}
=S^2-S(\hat n_n+\hat n_{n+1})
+S(a_n^\dagger a_{n+1}+a_{n+1}^\dagger a_n)+O(S^0).
$$
Periodic boundaries count each occupation in two bonds. The <linear spin-wave approximation> is therefore
$$
\boxed{H=-NJS^2+JS\sum_n[2a_n^\dagger a_n-a_n^\dagger a_{n+1}-a_{n+1}^\dagger a_n]+O(S^0).}
$$
The displayed fluctuation Hamiltonian is bilinear and conserves <boson> number. The omitted terms are magnon interactions, quartic at the first omitted order. With occupation held of order one, they are smaller by $1/S$ than the order-$S$ quadratic terms. The polarized vacuum has no zero-point depletion in this ferromagnetic expansion.
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