= Solution
Let $u_p$ be the longitudinal displacement of atom $p$. Translational invariance and a symmetric harmonic pair force give the <long-range harmonic-chain dispersion> equations
$$
M\ddot u_p=\sum_{n\ne0}C_{|n|}(u_{p+n}-u_p)
=\sum_{n>0}C_n(u_{p+n}+u_{p-n}-2u_p).
$$
Use a traveling <normal mode> $u_p=u_0e^{i(kpa-\omega t)}$. Dividing by its nonzero amplitude gives
$$
-M\omega^2=\sum_{n>0}C_n(e^{inka}+e^{-inka}-2)
=-4\sum_{n>0}C_n\sin^2(nka/2).
$$
Hence
$$
\boxed{\omega(k)=\frac2{\sqrt M}\left[\sum_{n>0}C_n\sin^2(nka/2)\right]^{1/2}}.
$$
Positive pair constants give a nonnegative squared <frequency> and a uniform-translation zero mode. More general signed force constants require checking stability of the entire sum. The <frequency> is even in $k$ and periodic with $2\pi/a$, reflecting reciprocal-lattice equivalence. The force sum must converge in the model. A finite ordinary acoustic slope additionally requires convergence of $\sum n^2C_n$, in which case $v_s=a\sqrt{\sum n^2C_n/M}$. Both finite-range examples below satisfy this condition.
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