= Solution
Write $y=g\mu_BB/(k_BT)$. Summing the <Boltzmann factors> gives the one-atom <partition function>
$$
Z_1=\sum_{m=-J}^Je^{-ym}=\frac{\sinh[(J+1/2)y]}{\sinh(y/2)}.
$$
Since $\langle m\rangle=-\partial_y\log Z_1$, the <Brillouin magnetization formula> is
$$
\boxed{\langle\mu_z\rangle=g\mu_B\left[(J+\tfrac12)\coth((J+\tfrac12)y)-\tfrac12\coth(y/2)\right],\qquad M=N\langle\mu_z\rangle.}
$$
Here $M$ is the total <magnetic moment> used by the question; the <magnetization> per unit volume is $M/V$. For independent distinguishable atoms, $Z_N=Z_1^N$, with no indistinguishability factor.
Putting $J=1/2$ yields
$$
\boxed{M=Ng\mu_B[\coth y-\tfrac12\coth(y/2)]=\frac{Ng\mu_B}{2}\tanh(y/2).}
$$
The apparent singularities of the hyperbolic cotangents at $B=0$ cancel; the finite sum makes the zero-field limit explicit.<Statistical Physics by David Tong>[Chapter 1: canonical partition functions and independent two-state systems]{ref}
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