= Solution
The $1/x$ terms in the <Brillouin function> cancel. Keeping the linear terms gives
$$
B_J(x)=\frac{(2J+1)^2-1}{12J^2}x+O(x^3)=\frac{J+1}{3J}x+O(x^3).
$$
Consequently,
$$
M\simeq\frac{n(g\mu_B)^2J(J+1)}{3k_BT}B.
$$
Use the dimensionless SI <magnetic susceptibility> $\chi=\partial M/\partial H_{\rm applied}$ and $B_{\rm applied}=\mu_0H_{\rm applied}$, as in the source's molecular-field convention. The <Curie law> is
$$
\boxed{\chi=\frac CT,\qquad C=\frac{\mu_0n(g\mu_B)^2J(J+1)}{3k_B}.}
$$
If susceptibility is instead defined as $\partial M/\partial B$, its coefficient is $C/\mu_0$ and is dimensionful. The SI convention above makes the subsequent <Curie–Weiss law> and $\lambda$ consistent.
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