= Solution
Put
$$
x=\beta\mu B.
$$
The one-spin partition function is
$$
z_1=e^{-x}+1+e^x=1+2\cosh x.
$$
The spins are independent, so the <spin-1 paramagnet> has
$$
Z=z_1^N
$$
and free energy
$$
\boxed{
F=-\frac N\beta\log(1+2\cosh(\beta\mu B))}.
$$
The canonical heat capacity can be obtained from the energy variance:
$$
C=k_B\beta^2\frac{\partial^2}{\partial\beta^2}\log Z.
$$
Because
$$
\frac{d^2}{dx^2}\log(1+2\cosh x)
=\frac{2(\cosh x+2)}{(1+2\cosh x)^2},
$$
we obtain
$$
\boxed{
C=Nk_B(\beta\mu B)^2
\frac{2(\cosh(\beta\mu B)+2)}
{(1+2\cosh(\beta\mu B))^2}}.
$$
At high temperature, $x\to0$, and hence
$$
\boxed{-\beta F\longrightarrow N\log3},
\qquad
\boxed{C\sim\frac23Nk_B(\beta\mu B)^2\longrightarrow0}.
$$
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