= Solution
{c}
Apply the given small-$x$ expansion with $x=T_E/T$ to the internal energy, including its <zero-point energy>:
$$
\begin{aligned}
U&=3Nk_BT_E\left[\frac12+\frac{T}{T_E}-\frac12+\frac{T_E}{12T}+O(T^{-3})\right]\\
&=3Nk_B\left[T+\frac{T_E^2}{12T}+O(T^{-3})\right].
\end{aligned}
$$
The temperature-independent terms cancel. Taking the derivative gives the <Einstein high-temperature heat-capacity expansion>
$$
C_V=3Nk_B\left[1-\frac{T_E^2}{12T^2}+O(T^{-4})\right].
$$
Thus
$$
\boxed{A_0=3Nk_B,\qquad A_1=0,\qquad A_2=-\frac{Nk_BT_E^2}{4}=-\frac{A_0T_E^2}{12}.}
$$
There is no $1/T$ correction to the <heat capacity> even though the internal energy has a $1/T$ correction.
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