= Solution
In the <canonical ensemble>, put $\beta=(k_BT)^{-1}$ and $Z=\sum_j e^{-\beta E_j}$. For a temperature-independent observable $A$, differentiating its normalized expectation gives $\partial_\beta\langle A\rangle=-\langle AE\rangle+\langle A\rangle\langle E\rangle$. Taking $A=E$ yields $\partial_\beta\langle E\rangle=-\operatorname{Var}E$. Since $\partial_\beta=-k_BT^2\partial_T$ at fixed volume,
$$
\boxed{\langle(E-\langle E\rangle)^2\rangle=k_BT^2C_V.}
$$
Differentiating $\langle E^2\rangle-\langle E\rangle^2$ with the same rule gives minus $\langle E^3\rangle-3\langle E\rangle\langle E^2\rangle+2\langle E\rangle^3$, namely minus the third <central moment>. Thus
$$
\boxed{\langle(E-\langle E\rangle)^3\rangle=k_B^2\left[T^4\left(\frac{\partial C_V}{\partial T}\right)_V+2T^3C_V\right].}
$$
These identities exhibit the second and third energy <cumulants> as successive derivatives of $\log Z$.
For the monatomic <ideal gas>, $\langle E\rangle=3Nk_BT/2$, $C_V=3Nk_B/2$ and $\partial_TC_V=0$. The <variance> is $3N(k_BT)^2/2$ and the third <central moment> is $3N(k_BT)^3$. Division by the corresponding mean-energy powers gives
$$
\boxed{\frac{\langle(E-\langle E\rangle)^2\rangle}{\langle E\rangle^2}=\frac2{3N},\qquad\frac{\langle(E-\langle E\rangle)^3\rangle}{\langle E\rangle^3}=\frac8{9N^2}.}
$$
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