I'm trying to show that for an ideal gas in a uniform gravitational field $\vec{g} = (0, 0, -g)$, confined to a box of height $L$ and base area $A$, the mean potential energy per particle is $$\langle v (\vec{r_i}) \rangle \approx \frac{mgL}{2},$$ provided that $k_B T \gg mgL$.
Thus far I've found that the distribution for the height of a particle is
$$P_z(z_i) = \frac{\beta mg}{1 - \exp(-\beta mgL)} \exp(-\beta mgz_i),$$
and so, the mean potential energy is given by
$$\langle v (\vec{r_i}) \rangle = \langle mgz_i \rangle = \frac{1}{\beta} - \frac{mgL}{\exp(\beta mgL) - 1}.$$
Now, that the mean potential energy should be $mgL/2$ when $k_BT \gg mgL$ seems very intuitive to me, however, I'm having trouble demonstrating this using the expression I've found. Any help would be greatly appreciated!