Meaning of critical density in radiative transfer
Your misunderstanding is in the second paragraph. If $n&lt;n_c$ then there are insufficient collisions to excite the molecule into the upper level. Thus in equilibrium there are very few excited molecules and since the luminosity is proportional to the number of molecules in the excited level then little radiation is seen.

The critical density is defined as that (temperature-dependent) density at which the rate of collisional excitation to the upper level of a molecular (or atomic, I suppose) transition, equals the rate of radiative de-excitation.

The rate of de-excitation is given by $n_2\ A_{21}$, where $n_2$ is the number density of the molecular species in the excited state and $A_{21}$ is the relevant spontaneous emission coefficient.

The rate of excitation is given by $\gamma_{12} n_1 n$,
where $\gamma_{12}$ is the collision probability (which depends on the speed of molecules in the gas and hence temperature), $n_1$ is the number density of the molecule of interest in the lower energy state and $n$ is the number density of all molecules in the gas.

There is also a rate of collisional de-excitation given by $\gamma_{21}\ n_2 n$. However there must be a relationship between $\gamma_{12}$ and $\gamma_{21}$ because for a gas in thermodynamic equilibrium, we know that 
$$n_2/n_1 = \exp(-E_{21}/k_BT),$$ and so if collisions are in detailed balance then
$$\gamma_{21}\ n_1 \exp(-E_{21}/k_BT) n = \gamma_{12} n_1 n$$
$$ \gamma_{21} = \gamma_{12} \exp(E_{21}/k_BT)$$

OK, so now consider a gas where there are radiative losses from state 2 and allow it to come into equilibrium such that
$$\frac{dn_2}{dt} = 0$$
$$ -A_{21}n_2 - \gamma_{21}n_2 n + \gamma_{12} n_1 n = 0$$
$$ \frac{n_2}{n_1} = \exp(-E_{21}/k_BT)\ \frac{1}{1 + A_{21}/\gamma_{21}n}$$

If now we say that $A_{21} \ll \gamma_{21}n$, then radiative de-excitation is (relatively) negligible compared with collisional de-excitation then the $n_2/n_1$ ratio approaches the level expected by the Boltzmann ratio.

If on the other hand $A_{21} \gg \gamma_{21}n$, then the ratio of $n_2/n_1$ becomes much lower and radiation removes excited molecules faster than collisions create them.

Now we ask how much radiation we are going to see from this gas.

(1) If $A_{21} \ll \gamma_{21}n$, which we can restate as $n \gg n_c$, the "critical density".

The luminosity from the molecules will be 
$$L \propto n_2 A_{21} \simeq n_1 A_{21} \exp(-E_{21}/k_BT)$$
If $k_BT &lt; E_{21}$ this means that $n_1 \gg n_2$ and the luminosity per molecule in the gas has a fixed value that is independent of density $n$. There is just a fixed amount of radiative losses per molecule of the species in question.

(2) If $A_{21} \gg \gamma_{21}n$, which we can restate as $n \ll n_c$, The luminosity from the molecules will be 
$$L \propto n_2 A_{21} \simeq n_1 A_{21} \exp(-E_{21}/k_BT) \frac{\gamma_{21} n}{A_{21}}$$
and so this is smaller than for the first case by a factor of $n/n_c$. So now there is less radiation per molecule of the interesting species because fewer of those molecules are in an excited state.

NB This is a different thing to the "quenching" density for forbidden lines that I refer to in my comment.