One way of calculating the critical temperature (or the critical number of particles) is (as seen in Statistical Mechanics, Pathria, p. 184) this:
Starting with the equation:
$\frac{N}{V} =\frac{1}{V}\frac{z}{1-z} + \frac{1}{\lambda^3}g_{\frac{3}{2}}(z)$
where $N$ is the total number of particles, $V$ the volume, $z$ the fugacity, $\lambda$ the thermal wavelength and $g_\frac{3}{2}$ one of the Bose-Einstein functions. One then associates:
$N_0=\frac{z}{1-z}$ and $N_e=\frac{V}{\lambda^3}g_{\frac{3}{2}}(z)$
with the number of particles in the ground and the excited states, respectively. Then one can replace $g_{\frac{3}{2}}(z)$ for its maximum $g_{\frac{3}{2}}(1)=\zeta(\frac{3}{2}) =2.61$ so that $N_e$ has a maximum too:
$N_e<\frac{V}{\lambda^3}\zeta(\frac{3}{2})=N_{cr}$
This means that if $N>N_{cr}$ there has to be particles in the ground state. My question is: why does this also mean that if $N<N_{cr}$ there are no particles in the ground state at all?
I believe it has something to do with the fact that if z is close to unity (so that $N_0=\frac{z}{1-z}$ diverges) then $N_0$ is comparable to $N$ but for $z<1$ $N_0$ is finite so it gives zero contribution in the thermodynamic limit of $N \rightarrow \infty$.