BEC question about number of excited particles

[Coffee_]

Assuming no interactions, and the energy levels $\epsilon_j$ for a single particle state, the occupation number for this state for a system of particles is given by:
à
$n_j=\frac{1}{exp(\frac{\epsilon_j-µ}{kT}) - 1}$ (1)

(Here comes argument 1)

We conclude that the maximal value that µ can take is equal to $µ=\epsilon_0$. This means that the distribution

$n_j=\frac{1}{exp(\frac{\epsilon_j-\epsilon_0}{kT})- 1}$ (2)

Is an upper limit on any real distribution.

So using (2) we can calculate a statistical upper limit on the expected amount of particles in the excited states $N_{exc}$.

Questions:

1) What explanation should come in the placeholder for 'argument 1' to conclude that µ is between 0 and $\epsilon_0$? I've seen one on the net that already uses the result of having a macroscopic amount of particles in the groundstate to conclude this, which is kind of a circular reasoning.

2) Am I correct in thinking about this in terms of (2) being an upper limit? The problem that I then have is this - the upper limit on the state $\epsilon_0$ is infinite right? How to conclude that this state will only start getting a macroscopic occupation after the number of particles has exceede $N_{exc}$ - if we see an upper limit of infinity here how can we know that we won't get a big occupation in the groundstate from the very beginning?

 

[Cadaei]

1) If µ were > than ϵ0, the occupation number would be negative for ground states.

2) Your question here is unclear to me. As µ -> ϵ0, n_j -> infinity for j = 0, if that's what you mean by "macroscopic number of particles."