The Bose-Einstein distribution

Particles that are described by symmetric wavefunctions are called bosons. Free of the limitations discussed for fermions, each state can be occupied by any number of bosons. The probability of a state being occupied $n_r$ times is:

$$p_r(n_r) = \frac{1}{\mathcal{Z}^{BE}_r} e^{\beta n_r (\mu-\epsilon_r)},$$ (5.38)

from which we get the Bose-Einstein (BE) single state grand partition function

$$\mathcal{Z}^{BE}_r = \sum_{n_r = 0}^\infty e^{\beta n_r (\mu-\epsilon_r)}.$$ (5.39)

This sum can only converge if $\mu - \epsilon_r < 0$, and since the grand partition function of the gas is equal to the product of the single state partition functions, $\mathcal{Z}^{BE} = \mathcal{Z}^{BE}_1\mathcal{Z}^{BE}_2\dots\mathcal{Z}^{BE}_r\dots$, then we must have $\mu - \epsilon_1 < 0$, where $\epsilon_1$ is the energy of the ground state. For a gas of non-interacting bosons the energy is the kinetic energy and so $\epsilon_1 = 0$, which means that we must have $\mu < 0$. If this is the case,  can be summed, and the result is:

$$\mathcal{Z}^{BE}_r = \frac{1}{1- e^{\beta (\mu-\epsilon_r)}}.$$ (5.40)

The average occupation number is:

$$\bar{n}_r = \frac{1}{\beta}\frac{\partial \ln \mathcal{Z}^{BE}_r}{\partial \mu} = \frac{e^{\beta (\mu-\epsilon_r)}}{1- e^{\beta (\mu-\epsilon_r)}},$$ (5.41)

which we write as:

$$\bar{n}_r = \frac{1}{e^{\beta (\epsilon_r-\mu)}-1}.$$ (5.42)

Note that the ground state occupation number for the BE distribution becomes very large as the chemical potential approaches zero. Similarly to the FD distribution, at energies that are large compared to $k_{\rm B}T$ the average occupation numbers are small and the BE distribution can be approximated by the MB distribution:

$$\bar{n}_r \simeq e^{\beta (\mu - \epsilon_r)}, \quad \quad \epsilon_r - \mu \gg k_{\rm B}T.$$