In the following Wiki derivation of the Bose-Einstein distribution, a geometric sum is used to make the following step
$$ \sum_{n=0}^\infty\left (\exp \left (\frac{\mu -\epsilon}{k_B T}\right)\right)^n = \frac{1}{1-\exp\left(\frac{\mu -\epsilon}{k_B T}\right)} $$
but using a geometric series requires the absolute value of the argument to be less than 1.
Can I have some assistance in understanding why
$$ \exp \left(\frac{\mu -\epsilon}{k_B T}\right ) $$
is necessarily less than 1?
The chemical potential of a Boson gas is always negative or (at worst) zero. (See here for why.)
Since the system is considered to be non-interacting (ideal), the Hamiltonian is just the kinetic energy. The lowest energy is thus the $\epsilon_0=0$ state. This means that in general $\epsilon\geq0$, and so $$\exp{\left(\frac{\mu-\epsilon}{k_B T}\right)}\leq\exp{\left(-\frac{\epsilon}{k_B T}\right)}\leq1.$$
(The only case when the sum diverges is when this term is equal to 1, and when this happens the occupation of the lowest energy level diverges, which is in fact what happens in Bose-Einstein condensation.)
