 Solution

Measure energy from the single-particle ground state and write $z=e^{\mu /(k_{B}T)}\le 1$. The Bose-Einstein distribution gives the number in excited levels asHere ${\mathrm{Li}}_s(z)={\sum }_{j\ge 1}{z}^j/j^s$, obtained directly from the supplied integral identity. At the Bose-Einstein condensation threshold, $z\uparrow 1$ and the excited states are just able to hold all $N$ particles. ThusThe ground-state population is treated separately from the continuum density of states.