 Solution

The chemical potential is $\mu =(\partial E/\partial N{)}_{S,V}$, equivalently $(\partial F/\partial N{)}_{T,V}$: it is the free-energy cost of adding a particle at fixed temperature and volume. With units $k_B=1$, a large reservoir has entropy $S_R(E_{\mathrm{tot}}-E,N_{\mathrm{tot}}-N)=S_R^0-E/T+\mu N/T$ to first order. Since the number of compatible reservoir states is proportional to its exponential, the grand canonical ensemble assignsFor independent fermions, a one-particle state of energy $\epsilon$ has occupation zero or one. Its partition factor is $1+e^{-(\epsilon -\mu )/T}$, so the mean occupation is the Fermi-Dirac distribution $[e^{(\epsilon -\mu )/T}+1{]}^{-1}$. Multiplication by the given density of states yields $dN=C{\epsilon }^{1/2}d\epsilon /[e^{(\epsilon -\mu )/T}+1]$.

At zero temperature the occupation is a step function, filled below $\mu$ and empty above it, so $\mu ={\epsilon }_F$ and $N=(2C/3){\epsilon }_F^{3/2}$. At fixed particle number, the low-temperature shift of $\mu$ is $O(T^2/{\epsilon }_F)$ and does not affect the leading estimate. In the narrow interval of width $T$ above ${\epsilon }_F$, replace the density of states by $C\sqrt{{\epsilon }_F}$ and put $u=(\epsilon -{\epsilon }_F)/T$. ThenEquivalently, the fraction above the Fermi energy is asymptotic to $(3\log 2/2)T/{\epsilon }_F$.