Apparently it is to be shown that for electrons under an external magnetic field, in the limit as $B\to 0 $
$$
\chi = \frac{dM}{dB} \approx \frac{n\,\mu^{*^2}}{k\,T}\,\frac{f_{1/2}(z)}{f_{3/2}(z)}
$$
So this is the Pauli paramagnetism problem and I understand the effect and the physics, but cannot seem to get the requested ratio of $f_{\nu}(z)$ functions.
Here is how I have been doing it:
With the grand canonical ensemble formalism I have
$$
\langle N_{\pm} \rangle = \int_0^{\infty} d\epsilon\, g(\epsilon)\,\frac{1}{\exp^{\beta(\epsilon-\mu\pm \mu_B\,B)}}
$$
where
$$
g(\epsilon) = \frac{2}{\sqrt{\pi}}\,\left(\frac{2\,\pi\,m}{h^2}\right)^{3/2}\;V\,\sqrt{\epsilon}
$$
and is the density of states from translational motion.
I then say that
$$
\langle M \rangle = \mu_B\,(\langle N_- \rangle – \langle N_+ \rangle ) = \cdots = Const*V*[f_{1/2}(\beta\,\mu+\mu_B\,B)-f_{1/2}(\beta\,\mu-\mu_B\,B)]
$$
This is fine I think except that when I calculate $\chi$ I will get fermi-dirac functions with -1/2 and not the ratio I need.
$$
\chi = \lim_{B\to 0} \frac{\partial\,M}{\partial\,B} = \cdots ?
$$
Best Answer
I believe that you need to shift not Fermi distribution, but density of states. When electron with parallel spin feels magnetic field it's density of states shifts downward by energy $\mu_B B$.You also need to take integral with other limits. So the real quantities will be: $$ N_+ = \frac{1}{2}\int\limits^{\epsilon_F}_{-\mu_B B}d\epsilon f(\epsilon)D(\epsilon+\mu_B B)\\ N_- = \frac{1}{2}\int\limits^{\epsilon_F}_{\mu_B B}d\epsilon f(\epsilon)D(\epsilon-\mu_B B) $$ Then use equation: $M = \mu_{B}(N_+-N_-)$ and you'll get what you needed.
Information from C.Kittel "Introduction to Solid State Physics".
P.S. You need to evaluate integrals approximately, because you can't analytically evaluate them.