= Solution
The bosonic <coherent-state resolution of identity> gives the ordinary trace by insertion into a number-state basis. For fermions, the <fermionic coherent-state trace> requires a sign in the bra. In one mode, for an even operator $A$,
$$
\langle-\psi|A|\psi\rangle=A_{00}-\bar\psi\psi A_{11},\qquad \int d\bar\psi\,d\psi\,e^{-\bar\psi\psi}\langle-\psi|A|\psi\rangle=A_{00}+A_{11}.
$$
Without that sign the result would be the <supertrace> $A_{00}-A_{11}$. Applying this identity mode by mode to the even operator $A=e^{-\beta(H-\mu N)}$ gives the <grand canonical partition function>
$$
\boxed{Z=\int d[\bar\psi,\psi]e^{-\sum_n\bar\psi_n\psi_n}\langle\zeta\psi|e^{-\beta(H-\mu N)}|\psi\rangle}.
$$
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