In a Fermi liquid the momentum distribution shows a jump at the Fermi surface, i.e.\begin{equation}\langle n_{k_F-\delta k} - n_{k_F+\delta k}\rangle = Z_{k_F}\end{equation}with $Z_k$ the strength of the pole in the Green's function $G\left(k,\omega\right) = \frac{Z_{k}}{\omega - \epsilon_k + i \delta \mathrm{sgn}\left(k-k_F\right)} + g\left(k,\omega\right)$ and where $g$ is supposed to be regular near $k_F$. By relating $\langle n_k \rangle$ to the Green's function and closing the contour in the complex plane this is straightforward to show - the integral over $g$ cancels and the only remaining contribution is $Z_k$ for $\langle n_{k_F-\delta k}\rangle$.
However I don't see how this is reflected in writing the spectral representation (I'm using Coleman's book)\begin{equation}G\left(k,\omega\right) = \sum_\lambda \frac{|M_{\lambda} \left(k\right)|^2}{w- \epsilon_{\lambda} + i \delta \mathrm{sgn} \left(\epsilon_\lambda\right)}\end{equation}which would contribute $\sum_{\lambda'} |M_{\lambda'} \left(k\right)|^2$ to $\langle n_k \rangle$, where $\lambda'$ is such that $\epsilon_{\lambda'} < 0$. This suggests
\begin{equation}\langle n_{k_F-\delta k} - n_{k_F+ \delta k} \rangle = \sum_{\lambda'} |M_{\lambda'}\left(k_F-\delta k\right)|^2 - |M_{\lambda'} \left(k_F+ \delta k\right)|^2.\end{equation}
I wonder if we can actually use the spectral representation, because it seems to suggest a series of sharp poles, but I would expect them to broaden into a continuum especially away from the Fermi surface.
So: can we actually use the spectral representation written this way? If so, how does the last equation simplify to $Z_{k_F}$?
