Eq.(\[gineq1\]) $$\label{roexp}\rho ({\bf k},\omega )=\frac 1{K_{{\bf k}}}N^{-1}\sum_{{\bf q}
}M^2({\bf k},{\bf q})\left[ (1+\nu _{{\bf q}})A_p({\bf k}-{\bf q},z-\omega _{
{\bf q}})+\nu _{{\bf q}}A_p({\bf k}-{\bf q},z+\omega _{{\bf q}})\right] .$$

If we shall put the expression for $\rho ({\bf k},\omega )$ in Eq.(\[polab\]) then the coefficients $a_{n+1}$ and $b_{n+2}$ will be expressed through the integrals of the form $$\label{intw}\int_{-\infty }^\infty (\omega \pm \omega _{{\bf q}})^m\Pi
_i^2A_p({\bf k}-{\bf q},\omega )dw,\quad i\leq n,\quad m=0,1,$$

Now, the trick is that the polynomials in $\omega $ in the integrals (\[intw\]) have the degree less than or equal $2n+1$. As it was proved by Nex [@rm5] such integrals may be expressed through the coefficients $ \{a_0,\ldots ,a_n,b_0\ldots 
,b_n\}$.The details of such a procedure are presented in the Appendix. So, it turns out that in SCBA we can recursively calculate pairs of coefficients $a_{n+1},b_{n+1}$and to obtain $\Sigma ({\bf
k},\omega )$ in the continued fraction form. Of course we must calculate the coefficients at all the chosen ${\bf k}+{\bf q}$ points in the first Brillouin zone simultaneously. Below the chosen points correspond to a lattice of $32\times 32$ unit cells.The presented procedure provides the opportunity to avoid the iterative solution of Eq.(\[gineq\]) for complex energies.

Termination of the continued fraction
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