has the CF expansion of incomplete Gamma function that is written as [@abram] $$\label{gami1}\Gamma (\alpha ,x)=\frac{{\rm e}^{-x}x^\alpha }{x+}\frac{
1-\alpha }{1+}\frac 1{x+}\frac{2-\alpha }{1+}\cdots$$ This circumstance we shall use for the construction of the terminator $
\tilde T_N({\bf k},z)$ for $G({\bf k},z)$ (\[cf2\])

Let us introduce the function $$\label{gami2}\tilde g(\alpha ,x)=-\frac{\Gamma (\alpha ,-x)}{{\rm e}
^x(-x)^\alpha }=1/\left( x-\frac{1-\alpha }{1-\theta _1}\right) ,$$ where $$\label{gami3}\theta _n=n/\left( x-\frac{n+1-\alpha }{1-\theta _{n+1}}\right)
.$$ In order to rewrite the CF (\[gami2\]) in the form analogous to Eq.(\[cf1\]) we denote $$\label{gami4}\frac 1{1-\theta _n}\equiv 1+nt_n.$$ Then it is not difficult to obtain the relations $$\label{gami5}t_n=\frac 1{x-(2n+1-\alpha )-(n+1)(n+1-\alpha )t_{n+1}},$$ so that $$\label{gami6}\tilde g(\alpha ,x)=t_0$$ has the form (\[cf2\]) with the coefficients $\tilde b_0^2=1$ and $$\label{gami7}\tilde a_n=2n+1-\alpha ,\quad \tilde b_n^2=n(n-\alpha ).$$ Comparing Eqs. (\[asab\]) and (\[gami7\]) for large $n$, when $\sqrt{ n(n-\alpha )}\approx n-\alpha /2$, we see that the substitution $$\alpha =-\frac{2\lambda _2}{\lambda 
_1},x=\frac{z+2\lambda _2-\lambda _3+\lambda _1}{\lambda _1}$$ leads to the function $\tilde G$ $$\label{gami8}\tilde G({\bf k},z)=\frac 1{\lambda _1}\tilde g(-\frac{2\lambda
_2}{\lambda _1},\frac{z+2\lambda _2-\lambda _3+\lambda _1}{\lambda _1})$$ that has the same asymptotic as $G({\bf k},z)$ (\[cf1\]). It means that $
\tilde G({\bf k},z)$ can be used as the terminator for $G({\bf k},z)$. , i.e. we can express $\tilde T_{n_0}({\bf k},z)=\tilde b_{n_0+1}t_{n_0+1}$ through $\tilde G({\bf 