I am trying to approximate $$\frac{p\left( n+a,\bar\lambda T \right){{\left( 1-p \right)}^{n}}_{1}{{F}_{1}}\left( n+1,n+a+1,\lambda T \right)}{p\left( n+a,\bar\lambda T \right)}={{\left( 1-p \right)}^{n}}{{\text{ }}_{1}}{{F}_{1}}\left( n+1,n+a+1,\lambda T \right)$$
If I use the asymptotic behavior of confluent function, I can rewrite it as:
$${{\left( 1-p \right)}^{n}}{{\text{ }}_{1}}{{F}_{1}}\left( n+1,n+a+1,\lambda T \right)\approx \frac{b\left( n+a,a,p \right)}{p\left( a,\lambda T \right){{p}^{a}}}$$
That didn't help me, but maybe gives you an idea.
I could rewrite numerator differently so that I get the following ratio: $$\frac{\sum\limits_{k=0}^{\infty }{p\left( n+a+k,\bar\lambda T \right)}b\left( n+k,k,p \right)}{p\left( n+a,\bar\lambda T \right)}$$ It didn't help me either but i though it might give you an idea.
when $(n>0)$ $(a>0)$ $(0<p<1)$ $(\bar\lambda>\lambda)$ $$b\left( n+k,k,p \right)=\left( \begin{matrix} n+k \\ k \\ \end{matrix} \right){{\left( p \right)}^{k}}{{\left( 1-p \right)}^ {n}}$$ $$p\left( n+a,\lambda T \right)=\frac{{{e}^{-\lambda T}}{{\left( \lambda T \right)}^{n+a}}}{\left( n+a \right)!}$$
Thanks for your help!