I have just asked the calculation of the following summation [see here](https://mathoverflow.net/questions/431474/how-to-calculate-this-summation-sum-k-0m-ak-bm-k-n-1-choose-k-n-2-ch) $$S(a,b,m,n\_1,n\_2)=\sum\_{k=0}^m a^k b^{m-k} {n\_1\choose k} {n\_2\choose m-k}, $$
which is motivated by the calculation of the following limit (if exist)
$$L\_0=\lim\_{m \to \infty}\frac{\sqrt{ab}S(a,b,m-1,n\_1-1,n\_2-1)}{S(a,b,m,n\_1,n\_2)}$$
where $a>0, b>0, \frac{m}{n\_1},\frac{m}{n\_2}$ are kept fixed.


**My question is**: Whether the above limit exists ($a\neq b$)? If it exists, then how to calculate it? When $a=b$, the result is simple $\lim\_{m \to \infty}\frac{a\cdot a^{m-1} {n\_1+n\_2-2\choose m-1}}{a^{m} {n\_1+n\_2\choose m}}=\lim\_{m \to \infty}\frac{m(n\_1+n\_2-m)}{(n\_1+n\_2)(n\_1+n\_2-1)}$ and this limit obviously exists, where we have used $S(a,b,m,n\_1,n\_2)=a^m {n\_1+n\_2\choose m}$ for $a=b$.


Despite the nice comments and answers in the [previous question](https://mathoverflow.net/questions/431474/how-to-calculate-this-summation-sum-k-0m-ak-bm-k-n-1-choose-k-n-2-ch), I still don't know how to calculate the limit.



How to calculate this limit (if exist)?
$\newcommand\ka\kappa\newcommand{\de}{\delta}\newcommand\ep\varepsilon$The problem obviously reduces to this one: find
\begin{equation}
 L:=\lim\_{m\to\infty}\frac{S(c,m,n\_1,n\_2)}{S(c,m-1,n\_1-1,n\_2-1)},
\end{equation}
where $c:=a/b\ne1$ and
\begin{equation}
 S(c,m,n\_1,n\_2):=\sum\_{k=0}^m c^k A\_k(m,n\_1,n\_2), 
\end{equation}
where
\begin{equation}
 A\_k(m,n\_1,n\_2):=\binom{n\_1}k \binom{n\_2}{m-k}. 
\end{equation}


In accordance with the linked post, assume that $m<n\_1$ and $m<n\_2$, so that $m\to\infty$,
\begin{equation}
 n\_1/m\to\nu\_1\in[1,\infty),\quad n\_2/m\to\nu\_2\in[1,\infty). 
\end{equation}


By considering the ratios $r\_k:=c^{k+1}A\_{k+1}(m,n\_1,n\_2)/(c^k A\_k(m,n\_1,n\_2))$, one sees that
\begin{equation}
 S(c,m,n\_1,n\_2)\sim\sum\_{k\sim \ka m} c^k A\_k(m,n\_1,n\_2), 
\end{equation}
where
\begin{equation}
 \ka:=\frac{c \nu \_1-\sqrt{\left(c \nu \_1+c+\nu \_2-1\right){}^2-4 (c-1) c \nu \_1}+c+\nu \_2-1}{2 (c-1)}\in(0,1). 
\end{equation}
(More specifically, note that $r\_k\ge1$ for $k\le k\_\*$ and $r\_k\le1$ for $k\ge k\_\*$, for a certain integer $k\_\*\sim\ka m$. Next, note that for each real $\ep>0$ there is some $\de>0$ such that for all large enough $m$ and all $k=0,\dots,m-1$ we have the implication
if $k/m-\ka\ge\de\implies r\_k<1-\ep$. So, for integers $k\ge(\ka+3\de)m$ we will have $c^k A\_k(m,n\_1,n\_2)\le(1-\ep)^{\de m}c^{k\_\*+1} A\_{k\_\*+1}(m,n\_1,n\_2)$ and hence  

$\sum\_{k\ge (\ka+3\de)m} c^k A\_k(m,n\_1,n\_2)\le m (1-\ep)^{\de m} c^{k\_\*+1} A\_{k\_\*+1}(m,n\_1,n\_2)
=o(\sum\_{|k-\ka m|\le3\de m} c^k A\_k(m,n\_1,n\_2))$.


Similarly, $\sum\_{k\le (\ka-3\de)m} c^k A\_k(m,n\_1,n\_2)
=o(\sum\_{|k-\ka m|\le3\de m} c^k A\_k(m,n\_1,n\_2))$.


Finally here, note that we can choose $\de>0$ to be however small.)


Also, for $k\sim \ka m$,
\begin{equation}
 \frac{A\_k(m-1,n\_1-1,n\_2-1)}{A\_k(m,n\_1,n\_2)}
 =\frac{(m-k)(n\_1-k)}{n\_1 n\_2}\sim\frac{(1-\ka )(\nu\_1-\ka )}{\nu\_1\nu\_2}. 
\end{equation}
So,
\begin{equation}
 L=\frac{(1-\ka )(\nu\_1-\ka )}{\nu\_1\nu\_2}.
\end{equation}


