<s>[INST] A sum involving a ratio of two binomial factors. Let $a\ge 0$, $a_1\ge 0$ ,$b \ge 0$ and $b_1\ge0$ be real numbers subject to $1+b+a_1-b_1-a >0$. Let $m$ be a positive integer. Then using methods similar to those in Another sum involving binomial coefficients.  I have shown that the following identity holds:
\begin{eqnarray}
\sum\limits_{i=0}^{m-1}\frac{\binom{i+a}{b}}{\binom{i+a_1}{b_1}} &=& 
\frac{\binom{a+m}{b+1}}{\binom{a_1+m}{b_1}} 
F_{3,2}\left[
\begin{array}{rrr} 1 & b_1 & 1+ a +m \\ 2+b & 1+a_1+m \end{array}; 1
\right]\\
&-&
\frac{\binom{a}{b+1}}{\binom{a_1}{b_1}} 
F_{3,2}\left[
\begin{array}{rrr} 1 & b_1 & 1+ a  \\ 2+b & 1+a_1 \end{array}; 1
\right]
\end{eqnarray}
Now, what is the asymptotic behaviour of this sum when $m \rightarrow \infty$ ?
 [/INST] We need to analyze a certain rational function of $m$. We decompose that rational function into simple fractions and we have:
\begin{equation}
\frac{(1+a+m)^{(n)}}{(1+a_1+m)^{(n)}} = 1 + \sum\limits_{l=1}^n \frac{{\mathcal A}_l}{m+a_1+l}
\end{equation}
where ${\mathcal A}_l = (a-a_1-l+1)^{(n)} (-1)^{l-1}/((l-1)!(n-l)!)$. Therefore we have:
\begin{equation}
F_{3,2}\left[
\begin{array}{rrr} 1 & b_1 & 1+a+m\\ 2+b & 1+a_1+m\end{array}; 1
\right]
=
F_{2,1} \left[
\begin{array}{rr} 1 & b_1 \\ 2+b\end{array};1
\right]
+
\sum\limits_{l=1}^\infty \frac{(-1)^{l-1}}{(l-1)!} 
\left.
\frac{d^l}{d x^l}
 F_{2,1} \left[
\begin{array}{rr} b_1 & a-\theta-l+1 \\ 2+b\end{array};x
\right]
\right|_{x=1} \cdot \frac{1}{m+\theta+l}
\end{equation}
We use the ``generalized Gauss' theorem'':
\begin{equation}
\left.
\frac{d^l}{d x^l}
F_{2,1}\left[
\begin{array}{rr} a & b \\ c \end{array};x
\right]
\right|_{x=1} = \frac{a^{(l)} b^{(l)} (-1)^l}{(1+a+b-c)^{(l)}} \cdot \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)}
\end{equation}
and we easily get:
\begin{eqnarray}
F_{3,2}\left[
\begin{array}{rrr} 1 & b_1 & 1+a+m\\ 2+b & 1+a_1+m\end{array}; 1
\right]
=\\
\frac{1+b}{1+b-b_1} + 
\frac{(a-a_1) b_1}{1+a_1+m} \frac{\Gamma(2+b) \Gamma(1-a+a_1+b-b_1)}{\Gamma(2+b-b_1) \Gamma(2-a+a_1+b) } \cdot \\
F_{3,2}\left[
\begin{array}{rrr} 1-a+a_1 & 1+b_1 & 1+a_1+m \\ 2-a+a_1+b & 2+a_1+m\end{array};1
\right]
\end{eqnarray}
The first term on the right hand side is the large $m$ limit and the second term is of the order of $O(1/m)$ . Repeating the above procedure $p$ times we obtain a following large-$m$ expansion of the hypergeometric function:
\begin{eqnarray}
F_{3,2}\left[
\begin{array}{rrr} 1 & b_1 & 1+a+m\\ 2+b & 1+a_1+m\end{array}; 1
\right]
=\\
\frac{1+b}{1+b-b_1} + 
\sum\limits_{l=1}^p \frac{(a-a_1)_{(l)} b_1^{(l)} (1+b)}{(1+a_1+m)^{(l)} (1+b-b_1)_{(l+1)}} + 
\frac{(a-a_1)_{(p+1)} b_1^{(p+1)}}{(1+a_1+m)^{(p+1)}}
\frac{\Gamma(2+b) \Gamma(1-a+a_1+b-b_1)}{\Gamma(2+b-b_1) \Gamma(p+2-a+a_1+b)}\cdot \\
F_{3,2}\left[
\begin{array}{rrr} p+1-a+a_1 & p+1+b_1 & p+1+a_1+m\\ p+2-a+a_1+b & p+2+a_1+m\end{array}; 1
\right]
\end{eqnarray}
</s>