A sum involving a ratio of two binomial factors.

Question

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 $m \rightarrow \infty$ ?

Answer 1

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}