A sum expressed by a Kampe de Feriet function.

Question

Let $a_1$,$a_2$, $a_3$ and $b_1$,$b_2$, $b_3$ be real numbers subject to $1+b_1+b_2 - b_3 > 0 $. By generalizing the result from A sum involving a ratio of two binomial factors. we have shown that the following identity holds: \begin{eqnarray} &&S:= \sum\limits_{i=0}^{m-1} \frac{\binom{i+a_1}{b_1} \binom{i+a_2}{b_2}}{\binom{i+a_3}{b_3}} =\\ && \left. \frac{\binom{m+a_1}{1+b_1} \binom{m+a_2}{b_2}}{\binom{m+a_3}{b_3}} \sum\limits_{(q_2,q_3) \in {\mathbb N}_+^2} \binom{q_2+q_3}{q_3} \frac{b_3^{(q_3)} (-b_2)^{(q_2)}}{(2+b_1)^{(q_2+q_3)}} \frac{(1+a_1+m)^{(q_2+q_3)} (1+a_2+m)^{(q_3)}}{(1+a_2-b_2+m)^{(q_2+q_3)}(1+a_3+m)^{(q_3)}} \right|_0^m =\\ && \left. \frac{\binom{m+a_1}{1+b_1} \binom{m+a_2}{b_2}}{\binom{m+a_3}{b_3}} \sum\limits_{q=0}^\infty \frac{(1+a_1+m)^{(q)}(1+a_2+m)^{(q)}b_3^{(q)}}{(1+a_3+m)^{(q)}(1+a_2-b_2+m)^{(q)}(2+b_1)^{(q)}}\\ F_{3,2}\left[ \begin{array}{rrr} -b_2 & -q & -a_3-m-q\\ 1-b_3-q & -a_2-m-q\end{array};1 \right] \right|_0^m \\ \end{eqnarray} The right hand side of the above equation can be formally expressed as a Kampe de Feriet function. It converges whenever $1+b_1+b_2 - b_3 > 0 $. Now, the question is how does the right hand side behave when $m\rightarrow \infty$?

Answer 1

All we need to do is to analyze the large-$m$ behavior of a certain rational function. We decompose the rational function into simple fractions. We have: \begin{equation} \frac{(1+a_1+m)^{(q_3+q_2)}(1+a_2+m)^{(q_3)}}{(1+a_2-b_2+m)^{(q_3+q_2)}(1+a_3+m)^{(q_3)}} = 1+ \sum\limits_{l_1=1}^{q_3} \frac{{\mathcal A}_{l_1}}{(l_1+a_3+m)} + \sum\limits_{l_2=1}^{q_3+q_2} \frac{{\mathcal B}_{l_2}}{(l_2+a_2-b_2+m)} \end{equation} where \begin{eqnarray} &&{\mathcal A}_{l_1} = \\ &&(a_3-a_2+l_1-1)_{(q_3)} (a_3-a_1+l_1-1)^{(q_3+q_2)} \cdot \frac{(-1)^{l_1-1}}{(a_2-b_2-a_3-l_1+1)^{(q_3+q_2)}} \frac{(-1)^{q_2}}{(l_1-1)!(q_3-l_1)!}\\ &&{\mathcal B}_{l_2} = \\ &&(a_2-b_2-a_2+l_2-1)_{(q_3)}(a_2-b_2-a_1+l_2-1)_{(q_3+q_2)} \cdot \frac{(-1)^{l_2-1}}{(a_2-b_2-a_3+l_2-1)_{(q_3)}} \frac{(-1)^{q_3+q_2}}{(l_2-1)!(q_3+q_2-l_2)!} \end{eqnarray} Inserting the above into the right hand side of the identity in question and using A multivariate sum that yields a closed form expression we get : \begin{equation} rhs = \frac{\binom{m+a_1}{1+b_1} \binom{m+a_2}{b_2}}{\binom{m+a_3}{b_3}} \frac{1+b_1}{1+b_1+b_2-b_3} + O\left(\frac{1}{m}\right) \end{equation} We will deal with the higher order corrections later on.

Answer 2

We will compute the sum in question in a slightly different way and only then analyze the large-$m$ behavior of the result. The idea is to take the second binomial factor in the numerator and rewrite it as a linear combination of factors that can be absorbed into the first binomial factor. We start with the following identity: \begin{equation} \tag{1} \binom{i+a_2}{b_2} = \frac{\sin(\pi (a_2+b_2))}{\sin(\pi a_2)} \cdot \sum\limits_{l=0}^\infty \binom{i+a_1+l}{l} \binom{a_1-a_2+b_2}{b_2-l} (-1)^l \end{equation} The identity is a simple consequence of the Chu-Vandermonde identity and the identity $\binom{n}{k} = (-1)^k \binom{k-n-1}{k}$. If we multiply the said identity by $\binom{i+a_1}{b_1}$ we can absorb the terms involving $i$ into the binomial factor and we get: \begin{equation} \tag{2} \binom{i+a_1}{b_1} \binom{i+a_2}{b_2} = \frac{\sin(\pi (a_2+b_2))}{\sin(\pi a_2)} \cdot \sum\limits_{l=0}^\infty \binom{b_1+l}{l} \binom{a_1-a_2+b_2}{b_2-l} (-1)^l \binom{i+a_1+l}{b_1+l} \end{equation} Now we divide both sides of the above equation by $\binom{i+a_3}{b_3}$ and we sum over $i$ using A sum involving a ratio of two binomial factors. . We get: \begin{eqnarray} \tag{3} &&S = \frac{\sin(\pi (a_2+b_2))}{\sin(\pi a_2)} \cdot \sum\limits_{l=0}^\infty \binom{b_1+l}{l} \binom{a_1-a_2+b_2}{b_2-l} (-1)^l \cdot \\ &&\left\{\left. \frac{\binom{a_1+l+m}{b_1+l+1}}{\binom{a_3+m}{b_3}} F_{3,2}\left[ \begin{array}{rrr} 1 & b_3 & 1+a_1+l+m \\ 2+b_1+l & 1+a_3+m\end{array};1 \right] \right|_0^m \right\} \end{eqnarray} So far all the results above are exact. Now we will take the limit $m \rightarrow \infty$ and try to see if we get the same limit as in the first answer to this question. We have: \begin{equation} \tag{4} F_{3,2}\left[\cdots ;1\right] \underset{m\rightarrow \infty}{=} F_{2,1} \left[\begin{array}{rr} 1 & b_3 \\ 2+b_2+l\end{array};1\right]+ O(\frac{1}{m}) = \frac{1+b_1+l}{1+b_1-b_3+l} + O(\frac{1}{m}) \end{equation} Now, inserting $(4)$ into $(3)$, absorbing the $1+b_1+l$ factor into the first binomial coefficient under the sum and then using identity $(2)$ along with the mean value theorem we easily get : \begin{equation} S \underset{m\rightarrow \infty}{=} \frac{\binom{m+a_1}{1+b_1} \binom{m+a_2}{b_2}}{\binom{m+a_3}{b_3}} \cdot \frac{1+b_1}{1+b_1+\theta b_2 - b_3} \end{equation} where $ \theta \in (0,1)$. So as we can see two different approaches lead to the same result.