0
$\begingroup$

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 $m \rightarrow \infty$ ?

$\endgroup$
0
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.