Hypergeometric 2F1 with negative c

Question

I've got this hypergeometric series

$_2F_1 \left[ \begin{array}{ll} a &-n \\ -a-n+1 & \end{array} ; 1\right]$

where $a,n>0$ and $a,n\in \mathbb{N}$

The problem is that $-a-n+1$ is negative in this case. So when I try to use Gauss's identity

$_2F_1 \left[ \begin{array}{ll} a & b \\ c & \end{array} ; 1\right] = \dfrac{\Gamma(c-a-b)\Gamma(c)}{\Gamma(c-a)\Gamma(c-b)}$

I give negative parameters to the $\Gamma$ function.

What other identity can I use?

I'm trying to find a closed form to this: $\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$

Wolfram Mathematica answered this as a closed form: $\frac{2^{-2 a} \Gamma \left(\frac{1}{2} (1-2 a)\right) \binom{a+n-1}{n} \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$

But I would like to have a manual solution with proof.

This is how I got that hypergeometric series:

$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$

$\dfrac{t_{i+1}}{t_{i}} = \frac{\binom{a+i+1-1}{i+1} \binom{a-i+n-2}{-i+n-1}}{\binom{a+i-1}{i} \binom{a-i+n-1}{n-i}} = \frac{(a+i) (n-i)}{(i+1) (a-i+n-1)} = \frac{(a+i) (i-n)}{ (i-a-n+1)(i+1)}$

$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i} = _2F_1 \left[ \begin{array}{ll} a &-n \\ -a-n+1 & \end{array} ; 1\right]$

UPDATE

Thanks to David H, I got closer to the solution.

$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i} = _2F_1 \left[ \begin{array}{ll} a &-n \\ -a-n+1 & \end{array} ; 1\right]$

$\lim\limits_{\epsilon \to0} \frac{\Gamma (-2 a-2 \epsilon +1) \Gamma (-a-n-\epsilon +1)}{\Gamma (-a-\epsilon +1) \Gamma (-2 a-n-2 \epsilon +1)} = \frac{4^{-a} \Gamma \left(\frac{1}{2}-a\right) \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$

As you can see this result is close to the expected $\frac{2^{-2 a} \Gamma \left(\frac{1}{2} (1-2 a)\right) \binom{a+n-1}{n} \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$ formula. But the $\binom{a+n-1}{n}$ factor is still missing and I don't really understand why.

Answer 1

Using this identity to express the hypergeometric function as a Gegenbauer function, and this identity which gives the value of the Gegenbauer function evaluated at $1$, the hypergeometric function in question may then be expressed as a ratio of gamma functions whose arguments are each positive integers. These can be reorganized into binomial terms for a compact final expression:

$$\begin{align} {_2F_1}{\left(a,-n;1-a-n;1\right)} &=\frac{n!}{(a)_{n}}C_{n}^{a}{\left(1\right)}\\ &=\frac{\Gamma{\left(n+1\right)}\,\Gamma{\left(a\right)}}{\Gamma{\left(a+n\right)}}\cdot\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(2a\right)}\,\Gamma{\left(n+1\right)}}\\ &=\frac{\Gamma{\left(a\right)}\,\Gamma{\left(2a+n\right)}}{\Gamma{\left(a+n\right)}\,\Gamma{\left(2a\right)}}\\ &=\frac{\binom{2a+n-1}{2a-1}}{\binom{a+n-1}{a-1}}.\\ \end{align}$$

There's likely a much more direct way to derive this without this absurd detour into exotic special functions, but maybe this response will tide you over until someone more knowledgeable in combinatorial comes around. =)

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Edit:

Here's a much nicer way of evaluating the sum using beta function machinery.

$$\begin{align} s{(a,n)} &=\sum_{k=0}^{n}\binom{a+k-1}{k}\binom{a+n-k-1}{n-k}\\ &=\sum_{k=0}^{n}\frac{\Gamma{\left(a+k\right)}}{\Gamma{\left(k+1\right)}\,\Gamma{\left(a\right)}}\cdot\frac{\Gamma{\left(a+n-k\right)}}{\Gamma{\left(n-k+1\right)}\,\Gamma{\left(a\right)}}\\ &=\frac{1}{\Gamma{\left(a\right)}^2}\sum_{k=0}^{n}\frac{\Gamma{\left(a+k\right)}}{\Gamma{\left(k+1\right)}}\cdot\frac{\Gamma{\left(a+n-k\right)}}{\Gamma{\left(n-k+1\right)}}\\ &=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(a\right)}^2\,\Gamma{\left(n+1\right)}}\sum_{k=0}^{n}\frac{\Gamma{\left(n+1\right)}}{\Gamma{\left(k+1\right)}\,\Gamma{\left(n-k+1\right)}}\cdot\frac{\Gamma{\left(a+k\right)}\,\Gamma{\left(a+n-k\right)}}{\Gamma{\left(2a+n\right)}}\\ &=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(a\right)}^2\,\Gamma{\left(n+1\right)}}\sum_{k=0}^{n}\binom{n}{k}\cdot\operatorname{B}{\left(a+k,a+n-k\right)}\\ &=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(a\right)}^2\,\Gamma{\left(n+1\right)}}\sum_{k=0}^{n}\binom{n}{k}\int_{0}^{1}t^{a+k-1}(1-t)^{a+n-k-1}\,\mathrm{d}t\\ &=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(a\right)}^2\,\Gamma{\left(n+1\right)}}\int_{0}^{1}t^{a-1}(1-t)^{a-1}\sum_{k=0}^{n}\binom{n}{k}t^{k}(1-t)^{n-k}\,\mathrm{d}t\\ &=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(a\right)}^2\,\Gamma{\left(n+1\right)}}\int_{0}^{1}t^{a-1}(1-t)^{a-1}\cdot1\,\mathrm{d}t\\ &=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(a\right)}^2\,\Gamma{\left(n+1\right)}}\operatorname{B}{\left(a,a\right)}\\ &=\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(2a\right)}\,\Gamma{\left(n+1\right)}}\\ &=\binom{2a+n-1}{2a-1}.~~\blacksquare\\ \end{align}$$

Answer 2

Here I post the full solution

The problem: we are looking for the closed form of this sum:

$\sum\limits_{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$

The first term of the sum for $i=0$ is $\binom{a+n-1}{n}$

The ratio two consecutive terms:

$\dfrac{t_{i+1}}{t_i} = \dfrac{P(i)}{Q(i)} = \dfrac{\binom{a+i}{i+1} \binom{a-i+n-2}{-i+n-1}}{\binom{a+i-1}{i} \binom{a-i+n-1}{n-i}} = \dfrac{(a+i) (n-i)}{(i+1) (a-i+n-1)} = \dfrac{(i+a)(i-n)}{(i-a-n+1)(i+1)}$

So our hypergeometric series can be described this way:

$\, _2F_1(a,-n;-a-n+1;1)$

Using this and this identities:

$\begin{align} {_2F_1}{\left(a,-n;1-a-n;1\right)} &=\frac{n!}{(a)_{n}}C_{n}^{a}{\left(1\right)}\\ &=\frac{\Gamma{\left(n+1\right)}\,\Gamma{\left(a\right)}}{\Gamma{\left(a+n\right)}}\cdot\frac{\Gamma{\left(2a+n\right)}}{\Gamma{\left(2a\right)}\,\Gamma{\left(n+1\right)}}\\ &=\frac{\Gamma{\left(a\right)}\,\Gamma{\left(2a+n\right)}}{\Gamma{\left(a+n\right)}\,\Gamma{\left(2a\right)}}\\ &=\frac{\binom{2a+n-1}{2a-1}}{\binom{a+n-1}{a-1}}.\\ \end{align}$

The final solution:

$\begin{align}\sum\limits_{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i} &= \binom{a+n-1}{n} \cdot \, _2F_1(a,-n;-a-n+1;1)\\ &= \binom{a+n-1}{n} \cdot \dfrac{\binom{2a+n-1}{2a-1}}{\binom{a+n-1}{a-1}}\\ &=\binom{2a+n-1}{n}\end{align}$

Answer 3

We may try to keep it simple. Suppose we seek to evaluate $$\sum_{k=0}^n {a-1+k\choose k} {a-1+n-k\choose n-k}.$$ It is immediately apparent that this is a convolution of two ordinary generating functions. To see what they are re-write the sum as $$\sum_{k=0}^n {a-1+k\choose a-1} {a-1+n-k\choose a-1}.$$ By the Newton binomial this is a convolution of $$\frac{1}{(1-z)^a}$$ with itself, giving $$\frac{1}{(1-z)^{2a}}.$$ Extracting coefficients we obtain $${n+2a-1\choose 2a-1} = {n+2a-1\choose n}.$$