Prove the identity $ \sum\limits_{i=0}^{n} (-1)^{i} (a+2i) \binom{n}{i} \binom{a+n+i}{n+1}^{-1}=(n+1) \delta_{a,0}$

Question

Playing with hypergeometric series I come across the identity $$ \sum_{i=0}^{n} (-1)^{i} (a+2i) \binom{n}{i} \binom{a+n+i}{n+1}^{-1}=\begin{cases} n+1, a=0, \\ 0, a\neq 0 \end{cases} $$ and $a \notin \{-1, -2,-3, \ldots \}.$ My attempt to prove it by rewritting \begin{align*} \binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz \end{align*} and reduce it to an integral like as here was unsuccessful.

Answer 1

Without use of integrals one can take the following path. Let $$ S_{n} = \sum_{i=0}^{n} (-1)^{i} (a+2i) \binom{n}{i} \binom{a+n+i}{n+1}^{-1} $$ and $$ S_{n}^{(1)} = 2 \, \sum_{i=0}^{n} (-1)^{i} (i) \binom{n}{i} \binom{n+i}{n+1}^{-1}. $$ Now, \begin{align} S_{n}^{(1)} &= 2 \, \sum_{k=0}^{n} (-1)^{k} (k) \binom{n}{k} \binom{n+k}{n+1}^{-1} \\ &= 2 \, \sum_{k=0}^{n} \frac{(-1)^k \, n! (n+1)! (k-1)!}{(k-1)! (n-k)! (n+k)!} \\ &= 2 \, (n+1) \, \sum_{k=0}^{n} \frac{(-1)^{k}}{(n+1)_{-k} \, (n+1)_{k}} \\ &= 2 \, (n+1) \, \sum_{k=0}^{n} \frac{(-n)_{k}}{(n+1)_{k}} = 2 \, (n+1) \, {}_{2}F_{1}(-n,1; n+1; 1) \\ &= 2 \, (n+1) \, \frac{\Gamma(n+1) \Gamma(2n)}{\Gamma(2n+1) \Gamma(n)} = (n+1). \end{align}

\begin{align} S_{n} &= \sum_{k=0}^{n} \frac{(-1)^{k} (a+2k) n! (n+1)! (a+k-1)!}{k! (n-k)! (a+n+k)!} \\ &= B(a, n+1) \, \sum_{k=0}^{n} \frac{(a+2k) (-n)_{k} (a)_{k}}{k! (a+n+1)_{k}} \\ &= B(a,n+1) \, \left[ a \, {}_{2}F_{1}(-n, a; a+n+1; 1) + 2 \, \sum_{k=1}^{n} \frac{(-n)_{k} (a)_{k}}{(k-1)! (a+n+1)_{k}} \right] \\ &= B(a,n+1) \, \left[a \, {}_{2}F_{1}(-n, a; a+n+1; 1) + \frac{2 (a) (-n)}{(a+n+1)} \, {}_{2}F_{1}(1-n, a+1; a+n+2; 1) \right] \\ &= a \, B(a, n+1) \, \left[ \frac{\Gamma(a+n+1) \Gamma(2n+1)}{\Gamma(2n+a+1) \Gamma(n+1)} - \frac{\Gamma(a+n+1) \Gamma(2n+1)}{\Gamma(2n+a+1) \Gamma(n+1)} \right] \\ &= 0. \end{align}

From these results it can be stated that $$\sum_{i=0}^{n} (-1)^{i} (a+2i) \binom{n}{i} \binom{a+n+i}{n+1}^{-1} = (n+1) \delta_{a,0}$$

Answer 2

Idea: reciprocal of binomial coefficients are related to Euler's Beta function, which has a simple integral representation. Implementation:

$$\sum_{i=0}^{n}(-1)^i (a+2i)\binom{n}{i}\frac{\Gamma(n+1)\Gamma(a+i)}{\Gamma(a+n+i+1)}(n+1)\\=\sum_{i=0}^{n}\binom{n}{i}(-1)^i(a+2i)(n+1)\int_{0}^{1}u^{n}(1-u)^{a+i-1}\,du\\=(n+1)\int_{0}^{1}\frac{u^{2n}(1-u)^a\left[(2n+a)u-2n\right]}{u(1-u)}\,du$$ and the claim follows by noticing that $\frac{1}{u(1-u)}=\frac{1}{u}+\frac{1}{1-u}$.