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}$ 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.

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
$$\sum_{i=0}^{n}  (-1)^{i} (a+2i) \binom{n}{i}  \binom{a+n+i}{n+1}^{-1} = (n+1) \delta_{a,0}$$
