Prove the identity $ \sum\limits_{i=k}^{n} \frac{1}{(i-k)! (a+i)_{n-i}}={\frac {a+n}{ \left( a{+} k \right) (n{-}k)! } }.$ Playing with hypergeometric series I have got the identity 
$$
\sum_{i=k}^{n} \frac{1}{(i-k)! (a+i)_{n-i}}={\frac {a+n}{ \left( a{+}
k \right) (n{-}k)! } },
$$
here $(x)_s=x(x+1)\cdots (x+s-1),$ $n, k$ are  integers and  $a$ is not integer number.  Is there any direct way to prove it?
EDIT.  My attempt for the case $k=0.$ Let   $$S_n=\sum_{i=0}^{n} \frac{1}{i! (a+i)_{n-i}}. $$ Then by direct calculation one may verify that  $S_n$ satisfy the recurrence equation 
$$
    (a+1)(n+3)  S_{ n+3 }=S_{ n+1} -( n-a ) S_{
  n+2 } ,
$$
with the initial condition 
$$
S_0 =1,S_1={\frac{a+2}{a+1}},S_2 =\frac 1 2{\frac{a+3}{a+1}}.
 $$
By solving it we get that 
$$
S_n= {\frac {a+n}{ a n! } }.
$$
I think that it works for any $k$  but I don't like it.
 A: 
We obtain
  \begin{align*}
\color{blue}{\sum_{i=k}^n}&\color{blue}{\frac{(n-k)!}{(i-k)!(a+i)^{\overline{n+i}}}}\tag{1}\\
&=\sum_{i=k}^n\frac{(n-k)!}{(i-k)!(a+n-1)^{\underline{n-i}}}\tag{2}\\
&=\sum_{i=k}^n\binom{n-k}{i-k}\binom{a+n-1}{n-i}^{-1}\tag{3}\\
&=\sum_{i=0}^{n-k}\binom{n-k}{i}\binom{a+n-1}{n-k-i}^{-1}\tag{4}\\
&=(a+n)\int_0^1\sum_{i=0}^{n-k}\binom{n-k}{i}z^{n-k-i}(1-z)^{a-1+k+i}\,dz\tag{5}\\
&=(a+n)\int_0^1z^{n-k}(1-z)^{a-1+k}\sum_{i=0}^{n-k}\binom{n-k}{i}\left(\frac{1-z}{z}\right)^{i}\,dz\tag{6}\\
&=(a+n)\int_0^1z^{n-k}(1-z)^{a-1+k}\left(1+\frac{1-z}{z}\right)^{n-k}\,dz\tag{7}\\
&=(a+n)\int_0^1(1-z)^{a-1+k}\,dz\\
&\color{blue}{=\frac{a+n}{a+k}}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we use the rising factorial  $n^{\overline{k}}=n(n+1)\cdots (n+k-1)$.

*In (2) we use the falling factorial $n^{\underline{k}}=n(n-1)\cdots (n-k+1)$.

*In (3) we use the identity $\binom{n}{k}=\frac{n^{\underline{k}}}{k!}$.

*In (4) we shift the index $i$ to start with $i=0$.

*In (5) we write the reciprocal of a binomial coefficient using the beta function
\begin{align*}
\binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz
\end{align*}

*In (6) we do some rearrangements as preparation for the next step.

*In (7) we apply the binomial theorem.
A: First of all we change the variables
$$ S_n\mapsto(n-k)!S_n\\
i\mapsto i-k\\
n-k\mapsto n\\
a+k\mapsto a
$$
 to bring the expression into a simpler form:
$$
S_n:=\sum_{i=0}^n\frac{\binom{n}{n-i}}{\binom{a+n-1}{n-i}}=\frac{a+n}{a}.\tag{1}
$$
In what follows we assume $a\not\in\{1-n,2-n,\dots,0\}$.
The equality (1) is obviously valid for $n=0$ and for arbitrary $a$: $S_0=1=\frac{a+0}{a}$. Assuming that it is valid for $n-1$ one obtains that it is valid for $n$ as well:
$$
S_n=\sum_{i=0}^n\frac{\binom{n}{n-i}}{\binom{a+n-1}{n-i}}=1+\sum_{i=0}^{n-1}\frac{\binom{n}{n-i}}{\binom{a+n-1}{n-i}}=1+\sum_{i=0}^{n-1}\frac{\frac{n}{n-i}\binom{n-1}{n-i-1}}{\frac{a+n-1}{n-i}\binom{a+n-2}{n-i-1}}\\=1+\frac{n}{a+n-1}\sum_{i=0}^{n-1}\frac{\binom{n-1}{n-1-i}}{\binom{a+n-2}{n-1-i}}=
1+\frac{n}{a+n-1}S_{n-1}\\
\stackrel{I.H}{=}1+\frac{n}{a+n-1}\cdot\frac{a+n-1}{a}=\frac{a+n}{a}.
$$
Thus the equality is proved.
