Identity involving falling factorial Let $(x)_n$ denote the falling factorial,
$$(x)_n = x(x-1) \dots (x - n + 1).$$
I came across the following identity in trying to solve another problem.
$$\sum_{k=1}^m \frac{(a)_k}{(x)_k} = \frac{a}{x - a + 1}\left( 1 - \frac{(a-1)_m}{(x)_m}\right).$$
This identity allows you to find closed forms for sums such as
$$\sum_{k=0}^m \frac{\Gamma(x + k)}{\Gamma(y + k)}$$
or
$$\sum_{k = 0}^c \frac{a \choose k}{b \choose k}.$$
It is similar to a series mentioned here (Infinite sum of falling factorial quotients. ), except that my sum is finite.
I proved the identity by induction, which isn't too hard, as long as you know what you're looking for. I have a couple of questions, though.


*

*I tried to find this identity mentioned somewhere, but couldn't find it. Is it well-known, or a special case of a better known identity?

*I found the expression on the right-hand side through trial and error. Is there a more natural derivation of the identity, particularly if you know the left-hand side and are looking for the right-hand side? 
For example, after multiplying through by $(x - a + 1)(x)_m$, the identity becomes one of polynomials in $a$ and $x$, hence it would be sufficient to prove it for sufficiently large integers $a$ and $x$. Perhaps there is a combinatorial explanation. 

We  prove
\begin{align*}
\sum_{k=m}^n\frac{\binom{a}{k}}{\binom{b}{k}}=\frac{b+1}{b-a+1}\left(\frac{\binom{a}{m}}{\binom{b+1}{m}}-\frac{\binom{a}{n+1}}{\binom{b+1}{n+1}}\right)\tag{1}
\end{align*}

We  start with the right-hand side of (1) and obtain
  \begin{align*}
\color{blue}{\frac{b+1}{b-a+1}}&\color{blue}{\left(\frac{\binom{a}{m}}{\binom{b+1}{m}}-\frac{\binom{a}{n+1}}{\binom{b+1}{n+1}}\right)}\\
&=\frac{b+1}{b-a+1}\sum_{k=m}^n\left(\frac{\binom{a}{k}}{\binom{b+1}{k}}-\frac{\binom{a}{k+1}}{\binom{b+1}{k+1}}\right)\tag{2}\\
&=\frac{b+1}{b-a+1}\sum_{k=m}^n\frac{\binom{a}{k}}{\binom{b}{k}}\left(\frac{1}{\frac{b+1}{b-k+1}}-\frac{\frac{a-k}{k+1}}{\frac{b+1}{k+1}}\right)\tag{3}\\
&=\frac{b+1}{b-a+1}\sum_{k=m}^n\frac{\binom{a}{k}}{\binom{b}{k}}\left(\frac{b-k+1}{b+1}-\frac{a-k}{b+1}\right)\\
&=\frac{1}{b-a+1}\sum_{k=m}^n\frac{\binom{a}{k}}{\binom{b}{k}}\left(b-k+1-(a-k)\right)\\
&\,\,\color{blue}{=\sum_{k=m}^n\frac{\binom{a}{k}}{\binom{b}{k}}}
\end{align*}
  and the claim (1) follows.

Comment:


*

*In (2) we write  the  expression  as telescoping   sum.

*In (3)  we  factor out $\frac{\binom{a}{k}}{\binom{b}{k}}$  and   simplify in the following steps.
