Proving a sum that equals $y!/((y-z)\cdots(1-z))$ 
Question: How do you show that$$\frac {y!}{(y-z)(y-z-1)\cdots(1-z)}=\sum\limits_{n=0}^y\binom yn(-1)^n\frac n{z-n}$$

In the proof for a sum that converges to $\binom yz$, the solution made a jump from$$\frac {\sin\pi z}{\pi z}\frac {y!}{(y-z)(y-z-1)\cdots(1-z)}=\frac {\sin\pi z}{\pi z}\sum\limits_{n=0}^y\binom yn(-1)^n\frac n{z-n}=\sum\limits_{n=0}^y\binom yn\frac {\sin(\pi z-\pi n)}{\pi(z-n)}$$I’m having a hard time seeing how all three equations are equal. I don’t see where the summation sign comes in. Any ideas?
 A: I'll see if
I can derive it.
You want
$\sum\limits_{n=0}^y\binom yn(-1)^n\frac n{z-n}
=\dfrac {y!}{(y-z)(y-z-1)\cdots(1-z)}
=\dfrac{y!}{\prod_{n=0}^{y-1} (y-z-n)}
$.
Let's try to find a
partial fraction decomposition.
Suppose
$\dfrac{1}{\prod_{n=0}^{y-1} (y-z-n)}
=\sum_{n=0}^{y-1} \dfrac{a_n}{y-z-n}
$.
For $0 \le k \le y-1$,
multiply this by $y-z-k$.
The left side is
$\dfrac{1}{\prod_{n=0, n \ne k}^{y-1} (y-z-n)}
$.
Letting $z = y-k$,
this is
$\begin{array}\\
\dfrac{1}{\prod_{n=0, n \ne k}^{y-1} (y-(y-k)-n)}
&=\dfrac{1}{\prod_{n=0, n \ne k}^{y-1} (k-n)}\\
&=\dfrac{1}{\prod_{n=0}^{k-1} (k-n)\prod_{n=k+1}^{y-1} (k-n)}\\
&=\dfrac{1}{\prod_{n=0}^{k-1} (k-(k-1-n))\prod_{n=1}^{y-1-k} (k-(n+k))}\\
&=\dfrac{1}{\prod_{n=0}^{k-1} (n+1)\prod_{n=1}^{y-1-k} (-n)}\\
&=\dfrac{1}{k!(-1)^{y-1-k}(y-1-k)!}\\
\end{array}
$
The right side is
$\sum_{n=0}^{y-1} \dfrac{a_n(y-z-k)}{y-z-n}
$.
As $z \to y-k$,
all the terms
go to zero except
that where
$k = n$,
where the term goes to
$a_k$.
Therefore
$a_k
=\dfrac{1}{k!(-1)^{y-1-k}(y-1-k)!}
$
so
$\begin{array}\\
y!a_k
&=\dfrac{y!}{k!(-1)^{y-1-k}(y-1-k)!}\\
&=\dfrac{y!(y-k)(-1)^{y-1-k}}{k!(y-k)!}\\
&=(y-k)(-1)^{y-1-k}\binom{y}{k}\\
\end{array}
$
