# Proving Binomial identity involving algebraic expression

How did i prove

$$\frac{\binom{n}{0}}{x}-\frac{\binom{n}{1}}{x+1}+\frac{\binom{n}{2}}{x+2}-\cdots \cdots +(-1)^n\frac{\binom{n}{n}}{x+n}=\frac{n!}{x(x+1)(x+2)\cdots (x+n)}$$

what i try

$$\sum^{n}_{r=0}(-1)^r\frac{\binom{n}{r}}{x+r}=\int^{1}_{0}\sum^{n}_{r=0}(-1)^r\binom{n}{r}t^{x+r-1}dt$$

\begin{align}=\int^{1}_{0}t^{x-1}\sum^{n}_{r=0}(-1)^r\binom{n}{r}t^{r}dt =\int^{1}_{0}t^{x-1}(1-t)^ndt\end{align}

• Continuing further $$\int_0^1 t^{x-1} \left(\sum_{r=0}^n (-1)^r \binom nr t^r \right) dt=\int_0^1 t^{x-1}(1-t)^n dt =B(x,n+1)$$ – Rohan Shinde Feb 17 at 6:09
• And now you can finish by using the usual relation between the Beta and Gamma functions and the usual recursion for the Gamma function. – Ian Feb 17 at 6:17
• thanks Diagamma got it – jacky Feb 17 at 6:30

Multiplying by $$\prod_{i=0}^n(x+i)$$ gives
$$\sum_{k=0}^n(-1)^k\binom{n}{k}\prod_{\stackrel{i=0}{i\neq k}}^n(x+i) = n!$$
On the left side is a polynomial $$P(x)$$ of degree $$n$$. So we only need to check the identity at $$n+1$$ points. A good choice for that are the zeros of $$\prod_{i=0}^n(x+i)$$:
$$P(-k) = (-1)^k \binom{n}{k}\underbrace{\prod_{\stackrel{i=0}{i\neq k}}^n(i-k)}_{=(-1)^k\cdot k!\cdot (n-k)!} = n!$$