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


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

  • 4
    $\begingroup$ Continuing further $$\int_0^1 t^{x-1} \left(\sum_{r=0}^n (-1)^r \binom nr t^r \right) dt=\int_0^1 t^{x-1}(1-t)^n dt =B(x,n+1)$$ $\endgroup$ – Rohan Shinde Feb 17 at 6:09
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    $\begingroup$ And now you can finish by using the usual relation between the Beta and Gamma functions and the usual recursion for the Gamma function. $\endgroup$ – Ian Feb 17 at 6:17
  • $\begingroup$ thanks Diagamma got it $\endgroup$ – jacky Feb 17 at 6:30

A possible way is as follows:

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 $\prod_{i=0}^n(x+i)$:

$$P(-k) = (-1)^k \binom{n}{k}\underbrace{\prod_{\stackrel{i=0}{i\neq k}}^n(i-k)}_{=(-1)^k\cdot k!\cdot (n-k)!} = n!$$

So, we are done.


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