Prove $\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} = \prod_{k=1}^n \frac{k}{x+k}$ and more

Solution 1:

$$\sum_{k=0}^n \dbinom{n}k (-y)^{x+k-1} = (-y)^{x-1} (1-y)^n$$ Hence, $$\int_0^1\sum_{k=0}^n \dbinom{n}k (-y)^{x+k-1}dy = \int_0^1(-y)^{x-1} (1-y)^n dy$$ $$\sum_{k=0}^n \dbinom{n}k (-1)^{x+k-1}\int_0^1y^{x+k-1}dy = \sum_{k=0}^n \dbinom{n}k (-1)^{x+k-1} \dfrac1{x+k}$$ $$\int_0^1(-y)^{x-1} (1-y)^n dy = (-1)^{x-1} \beta(x,n+1)$$ Hence, $$\sum_{k=0}^n \dbinom{n}k (-1)^{k} \dfrac1{x+k} = \beta(x,n+1)$$

Solution 2:

$$\sum_{k=0}^n \dbinom{n}k p^k = (1+p)^n$$

$$\sum_{k=0}^n \dbinom{n}k (-1)^k = (1-1)^n=0$$

$$\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} =\sum_{k=0}^n \binom{n}{k}(-1)^k (1-\frac{k}{x+k} )= \sum_{k=0}^n \binom{n}{k}(-1)^k -\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{k}{x+k}= $$

$$\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} = -\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{k}{x+k}= \sum_{k=1}^n \binom{n}{k}(-1)^{k+1} \frac{k}{x+k} \tag1$$

$$ \prod_{k=1}^n \frac{k}{x+k}= \frac{A_1}{x+1}+\frac{A_2}{x+2}+....+\frac{A_n}{x+n} $$

$$ A_1=(x+1)\prod_{k=1}^n \frac{k}{x+k} \bigg|_{x=-1}=\frac{n!}{(n-1)!}=n $$

$$ A_2=(x+2)\prod_{k=1}^n \frac{k}{x+k} \bigg|_{x=-2}=\frac{n!}{(-1)(n-2)!}=-n(n-1) $$

$$ A_3=(x+3)\prod_{k=1}^n \frac{k}{x+k} \bigg|_{x=-3}=\frac{n!}{(-1)(-2)(n-3)!}=+\frac{n(n-1)(n-2)}{2!} $$

$$ A_k=(x+k)\prod_{m=1}^n \frac{m}{x+m} \bigg|_{x=-k}=\frac{n!}{(-1)(-2)(-3)..(-(k-1))(n-k)!}=(-1)^{k+1}\frac{n!k}{1.2.3..(k-1).k(n-k)!}=(-1)^{k+1}\frac{n!k}{k!(n-k)!}=(-1)^{k+1}k\dbinom{n}k $$

$$ \prod_{k=1}^n \frac{k}{x+k}= \frac{A_1}{x+1}+\frac{A_2}{x+2}+....+\frac{A_n}{x+n}=\sum_{k=1}^n \frac{A_k}{x+k}=\sum_{k=1}^n \binom{n}{k}(-1)^{k+1} \frac{k}{x+k} $$

If we use equation 1

$$ \prod_{k=1}^n \frac{k}{x+k}= \sum_{k=0}^n \binom{n}{k}(-1)^{k} \frac{x}{x+k} $$