Prove that the identity is true for all natural numbers

Question

for the identity: $$\frac{n!}{x(x+1)(x+2)...(x+n)} = \frac{A_0}{x+0} + \frac{A_1}{x+1}+...+ \frac{A_n}{x+n}$$ prove $$A_k= (-1)^kC(n,k)$$

I think this might work by induction, but i am not able to arrive at a final answer. please help!!!

Answer 1

As what you tagged, we prove it by induction. For $n=0$, then the result is trivial. Suppose that the result holds for some $n\ge 0$, then for $n+1$, \begin{align} &\frac{(n+1)!}{x(x+1)(x+2)\cdots[x+(n+1)]}\\ &\quad\quad=\frac{n!}{x(x+1)(x+2)\cdots(x+n)}\cdot\frac{n+1}{x+n+1}\\ &\quad\quad=\left(\frac{A_0}{x+0}+\frac{A_1}{x+1}+\cdots+\frac{A_n}{x+n}\right)\cdot\frac{n+1}{x+n+1}\\ &\quad\quad=\frac{(n+1)A_0}{(x+0)(x+n+1)}+\frac{(n+1)A_1}{(x+1)(x+n+1)}+\cdots+\frac{(n+1)A_n}{(x+n)(x+n+1)}\\ &\quad\quad=\left[\frac{(n+1)A_0}{x+0}-\frac{(n+1)A_0}{x+n+1}\right]\cdot\frac{1}{n+1} +\left[\frac{(n+1)A_1}{x+1}-\frac{(n+1)A_1}{x+n+1}\right]\cdot\frac{1}{n} \\&\quad\quad\,\,\,\,\,\,\,+\cdots+\left[\frac{(n+1)A_n}{x+n}-\frac{(n+1)A_n}{x+n+1}\right]\\ &\quad\quad=\frac{A_0}{x+0}+\frac{\frac{n+1}{n}A_1}{x+1}+\cdots+\frac{(n+1)A_n}{x+n} -\frac{A_0+\frac{n+1}{n}A_1+\cdots+(n+1)A_n}{x+n+1}\\ &\quad\quad=\frac{A_0'}{x+0}+\frac{A_1'}{x+1}+\cdots+\frac{A_n'}{x+n} +\frac{A_{n+1}'}{x+n+1}, \end{align} where $A_0',A_1',\ldots,A_{n+1}'$ denotes the new coefficient in terms of the case $n+1$. Now, by the assumption, we check that \begin{align*} A_k'=\frac{n+1}{n-k+1}A_k=\frac{n+1}{n-k+1}(-1)^k{n\choose k}=(-1)^k{n+1\choose k}\quad\mbox{for }0\le k\le n, \end{align*} and \begin{align*} A_{n+1}' &=-\left[A_0+\frac{n+1}{n}A_1+\cdots+(n+1)A_n\right]\\ &=-\sum_{k=0}^n(-1)^k\frac{n+1}{n-k+1}{n\choose k}\\ &=-\sum_{k=0}^n(-1)^{k}{n+1\choose k}\\ &=-\sum_{k=0}^{n+1}(-1)^{k}{n+1\choose k}+(-1)^{n+1}\\ &=(-1)^{n+1}. \end{align*} This completes the proof.

Answer 2

Multiply both sides by $x+k$ $$\frac{n!}{x(x+1)(x+2)\cdots(x+k-1)(x+k+1)\cdots(x+n)} = \frac{A_0 (x+k)}{x+0} + \frac{A_1 (x+k)}{x+1}+...+A_k+\cdots+ \frac{A_n (x+k)}{x+n}$$ and make the specification $x=-k$. Then

$$ \frac{n!}{(-k)(-k+1)(-k+2)\cdots(-1)\cdot 1 \cdots (n-k)} = \frac{n!}{(-1)^k k!\, (n-k)!}=(-1)^k \binom{n}{k} =A_k. $$