1

$$\frac{n!}{x(x+1)\cdots(x+n)} = \binom{n}{0}\frac{1}{x}-\binom{n}{1}\frac{1}{x+1}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n}, \quad \text{for } x \not \in \{0,-1,-2,\dots,-n\}$$

Can somebody please help me with this? I tried to multiply both sides with $x(x+1)\cdots(x+n)$.

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

Then to divide both sides with $n!$.

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

And then, assume it holds for $n$, and prove for $n+1$:

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

And this $\frac{x+n+1}{n+1-k}$ term is keeping me from applying the induction hypothesis.

EDIT: @Daniel's hint was pretty useful, thanks. After doing that, it boils down to this: $$\sum_{k=0}^{n+1}\frac{(-1)^k}{k!(n+1-k)!}\frac{x(x+1)\cdots(x+n+1)}{x+k} = 1 + x(x+1)\cdots(x+n)\sum_{k=0}^{n+1}\frac{(-1)^k}{k!(n+1-k)!}$$ So, proving that $\sum_{k=0}^{n+1}\frac{(-1)^k}{k!(n+1-k)!} = 0$ should do it.

eudoxyz
  • 128

1 Answers1

0

Hint. Use that fact that the result is an identity for all $x$ (with the stated exceptions). So, assume the result is true for a specific $n$. Then we can say $$\frac{n!}{x(x+1)\cdots(x+n)} = \binom{n}{0}\frac{1}{x}-\binom{n}{1}\frac{1}{x+1}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n}\ ,$$ and also $$\eqalign{&\frac{n!}{(x+1)(x+2)\cdots(x+n+1)}\cr &\qquad\qquad{}= \binom{n}{0}\frac{1}{x+1}-\binom{n}{1}\frac{1}{x+2}+\cdots+(-1)^n\binom{n}{n}\frac{1}{x+n+1}\ .\cr}$$ Now write down the RHS for $n+1$, simplify and use the above two equations. You should find it just drops out.

David
  • 82,662