Proof by induction, or without it if possible?

Question

I was given a task to prove: $$ \frac{1}{(x+1)(x+2)\ldots(x+n)}=\frac{1}{(n-1)!}\sum_{i=1}^n\binom{n-1}{i-1}\frac{(-1)^{i-1}}{x+i} $$ I am almost 100% sure this is best solved by induction but to be honest, I'm scared to start using induction for such an expression on a test where I am limited with time. This reminds me of partial fraction decomposition but that binomial coefficient is disturbing me. I'm not sure how difficult this is to prove, but what I am looking for is a way to do this without induction but it seems impossible to do so without it. Induction in general bothers me. I agree it's a fine way of proving something but I would always chose another way around. Unfortunately, most problems I encountered can't be solved any other way. So before you hit the "report as not real question" button, I would like to ask for a HINT on how to solve this, with or without induction?

Answer 1

Since every $k,\; k=-n,\ldots, -1$ is a simple pole of the given fraction then its decomposition take the form

$$\frac{1}{(x+1)(x+2)...(x+n)}=\sum_{k=1}^n\frac{a_k}{x+k}$$ and we have $$a_k=\lim_{x \to -k}\sum_{i=1}^n\frac{a_i(x+k)}{x+i} = \lim_{x \to -k} (x+k)\sum_{i=1}^n\frac{a_i}{x+i} = \lim_{x \to -k} \frac{x+k}{(x+1)(x+2)...(x+n)}=\frac{1}{(-k+1)(-k+2)...(-k+n)}=\frac{(-1)^{(k-1)}}{(k-1)!(n-k)!}$$ so yes it's true that $$\frac{1}{x(x+1)(x+2)...(x+n)}=\frac{1}{(n-1)!}\sum_{k=1}^n {n-1\choose k-1}\frac{(-1)^{(k-1)}}{x+k}$$

Answer 2

Hint: Multiply both sides of the equation by $(x+1)(x+2)\cdots(x+n)$. This gives the polynomial identity $$P(x) = \frac{1}{(n-1)!} \sum_{i=1}^{n} \binom{n-1}{i-1} (-1)^{i-1} (x+1)(x+2)\cdots (x+i-1)(x+i+1)\cdots (x+n) = 1$$ Since each term in the summation is a polynomial of degree $n-1$, $P(x)$ is of degree at most $n-1$. Accordingly, if $P$ is nonconstant, then $P$ can equal $1$ for at most $n-1$ distinct $x$-values. But we know $n$ distinct points $x$ at which $P(x) = 1$.