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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?

Transcendental
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  • Does $i$ really start at $0$ ? Negative entries are unusual in binomial coefficients. – Peter Dec 01 '14 at 22:17
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    Hint : How do you get from $n$ to $n+1$, only considering the left side ? – Peter Dec 01 '14 at 22:18
  • @Peter Sorry, it starts from 1. It was a typo.. Well considering the left side we just devide the expression with $(x+n+1)$ It's the right part of the expression that bothers me. – Transcendental Dec 01 '14 at 22:22
  • Compare the right side for $n$ and $n+1$ and look for simplifications. – Peter Dec 01 '14 at 22:25
  • Well the only alteration i can make to that expression is to represent it as a $\frac{1}{n} (\frac{1}{(x+1)(x+2)\ldots(x+n)}-\frac{1}{(x+2)(x+3)\ldots(x+n+1)})$ but i still can't figure out what to do with that sum..:( – Transcendental Dec 01 '14 at 22:32
  • Maybe, there is some useful theorem about sums of binomial coefficients. By the way, this exercise is far from easy. Who asked you to prove this ? – Peter Dec 01 '14 at 22:35
  • Well the only one I know of is Pascal's rule but I'm not sure if it's useful in this case. Well it was given on one of previous tests. – Transcendental Dec 01 '14 at 22:46
  • Duplicate of this. – Lucian Dec 06 '14 at 05:50

2 Answers2

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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 $$\frac{1}{x(x+1)(x+2)...(x+n)}=\frac{1}{(n-1)!}\sum_{k=1}^n {n-1\choose k-1}\frac{(-1)^{(k-1)}}{x+k}$$

alexjo
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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 $P(x) = 1$.

Sameer Kailasa
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