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I try to find a partial fraction expansion of $\dfrac{1}{\prod_{k=0}^n (x+k)}$ (to calculate its integral). After checking some values of $n$, I noticed that it seems to be true that $\dfrac{n!}{\prod_{k=0}^n (x+k)}=\sum_{k=0}^n\dfrac{(-1)^k{n \choose k}}{x+k}$. However, I can't think of a way to prove it. Can somebody please help me?

Michael Hardy
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asaelr
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  • See my answer over here: http://math.stackexchange.com/questions/444288/partial-fractions-of-frac-5x19x-1-2x1-3 Doesn't really help you prove what you want to, but it provides a very easy method for "partial-fractionating" such functions. – Parth Thakkar Mar 17 '14 at 17:21

2 Answers2

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Since every $k,\; k=-n,\ldots, 0$ is a simple pole of the given fraction then its decomposition take the form

$$\frac{1}{x(x+1)(x+2)...(x+n)}=\sum_{k=0}^n\frac{a_k}{x+k}$$ and we have $$a_k=\lim_{x \to -k}\sum_{i=0}^n\frac{a_i(x+k)}{x+i} = \lim_{x \to -k} (x+k)\sum_{i=0}^n\frac{a_i}{x+i}$$ $$= \lim_{x \to -k} \frac{x+k}{x(x+1)(x+2)...(x+n)}=\frac{1}{-k(-k+1)(-k+2)...(-k+n)}=\frac{(-1)^k}{k!(n-k)!}$$ so yes it's true that $$\frac{n!}{x(x+1)(x+2)...(x+n)}=\sum_{k=0}^n\frac{(-1)^k{n\choose k}}{x+k}$$

xyzzyz
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Notice that: $$ \frac{1}{x(x+1)\cdots(x+n)} = \frac{1}{n} \frac{(x+n) - x}{x(x+1)\cdots(x+n)} = \frac{1}{n} \left(\frac{1}{x(x+1)\cdots(x+n-1)} - \frac{1}{(x+1)\cdots(x+n)}\right) $$

This gives you a recursion formula for partial fraction expansion. You can use it to check your conjecture, and prove it by induction if it's indeed true.

See also this answer: Series of inverses of binomial coefficients for another approach.

Community
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xyzzyz
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