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In the identity $$\frac{n!}{x(x+1)(x+2)\cdots(x+n)}=\sum_{k=0}^n\frac{A_k}{x+k},$$prove that $$A_k=(-1)^k\binom nk.$$

My try: The given identity implies $$\frac{1\cdot2\cdots n}{x(x+1)(x+2)\cdots(x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1}+\dots+\frac{A_n}{x+n}.$$ Now putting $A_k=(-1)^k\binom nk,$$$\frac{1\cdot2\cdots n}{x(x+1)(x+2)\cdots(x+n)}=\frac1x-\frac{n}{x+1}+\dots+\frac{(-1)^n}{x+n}.$$ How to proceed further?

Asaf Karagila
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Silent
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3 Answers3

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Hint: A technique commonly used in partial fraction decomposition is the following:

Suppose you have $$\frac{n!}{x(x+1)\cdots(x+n)}=\sum_{k=0}^n\frac{A_k}{x+k}.$$ Now, fix $i$ between $0$ and $n$, multiply both sides by $(x+i)$, simplify, and set $x=-i$. This eliminates all terms on the right-hand side except $A_i$, leaving you with a value you can manipulate on the left-hand side. You can repeat this process for all $i=0,1,\ldots,n$, and so derive a formula for $A_i$.

Clayton
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  • I think there should be something between "multiply by $x+i$" and "set $x=-i$". Namely: simplify both sides. Only after simplification does setting $x=-i$ makes sense. – Marc van Leeuwen Dec 03 '13 at 13:43
  • @Clayton Sir, pardon me if I am wrong.After multiplying both sides by $x(x+1)(x+2)\cdots(x+n),$ we get $\frac{n!x(x+1)(x+2)\cdots(x+n)}{x(x+1)(x+2)\cdots(x+n)}=x(x+1)(x+2)\cdots(x+n) \sum_{k=0}^n\frac{A_k}{x+k}$, which means $$n!=(x+1)(x+2)\cdots(x+n)A_0+\dots+x(x+1)\cdots(x+k-1)(x+k+1)\cdots(x+n)A_k+x(x+1)\cdots(x+n-1)A_n$$. Am I on your path? How to o further? – Silent Dec 03 '13 at 13:47
  • @Sush: Why would you multiply both sides by that expression? Why not just multiply both sides by $x+i$ as I suggest? – Clayton Dec 03 '13 at 14:08
  • @MarcvanLeeuwen: I suppose I take the simplification to be implicit. I'll edit the phrase in to make it more clear. – Clayton Dec 03 '13 at 14:08
  • @Clayton: While editing, you might also move "for $0\leq i\leq n$" to a less confusing place. Clearly OP misunderstood you as asking to perform all those multiplications together (so to multiply by their product). In fact, I think OP took your phrase literally, but that was not what you intendended. – Marc van Leeuwen Dec 03 '13 at 14:11
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HINT: Multiply both side by $x+k$ and then put $x=-k.$

pritam
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Hint: The number $A_k$ is the residue of the meromorphic function $$\frac{n!}{x(x+1) \cdots (x+n)} $$ at the point $x=-k.$ Since it's a simple pole the residue is $$A_k=\frac{n!}{\frac{d}{dx}[x(x+1) \cdots (x+n)]} \bigg|_{x=-k}=\frac{n!}{(-k)(-k+1) \cdots (-k+n)}, $$where in the final product the term $(-k+k)$ is obviously omitted.

user1337
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