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I need to prove this equation $$\sum_{k} { {n}\choose{k}}\frac{(-1)^k}{x+k} = \frac{n!}{x(x+1)...(x+n)}$$

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    I am supposing the sum runs from k=0 to n? – 2017-01-29
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    It would be 0 for other values of k, so I think yes – 2017-01-29
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    Funny that I ran into this while Euler summing the alternating harmonic series. – 2017-01-29

2 Answers 2

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(I'll assume that: $k$ runs from $0$ to $n$)

We observe that, $$\frac{1}{x+k} = \int_0^1 t^{x+k-1} \mathrm{d}t$$ Thus, we have

\begin{align*} \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{x+k} &= \sum_{k=0}^n \binom{n}{k} (-1)^k \int_0^1 t^{x+k-1} \mathrm{d}t \\ &= \int_0^1 t^{x-1} \left(\sum_{k=0}^n \binom{n}{k} (-t)^k \right) \mathrm{d}t \\ &= \int_0^1 t^{x-1} (1-t)^n \mathrm{d}t \\ &= \beta(x, n+1) \end{align*}

where $\beta(x,y)$ denotes the beta function. Now, if given that $x \in \mathbb{R^+}$ and $n \in \mathbb{N}$ then

$$\beta(x,n+1) = \frac{\Gamma(x)\Gamma(n+1)}{\Gamma(x+n+1)} = \frac{\Gamma(x)n!}{\Gamma(x+n+1)} = \frac{n!}{x(x+1)\ldots (x+n)}$$

P.S. I am not very sure but I guess we can justify for the change of integral and summation because of uniform convergence.

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    To make sense, it has to be $x>0$, so the integral $\int_0^1 t^{x+k-1} \ dt$ for the minimum value of $k$, that is, $\int_0^1 t^{x-1} \ dt$ would be convergent (because $x-1>-1$). As you are changing the order with a finite sum, there would be no problem. – 2017-01-29
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    Thanks for completing the proof! – 2017-01-29
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Let us put $\frac{n!}{x(x+1)\cdots (x+n)}$ into partial fractions. We can write \begin{align*} \frac{n!}{x(x+1)\cdots (x+n)} &= \sum_{k=0}^n \frac{A_k}{x+k} \end{align*} Clearing the fractions and putting $x = -k$, we get \begin{align*} A_k(-k)(-k+1) \cdots (-k+k-1)(-k+k+1) \cdots (-k+n) = n! \end{align*} and hence $A_k (-1)^k k! (n-k)! = n!$ and $A_k = (-1)^k\frac{n!}{k!(n-k)!} = (-1)^k \binom{n}{k}$