I have to prove that $$ \forall n \in \mathbb{N}_{0}, ~ \forall x \in \mathbb{R} \setminus \mathbb{N}_{0}: \qquad \sum_{k = 0}^{n} \frac{(-1)^{k} \binom{n}{k}}{x + k} = \frac{n!}{(x + 0) (x + 1) \cdots (x + n)}. $$ However, I was unable to find a proof. I have tried to use the binomial expansion of $ (1 + x)^{n} $ to get the l.h.s., by performing a suitable multiplication followed by integration, but I was unable to obtain the required form. Please help me out with the proof. Thanks in advance.
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1$\begingroup$ I believe the right side should be $n! / [x(x+1) \cdots (x+n)]$. This actually does hold for $n = 0$. This appears to be a generalization of the partial fraction expansion $1/x(x+1) = 1/x - 1/(x+1)$. $\endgroup$– Yakov ShklarovSep 3, 2016 at 2:26
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$\begingroup$ You need another factor of $x$ in the denominator of the righthand side. $\endgroup$– Brian M. ScottSep 3, 2016 at 2:27
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$\begingroup$ (For the record, I typed out my answer before I saw @Winther's comment.) $\endgroup$– arctic ternSep 3, 2016 at 2:51
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1$\begingroup$ Note that the domain of both expressions is $ \mathbb{R} \setminus \mathbb{N}_{0} $, not $ \mathbb{R} $. $\endgroup$– TranscendentalSep 4, 2016 at 0:22
5 Answers
Write down the partial fraction decomposition
$$\frac{1}{x(x+1)\cdots(x+n)}=\sum_{k=0}^n \frac{c_k}{x+k}. \tag{$\circ$}$$
To determine the coefficient $c_k$, multiply by $x+k$ then compute the limit $x\to -k$, getting
$$c_k=\frac{1}{(-k)(1-k)(2-k)\cdots (-2)\cdot (-1)\cdot 1\cdot2\cdots(n-k)}=\frac{(-1)^k}{k!(n-k)!}$$
Substitute this into $(\circ)$, muliply both sides by $n!$, then use $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ to conclude.
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$\begingroup$ (+1) I know that I have proven this somewhere using the Heaviside method. $\endgroup$– robjohn ♦Dec 30, 2021 at 15:37
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\begin{align*} (1-t)^n = \sum_{k=0}^n (-1)^k \binom{n}{k} t^k \end{align*} Multiplying by $t^{x-1}$ and integrating between 0 and 1, we get \begin{align*} \int_0^1 t^{x-1}(1-t)^n dt = \sum_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{x+k} \end{align*} and the left hand side is \begin{align*} \beta(x,n+1) &= \frac{\Gamma(x)\Gamma(n+1)}{\Gamma(x+n+1)} \end{align*} This simplifies to \begin{align*} \frac{n!}{x(x+1)(x+2)\ldots(x+n)} \end{align*}
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$\begingroup$ @YakovShklarov, Yes, you are correct. $\endgroup$– user348749Sep 3, 2016 at 2:35
An induction on $n$ will work. Let
$$f(x,n)=\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n}k\;,$$
and for the induction step suppose that
$$f(x,n)=\frac{n!}{x(x+1)\ldots(x+n)}\;.$$
Then
$$\begin{align*} \frac{(n+1)!}{x(x+1)\ldots(x+n+1)}&=\frac{n+1}{x+n+1}f(x,n)\\ &=\frac{n+1}{x+n+1}\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n}k\\ &=(n+1)\sum_{k=0}^n(-1)^k\binom{n}k\frac1{(x+k)(x+n+1)}\\ &=(n+1)\sum_{k=0}^n\frac{(-1)^k}{n+1-k}\binom{n}k\left(\frac1{x+k}-\frac1{x+n+1}\right)\\ &=(n+1)\sum_{k=0}^n\frac{(-1)^k}{n+1}\binom{n+1}k\frac1{x+k}\\ &\qquad-\frac{n+1}{x+n+1}\sum_{k=0}^n\frac{(-1)^k}{n+1}\binom{n+1}k\\ &=\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n+1}k-\frac1{x+n+1}\color{red}{\sum_{k=0}^n(-1)^k\binom{n+1}k}\\ &=f(x,n+1)-\frac{(-1)^{n+1}}{x+n+1}-\frac1{x+n+1}\color{red}{\left(0-(-1)^{n+1}\right)}\\ &=f(x,n+1)\;, \end{align*}$$
as desired.
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$\begingroup$ There is a question at this MSE link that would appear to be a match to your activity profile. $\endgroup$ Sep 3, 2016 at 23:02
I thought it might be instructive to present a proof by induction.
First, we establish a base case. For $n=0$, it is straightforward to show that the expression holds.
Second, we assume that for some $n\geq 0$, we have
$$\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{x+k}=n!\prod_{k=0}^n\frac{1}{x+k}$$
Third, we analyze the sum $\sum_{k=0}^{n+1} \binom{n+1}{k}\frac{(-1)^k}{x+k}$. Clearly, we can write
$$\begin{align} \sum_{k=0}^{n+1} \binom{n+1}{k}\frac{(-1)^k}{x+k}&=\frac1x+\sum_{k=1}^{n} \binom{n+1}{k}\frac{(-1)^k}{x+k}+\frac{(-1)^{n+1}}{x+n+1}\\\\ &=\color{green}{\frac1x}+\sum_{k=1}^{n} \left(\color{green}{\binom{n}{k}}+\color{blue}{\binom{n}{k-1}}\right)\frac{(-1)^k}{x+k}+\color{red}{\frac{(-1)^{n+1}}{x+n+1}}\\\\ &=\color{green}{n!\prod_{k=0}^n\frac{1}{x+k}}+\color{blue}{\sum_{k=0}^{n-1} \binom{n}{k}\frac{(-1)^{k+1}}{(x+1)+k}}+\color{red}{\frac{(-1)^{n+1}}{x+n+1}}\\\\ &=\color{green}{n!\prod_{k=0}^n\frac{1}{x+k}}+\color{blue}{-n!\prod_{k=0}^n\frac{1}{x+1+k}-\frac{(-1)^{n+1}}{x+n+1}}+\color{red}{\frac{(-1)^{n+1}}{x+n+1}}\\\\ &=(n+1)!\prod_{k=0}^{n+1}\frac{1}{x+k} \end{align}$$
And we are done!
Remark. What follows is an exact duplicate of an earlier post of mine which I am unable to locate despite making a considerable effort.
We seek to verify that
$$\sum_{k=0}^n \frac{1}{x+k} (-1)^k {n\choose k} = n! \times \prod_{q=0}^n \frac{1}{x+q}.$$
Consider the function
$$f(z) = n! \frac{1}{x+z} \prod_{q=0}^n \frac{1}{z-q}$$
where clearly $x$ is not equal to $0,-1,-2,\ldots -n$ or the LHS of the target formula becomes singular.
We compute the residues of $f(z).$ We get for the poles at $p\in [0,n]$ the residue
$$\mathrm{Res}_{z=p} f(z) = n!\frac{1}{x+p} \prod_{q=0}^{p-1} \frac{1}{p-q} \prod_{q=p+1}^n \frac{1}{p-q} \\ = n! \frac{1}{x+p} \frac{1}{p!} \frac{(-1)^{n-p}}{(n-p)!} = \frac{1}{x+p} (-1)^{n-p} {n\choose p}.$$
The residue at $z=-x$ yields
$$\mathrm{Res}_{z=-x} f(z) = n! \prod_{q=0}^n \frac{1}{-x-q} = n! (-1)^{n+1} \times \prod_{q=0}^n \frac{1}{x+q}.$$
The residue at infinity is
$$\mathrm{Res}_{z=\infty} f(z) = - \mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z) \\ = - \mathrm{Res}_{z=0} \frac{1}{z^2} n! \frac{1}{x+1/z} \prod_{q=0}^n \frac{1}{1/z-q} \\ = - \mathrm{Res}_{z=0} \frac{1}{z^2} n! \frac{z}{zx+1} \prod_{q=0}^n \frac{z}{1-qz} \\ = - \mathrm{Res}_{z=0} \frac{z^{n+2}}{z^2} n! \frac{1}{zx+1} \prod_{q=0}^n \frac{1}{1-qz} \\ = - \mathrm{Res}_{z=0} z^n n! \frac{1}{zx+1} \prod_{q=0}^n \frac{1}{1-qz} = 0.$$
To conclude observe that the residues sum to zero and collecting everything we obtain
$$\sum_{p=0}^n\frac{1}{x+p} (-1)^{n-p} {n\choose p} + n! (-1)^{n+1} \times \prod_{q=0}^n \frac{1}{x+q} = 0.$$
This is
$$\sum_{p=0}^n\frac{1}{x+p} (-1)^{p} {n\choose p} - n! \times \prod_{q=0}^n \frac{1}{x+q} = 0.$$
or
$$\sum_{p=0}^n\frac{1}{x+p} (-1)^{p} {n\choose p} = n! \times \prod_{q=0}^n \frac{1}{x+q}$$
which is the claim.