Prove that $\sum^n_{k=0}\frac{(-1)^k}{k+x}\binom{n}{k}=\frac{n!}{x(x+1)\cdots(x+n)}$.

Question

Given the following formula $$ \sum^n_{k=0}\frac{(-1)^k}{k+x}\binom{n}{k}\,. $$ How can I show that this is equal to $$ \frac{n!}{x(x+1)\cdots(x+n)}\,? $$

Answer 1

Consider the (unique) polynomial $p(x)\in\mathbb{Q}[x]$ of degree at most $n$ such that $p(-k)=1$ for all $k=0,1,2,\ldots,n$. Clearly, $p(x)$ is the constant polynomial $1$.

However, using Lagrange interpolation, we have $$p(x)=\sum_{k=0}^n\,p(-k)\,\frac{\prod\limits_{j\in[n]\setminus\{k\}}\,(x+j)}{\prod\limits_{j\in[n]\setminus\{k\}}\,(-k+j)}\,,$$ where $[n]:=\{0,1,2,\ldots,n\}$. This means $$1=\sum_{k=0}^n\,\frac{\prod\limits_{j\in[n]\setminus\{k\}}\,(x+j)}{\prod\limits_{j\in[n]\setminus\{k\}}\,(-k+j)}=\sum_{k=0}^n\,(-1)^k\,\frac{\prod\limits_{j\in[n]\setminus\{k\}}\,(x+j)}{k!\,(n-k)!}\,.$$ Multiplying both sides by $\dfrac{n!}{\prod\limits_{j\in[n]}\,(x+j)}$ yields $$\frac{n!}{\prod\limits_{j=0}^n\,(x+j)}=\sum_{k=0}^n\,(-1)^k\,\left(\frac{n!}{k!\,(n-k)!}\right)\,\frac{1}{x+k}=\sum_{k=0}^n\,\binom{n}{k}\,\frac{(-1)^k}{x+k}\,.$$

Answer 2

Induction step:

$$\begin{align} \sum_{k=0}^{n+1}&\frac{(-1)^k}{x+k}\binom{n+1}k=\frac1x+\frac{(-1)^{n+1}}{x+n+1}+\sum_{k=1}^{n}\frac{(-1)^k}{x+k}\left[\binom nk+\binom n{k-1}\right] \\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}+\sum_{k=1}^{n}\frac{(-1)^k}{x+k}\binom{n}{k-1} \\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}-\sum_{k=0}^{n-1}\frac{(-1)^k}{(x+1)+k}\binom{n}{k} \\&=\frac{n!}{x(x+1)\cdots(x+n)}+\frac{(-1)^{n+1}}{x+n+1}-\frac{n!}{(x+1)(x+2)\cdots(x+n+1)}+\frac{(-1)^n}{x+n+1} \\&=\frac{n!(x+n+1)-n!x}{x(x+1)\cdots(x+n+1)}=\frac{(n+1)!}{x(x+1)\cdots(x+n+1)} \end{align}$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\Re\pars{x} > 0}$:

\begin{align} &\bbox[10px,#ffd]{\sum_{k = 0}^{n}{\pars{-1}^{k} \over k + x}{n \choose k}} = \sum_{k = 0}^{n}\pars{-1}^{k} \pars{\int_{0}^{1}t^{k + x - 1}\,\dd t}{n \choose k} \\[5mm] = &\ \int_{0}^{1}t^{x - 1}\sum_{k = 0}^{n} {n \choose k}\pars{-t}^{k}\,\dd t \\[5mm] = &\ \int_{0}^{1}t^{x - 1}\,\pars{1 - t}^{n}\,\dd t = \mrm{B}\pars{x,n + 1}\ \pars{~\mrm{B}:\ Beta\ Function~} \\[5mm] = &\ {\Gamma\pars{x}\Gamma\pars{n + 1} \over \Gamma\pars{x + n + 1}} \phantom{= \mrm{B}\pars{x,n + 1}\,\,\,\,\,\,\,\,\,\,\,\,} \pars{~\Gamma:\ Gamma\ Function~} \\[5mm] = &\ {n! \over \Gamma\pars{x + n + 1}/\Gamma\pars{x}} = {n! \over x^{\overline{n +1}}} = \bbx{n! \over x\pars{x + 1}\cdots\pars{x + n}} \end{align}