Show $\lim_{X \to -k}\:\prod_{j\neq k}\left(X+j \right)=(-1)^{k}k!\left(n-k \right).)$

Question

I would like to know how to show that : $$\lim_{X \to -k}\:\prod_{j\neq k}\left(X+j \right)=(-1)^{k}k!\left(n-k \right).$$

This is came from solution of exercise that he said :

• what is Partial fraction decomposition of :

$$F(X)=\dfrac{n!}{\prod_{k=0}^{n}\left(X+k \right)}$$

indeed,

PFD of F:

$$F(X)=\sum_{k=0}^{n}\dfrac{a_k}{X+k} $$

$$\left(X+k\right)F(X)=\dfrac{n!}{\prod_{j\neq k}\left(X+j \right)}$$ and $$\prod_{j\neq k}\left(X+j \right)=(-1)^{k}k!\left(n-k \right)$$ then :

$$a_k=(-1)^{k}{n \choose k} $$

Finaly:

$$F(X)=\sum_{k=0}^{n}(-1)^{k}{n \choose k}\dfrac{1}{X+k} $$

Answer 1

I guess you would rather like to prove

$$ \lim_{X \to -k}\:\prod_{j\neq k}\left(X+j \right)=(-1)^{k}k!\left(n-k \right). $$

If this is the case, one may observe that $$ \begin{align} \lim_{X \to -k}\:\prod_{j\neq k}\left(X+j \right)&=\left.\prod_{j\neq k}\left(X+j \right)\right|_{X=-k} \\\\&=\prod_{j\neq k}\left(-k+j \right) \\\\&=\prod_{0\le j \le k,\,j\neq k}\left(-k+j \right)\prod_{k\le j \le n,\,j\neq k}\left(-k+j \right) \\\\&=(-1)^{k}\prod_{0\le j \le k-1}\left(k-j \right)\prod_{k+1\le j \le n}\left(j-k \right) \\\\&=(-1)^kk!(n-k)! \end{align} $$ as announced.

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\mrm{F}\pars{X}} & = {n! \over \prod_{k = 0}^{n}\pars{X + k}} = {n! \over \prod_{k = 0}^{n}\pars{k + X}} = {n! \over \pars{X}_{n + 1}} \end{align} where $\ds{\pars{a}_{m}}$ is the Pochhammer Symbol.

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Also, \begin{align} \prod_{j \not= k}\pars{X + j} & = {n! \over \pars{X + k}\,\mrm{F}\pars{X}} = {n! \over \pars{X + k}\bracks{n!/\pars{X}_{n + 1}}} = {\pars{X}_{n + 1} \over X + k} = {\Gamma\pars{X + n + 1} \over \pars{X + k}\Gamma\pars{X}} \\[5mm] & = {\pars{X + n}! \over \pars{X + k}\pars{X - 1}!} = {\pars{X + n}\ldots X \over X + k} = \pars{X + n}\ldots\pars{X + k + 1}\pars{X + k - 1}\ldots X \end{align}

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\begin{align} \lim_{X \to -k}\,\,\,\prod_{j \not= k}\pars{X + j} & = \pars{-k + n}\pars{-k + n - 1}\ldots 1\bracks{\pars{-1}\ldots\pars{-k}} \\[5mm] & = \pars{n - k}!\bracks{\pars{-1}^{k}\,1 \times 2 \ldots k} = \color{#f00}{\pars{-1}^{k}\,k!\pars{n - k}!} \end{align}