how can I prove this equation from Binomial theorem??

Question

$$\sum_{i=0}^n \frac{ {\left(\begin{array}{c}n\\ i\end{array}\right)} {(-1)^i}}{k+i} = {\frac{1}{k\left(\begin{array}{c}k+n\\ k\end{array}\right)}}$$

Answer 1

Start with binomial expansion $$ (x-1)^n = \sum_{i=0}^n \left(\begin{array}{c}n\\ i\end{array}\right) (-1)^{n-i} x^{i}. $$ $$ (x-1)^n x^{k-1} = \sum_{i=0}^n \left(\begin{array}{c}n\\ i\end{array}\right) (-1)^{n-i} x^{i+k-1}. $$ Then integrate it from $0$ to $1$ and get $$ (-1)^n B(n+1,k) = \int_{0}^1 (x-1)^n x^{k-1} dx = \sum_{i=0}^n \left(\begin{array}{c}n\\ i\end{array}\right) (-1)^{n-i} \frac{1}{k+i}. $$ Now divide both sides by $(-1)^n$. This gives you a final result.

EDIT: Indeed, using porperties of $B$ and $\Gamma$ functions (see link) one can get $$ B(n+1,k) = \frac{\Gamma(n+1) \Gamma(k)}{\Gamma(n+1+k)} = \frac{n! (k-1)!}{(n+k)!} = \frac{1}{k\left(\begin{array}{c}k+n\\ k\end{array}\right)}. $$

Answer 2

Alternative approach: by residues or equivalent techniques it is not difficult to show that the function $$f_n(x) = \frac{1}{x(x+1)(x+2)\cdot\ldots\cdot(x+n)}=\frac{1}{(x)_{n+1}} $$ has the following partial fraction decomposition: $$ f_n(x) = \frac{1}{n!}\sum_{h=0}^{n}(-1)^h\binom{n}{h}\frac{1}{x+h} $$ from which it follows that $$ \sum_{h=0}^{n}(-1)^h\binom{n}{h}\frac{1}{k+h} = n!\cdot f_n(k) = \frac{n!}{k(k+1)\cdot\ldots\cdot(k+n)} $$ as wanted.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[15px,#ffc]{\sum_{i = 0}^{n} {{n \choose i}\pars{-1}^{i} \over k + i} = {1 \over k{k + n \choose k}}}:\ \ds{\large ?}}$.

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\begin{align} \sum_{i = 0}^{n}{{n \choose i}\pars{-1}^{i} \over k + i} & = \sum_{i = 0}^{n}{n \choose i}\pars{-1}^{i}\int_{0}^{1}t^{k + i - 1}\,\dd t = \int_{0}^{1}t^{k - 1}\sum_{i = 0}^{n}{n \choose i}\pars{-t}^{i}\,\dd t \\[3mm] & = \underbrace{\int_{0}^{1}t^{k - 1}\pars{1 - t}^{n}\,\dd t} _{\ds{\mrm{B}\pars{k,n + 1}}} = {\Gamma\pars{k}\Gamma\pars{n + 1} \over \Gamma\pars{k + n + 1}} \end{align}

$\ds{B}$ is the Euler Beta Integral which is written in terms of the Gamma Function $\ds{\Gamma}$.

In addition, \begin{align} \sum_{i = 0}^{n}{{n \choose i}\pars{-1}^{i} \over k + i} & = {\pars{k - 1}!\, n! \over \pars{k + n}!} = {1 \over k}\,{1 \over \pars{k + n}!/\pars{k!\, n!}} \\[3mm] & = \bbox[15px,#ffc,border:1px groove navy]{\large 1 \over \large k{k + n \choose k}} \end{align}