how can I prove this equation from Binomial theorem?? $$\sum_{i=0}^n \frac{ {\left(\begin{array}{c}n\\  i\end{array}\right)} {(-1)^i}}{k+i} =  {\frac{1}{k\left(\begin{array}{c}k+n\\ k\end{array}\right)}}$$
 A: 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
$$
B(n+1,k) = \frac{\Gamma(n+1) \Gamma(k)}{\Gamma(n+1+k)} = \frac{n! (k-1)!}{(n+k)!} = \frac{1}{k\left(\begin{array}{c}k+n\\ k\end{array}\right)}.
$$
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[15px,#ffc]{\sum_{i = 0}^{n}
{{n \choose i}\pars{-1}^{i} \over k + i} = {1 \over k{k + n \choose k}}}:\ \ds{\large ?}}$.

\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}
