How to show that $\sum\limits_{k=0}^n (-1)^k\tfrac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}} = \frac{x}{x+n}$ I'm trying to show that 

$$\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}} = \frac{x}{x+n}$$

I've tried expanding this using definition of binomial coefficient, then binomial expansion, now I'm using induction to simplify $\sum_{k=0}^m (-1)^k\frac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}}$, but it looks very inelegant, is there a better way?
 A: We recall the Melzak's identity $$f\left(x+y\right)=x\dbinom{x+n}{n}\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{f\left(y-k\right)}{x+k},\, x,y\in\mathbb{R},\, x\neq-k
 $$ where $f
 $ is an algebraic polynomial up to degree $n
 $. So taking $f\left(z\right)\equiv1
 $ we get $$\frac{1}{x\dbinom{x+n}{n}}=\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{1}{x+k}
 $$ hence, by the binomial inversion, we have  $$\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{1}{\dbinom{x+k}{k}}=\color{red}{\frac{x}{x+n}}.$$
A: Here is a  variation  based upon telescoping.

We obtain
  \begin{align*}
\sum_{k=0}^n&(-1)^k\frac{\binom{n}{k}}{\binom{x+k}{k}}\\
&=\frac{1}{\binom{x+n}{n}}\sum_{k=0}^n(-1)^k\binom{x+n}{n-k}\tag{1}\\
&=\frac{1}{\binom{x+n}{n}}\left[\sum_{k=0}^{n-1}(-1)^k\left(\binom{x+n-1}{n-k}+\binom{x+n-1}{n-k-1}\right)
+(-1)^n\right]\tag{2}\\
&=\frac{1}{\binom{x+n}{n}}\left[\sum_{k=0}^{n-1}(-1)^k\binom{x+n-1}{n-k}-\sum_{k=1}^n(-1)^{k}\binom{x+n-1}{n-k}+(-1)^n\right]\tag{3}\\
&=\frac{\binom{x+n-1}{n}}{\binom{x+n}{n}}\tag{4}\\
&=\frac{x}{x+n}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we use 
$
\binom{n}{k}\binom{x+k}{k}^{-1}=\binom{x+n}{n}^{-1}\binom{x+n}{n-k}
$

*In (2) we use the binomial identity $\binom{p}{q}=\binom{p-1}{q}+\binom{p-1}{q-1}$

*In (3) we shift the index of the right-hand series by one

*In (4) we apply the telescoping
A: A well-known identity related to the Beta function is
$$\int_0^1 t^{p-1} (1-t)^{q-1} \; dt = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)}$$
from which we can easily derive 
$$\frac{1}{\binom{n}{k}} = (n+1)\int_0^1 t^k (1-t)^{n-k} \;dt$$
so
$$\sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{\binom{x+k}{k}} = \sum_{k=0}^n (-1)^k \binom{n}{k} (x+k+1) \int_0^1 t^k (1-t)^x \;dt = I_1+I_2$$
where we define
$$I_1 = \int_0^1 (1-t)^x x \sum_{k=0}^n (-1)^k \binom{n}{k} t^k \; dt$$
and
$$I_2 = \int_0^1 (1-t)^x \sum_{k=0}^n (-1)^k \binom{n}{k} (k+1) t^k \; dt$$
In $I_1$, substitute $\sum_{k=0}^n (-1)^k \binom{n}{k} t^k = (1-t)^n$, with result
$$I_1 = \int_0^1 x (1-t)^{n+x} \;dt = \frac{x}{(n+x)(1+n+x)}$$
Differentiating  $t (1-t)^n =\sum_{k=0}^n (-1)^k \binom{n}{k} t^{k+1}$ with respect to $t$, we find $(1-t)^{n-1} [1 - (n+1)t] = \sum_{k=0}^n (-1)^k \binom{n}{k} (k+1) t^k$. Substituting into $I_2$,
$$I_2 = \int_0^1 (1-t)^{x+n-1}[1-(n+1)t] \;dt =\frac{x}{1+n+x}$$
Combining these results and simplifying, we have
$$I_1+I_2 = \frac{x}{x+n}$$
which is the desired result.
A: Another way to demonstrate this identity is to start from the expression of the Forward Delta
$$
\Delta _{\,x} ^{\,n} f(x) = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,n - k} \left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)f(x + k)} 
$$
Understanding the rising and falling factorials as actually expressed through the Gamma function, we have
$$
\Delta _{\,x} ^m \;x^{\,\underline {\,r\,} }  = r^{\,\underline {\,m\,} } x^{\,\underline {\,r - m\,} } 
$$
Then
$$
\begin{gathered}
  f(x,n) = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)/\left( \begin{gathered}
  x + k \\ 
  k \\ 
\end{gathered}  \right)}  = \left( { - 1} \right)^{\,n} \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,n - k} \left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)/\left( \begin{gathered}
  x + k \\ 
  k \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \left( { - 1} \right)^{\,n} \left. {\Delta _{\,y} ^{\,n} \left( {1/\left( \begin{gathered}
  x + y \\ 
  y \\ 
\end{gathered}  \right)} \right)\;} \right|_{\,y\, = \,0}  = \left( { - 1} \right)^{\,n} \left. {\Delta _{\,y} ^{\,n} \left( {1/\left( \begin{gathered}
  x + y \\ 
  x \\ 
\end{gathered}  \right)} \right)\;} \right|_{\,y\, = \,0}  =  \hfill \\
   = \left( { - 1} \right)^{\,n} \left. {\Delta _{\,y} ^{\,n} \left( {\frac{{x!}}
{{\left( {x + y} \right)^{\,\underline {\,x\,} } }}} \right)\;} \right|_{\,y\, = \,0}  = \left( { - 1} \right)^{\,n} x!\;\left. {\Delta _{\,y} ^{\,n} \left( {y^{\,\underline {\, - x\,} } } \right)\;} \right|_{\,y\, = \,0}  =  \hfill \\
   = \left( { - 1} \right)^{\,n} x!\;\left( { - x} \right)^{\,\underline {\,n\,} } \left. {y^{\,\underline {\, - x - n\,} } \;} \right|_{\,y\, = \,0}  = x!\;x^{\,\overline {\,n\,} } \left. {y^{\,\underline {\, - x - n\,} } \;} \right|_{\,y\, = \,0}  =  \hfill \\
   = x!\;x^{\,\overline {\,n\,} } \frac{1}
{{\left( {x + n} \right)!}}\; = \frac{{x\left( {x + n - 1} \right)!}}
{{\left( {x + n} \right)!}} = \frac{x}
{{x + n}} \hfill \\ 
\end{gathered} 
$$
A: This is an application of the Binomial Transform to the Partial Fraction Decomposition of
$$
\frac1{x\binom{x+n}{n}}=\frac{n!}{x(x+1)\cdots(x+n)}=\sum_{k=0}^n\frac{(-1)^k\binom{n}{k}}{x+k}\tag1
$$
Equation $(1)$ can be derived easily using the Heaviside Method.
Since the Binomial Transform is its own inverse, applying it to $(1)$ gives
$$
\sum_{n=0}^m(-1)^n\binom{m}{n}\frac1{x\binom{x+n}{n}}=\frac1{x+m}\tag2
$$
Multiplying $(2)$ by $x$ gives the desired result:
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{n=0}^m(-1)^n\frac{\binom{m}{n}}{\binom{x+n}{n}}=\frac{x}{x+m}}\tag3
$$

The Binomial Transform Is Its Own Inverse
Let $b_n$ be the Binomial Transform of $a_k$:
$$
b_n=\sum_{k=0}^n(-1)^k\binom{n}{k}a_k\tag4
$$
then the Binomial Transform of $b_n$ is $a_m$:
$$
\begin{align}
\sum_{n=0}^m(-1)^n\binom{m}{n}b_n
&=\sum_{n=0}^m(-1)^n\binom{m}{n}\sum_{k=0}^n(-1)^k\binom{n}{k}a_k\tag{5a}\\
&=\sum_{k=0}^m\sum_{n=k}^m(-1)^{n+k}\binom{m}{n}\binom{n}{k}a_k\tag{5b}\\
&=\sum_{k=0}^m\sum_{n=k}^m(-1)^{n+k}\binom{m}{k}\binom{m-k}{n-k}[m\ge k]\,a_k\tag{5c}\\
&=\sum_{k=0}^m\binom{m}{k}[m=k]\,a_k\tag{5d}\\[6pt]
&=a_m\tag5
\end{align}
$$
Explanation: [...] are Iverson brackets
$\text{(5a)}$: $(4)$
$\text{(5b)}$: change the order of summation
$\text{(5c)}$: $\binom{m}{n}\binom{n}{k}=\binom{m}{k}\binom{m-k}{n-k}[m\ge k]$
$\text{(5d)}$: $\sum_{n=0}^{m-k}(-1)^n\binom{m-k}{n}[m\ge k]=(1-1)^{m-k}[m\ge k]=[m=k]$
$\phantom{a}\text{(5)}$: evaluate
