Binomial Theorem expansion and proving an interesting identity? In the identity 
$$\frac {n!}{x(x+1)(x+2)...(x+n)}  = \sum ^n_{k=0}\frac {A_k}{x+k} $$
Prove that $$A_k =(-1)^{k}\:\cdot\: ^{n}C_k$$
Also from this deduce that,
$$ \;^{n}C_0\frac 1{1.2} - \:^{n}C_1\frac1{2.3} +\; ^{n}C_2\frac1{3.4} \;  ... \;{(-1)^n}\; ^{n}C_n\frac1{(n+1)(n+2)}\;=\frac1{(n+2)}$$
So I have to tried to use the binomial theorem on,
$(b-a)^n$, and then multiplied both sides by $a^{x-1}$.
 Now I integrated both sides with respect to $a$. This gives me the binomially expanded side as same as the right hand side of the identity that we have to prove when I substitute $a=1$. I dont know how to integrate $a^{x-1}\;(b-a)^n$ with respect to $a$.  I am not able to prove the identity and solve the deduction. Can someone please help me out ? Thanks a lot.
 A: There are many ways to demonstrate such an interesting identity.   
a) Induction
I do not know at what level you are, so let's start with what should be the simpler: Induction
Given the thesis
$$
F(x,n) = {{n!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)}} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
\left( { - 1} \right)^{\,k} \binom{n}{k}{1 \over {x + k}}} 
$$


*

*for $n=0$ it is true for whichever value of $x$ different from $0$
$$
n = 0\quad  \Rightarrow \quad F(x,0) = {1 \over x} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,0} 
 \binom{0}{k}{1 \over {x + k}}}  = {1 \over x}\;:\;TRUE
$$

*for $n+1$ the LHS is
$$
\eqalign{
  & F(x,n + 1) = {{\left( {n + 1} \right)!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)\left( {x + n + 1} \right)}} =   \cr 
  &  = {{\left( {x + n + 1 - x} \right)n!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)\left( {x + n + 1} \right)}} =   \cr 
  &  = {{n!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)}} - {{n!} \over {\left( {x + 1} \right) \cdots \left( {x + n} \right)\left( {x + n + 1} \right)}} =   \cr 
  &  = F(x,n) - F(x + 1,n) \cr} 
$$
the same as the RHS
$$
\eqalign{
  & F(x,n + 1) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n+1}{k}{1 \over {x + k}}}  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}{1 \over {x + k}}}
  + \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k-1}{1 \over {x + k}}}  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}{1 \over {x + k}}}
  - \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k - 1} \binom{n}{k-1}{1 \over {x + 1 + k - 1}}}  =   \cr 
  &  = F(x,n) - F(x + 1,n) \cr} 
$$
and the thesis is demonstrated.
b) Finite Difference
Writing the Forward Difference of the function $f(x)$ wrt to the variable $x$ as
$$
\Delta _{\,x} \,f(x) = f(x + 1) - f(x)
$$
its iteration gives
$$
\Delta _{\,x} ^{\,n} \,f(x) = \Delta _{\,x} \,\left( {\Delta _{\,x} ^{\,n - 1} f(x)} \right) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
  \left( { - 1} \right)^{\,n - k} \binom{n}{k}f(x + k)} 
$$
So
$$
F(x,n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}{1 \over {x + k}}} 
 = \left( { - 1} \right)^{\,n} \Delta _{\,x} ^{\,n} \,\left( {{1 \over x}} \right)
$$
We can easily see that
$$
\eqalign{
  & \Delta _{\,x} ^{\,0} \,\left( {{1 \over x}} \right)\mathop  \equiv \limits^{def} {1 \over x}  \cr 
  & \Delta _{\,x} ^{\,1} \,\left( {{1 \over x}} \right) = {1 \over {x + 1}} - {1 \over x} = {{\left( { - 1} \right)} \over {x\left( {x + 1} \right)}}  \cr 
  & \Delta _{\,x} ^{\,2} \,\left( {{1 \over x}} \right) =  - {1 \over {\left( {x + 1} \right)\left( {x + 2} \right)}} + {1 \over {x\left( {x + 1} \right)}} = {{\left( { - 1} \right)\left( { - 2} \right)} \over {x\left( {x + 1} \right)\left( {x + 2} \right)}}  \cr 
  & \quad \quad  \vdots   \cr 
  & \Delta _{\,x} ^{\,n} \,\left( {{1 \over x}} \right) = {{\left( { - 1} \right)^{\,n} n!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)}} \cr} 
$$
and that demontrates the thesis.
c) Falling / Rising Factorials
For a more general approach, it's advisable to resort to
the properties of the Rising and Falling Factorials, in order that we can write
$$
{{n!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)}} = {{n!} \over {x^{\,\overline {\,n + 1\,} } }}
 = n!\;\left( {x - 1} \right)^{\,\underline {\, - \,\left( {n + 1} \right)} } 
$$
where $x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} } $ represent respectively the Falling and Rising Factorial.
One of the basic properties of the falling factorial is that its Delta resembles the rule of differentiation of "normal" powers 
$$
\Delta _{\,x} \;x^{\,\underline {\,n\,} }  = \left( {x + 1} \right)^{\,\underline {\,n\,} }  - x^{\,\underline {\,n\,} }  = nx^{\,\underline {\,n - 1\,} } 
$$
Therefore one automatically derives that
$$ \bbox[lightyellow] {  
\eqalign{
  & F(x,n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}{1 \over {x + k}}}
    = \left( { - 1} \right)^{\,n} \Delta _{\,x} ^{\,n} \,\left( {{1 \over x}} \right) =   \cr 
  &  = \left( { - 1} \right)^{\,n} \Delta _{\,x} ^{\,n} \,\left( {\left( {x - 1} \right)^{\,\underline {\, - \,1} } } \right) = \left( { - 1} \right)^{\,n} \Delta _{\,x - 1} ^{\,n} \,\left( {\left( {x - 1} \right)^{\,\underline {\, - \,1} } } \right) =   \cr 
  &  = \left( { - 1} \right)^{\,n} \left( { - 1} \right)\left( { - 2} \right) \cdots \left( { - n} \right)\left( {x - 1} \right)^{\,\underline {\, - 1 - \,n\;} }  =   \cr 
  &  = \left( { - 1} \right)^{\,n} \left( { - 1} \right)^{\,\underline {\,\,n\;} } \left( {x - 1} \right)^{\,\underline {\, - 1 - \,n\,} }  = 1^{\,\overline {\,n\,} } \left( {x - 1} \right)^{\,\underline {\, - 1 - \,n\,} }  = {{n!} \over {x^{\,\overline {\,n + 1\,} } }} \cr} 
 }$$
Also,  the second part of your question is easily solved as
$$
\eqalign{
  & G(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
\left( { - 1} \right)^{\,k} \binom{n}{k}{1 \over {\left( {k + 1} \right)\left( {k + 2} \right)}}}  =   \cr 
  &  = \left. {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {
  \left( { - 1} \right)^{\,k} \binom{n}{k} {1 \over {\left( {x + k} \right)\left( {x + 1 + k} \right)}}\;} } \right|_{\,x\, = \,1} = \cr
  &  = \left. {G(x,n)} \right|_{\,x\, = \,1}  \cr} 
$$
therefore
$$ \bbox[lightyellow] {  
\eqalign{
  & G(x,n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}
  {1 \over {\left( {x + k} \right)\left( {x + 1 + k} \right)}}\;}  =   \cr 
  &  = \left( { - 1} \right)^{\,n} \Delta _{\,x} ^{\,n} \left( {{1 \over {\left( x \right)\left( {x + 1} \right)}}} \right)
 = \left( { - 1} \right)^{\,n} \Delta _{\,x} ^{\,n} \left( {{1 \over {x^{\,\overline {\,2\,} } }}} \right) =   \cr 
  &  = \left( { - 1} \right)^{\,n} \Delta _{\,x} ^{\,n} \left( {x - 1} \right)^{\,\underline {\, - 2\,} }
  = \left( { - 1} \right)^{\,n} \left( { - 2} \right)^{\,\underline {\,n\,} } \left( {x - 1} \right)^{\,\underline {\, - 2 - n\,} }  =   \cr 
  &  = {{2^{\,\overline {\,n\,} } } \over {x^{\,\overline {\,n + 2\,} } }}
 = {{\left( {n + 1} \right)!} \over {x^{\,\overline {\,n + 2\,} } }}\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad G(1,n) = {{\left( {n + 1} \right)!} \over {1^{\,\overline {\,n + 2\,} } }}
 = {{\left( {n + 1} \right)!} \over {\left( {n + 2} \right)!}} = {1 \over {\left( {n + 2} \right)}} \cr} 
 }$$
Finally, it is surely interesting to point out that the Inversion property of the Binomial Convolution
implies
$$
f(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}g(k)} 
\quad  \Leftrightarrow \quad
 g(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}f(k)} 
$$
A: For the second identity, start with the binomial theorem:
$$
(1-t)^n=\sum_{k=0}^n\binom{n}k(-1)^kt^k.
$$
Integrating both sides from $0$ to $x$ with respect to $dt$,
$$
\frac{1-(1-x)^{n+1}}{n+1}=\sum_{k=0}^n\binom{n}k(-1)^k\frac{x^{k+1}}{k+1}.
$$
Integrating both sides from $0$ to $1$ with respect to $dx$,
$$
\frac1{n+2}=\sum_{k=0}^n\binom{n}k(-1)^k\frac1{(k+1)(k+2)}.
$$
I cannot see how the second result follows immediately from the first.
A: Hint: Place the RHS over a common denominator. This gives in the numerator a sum of terms of the form
$$A_ix(x+1)\cdots(x+i-1)(x+i+1)\cdots(x+n).$$
Substitute successively $x=0$, $x=1,\dotsc, x=n$. This gives the result directly once you get the signs right: for each $i$, one gets $\pm i!(n-i)!A_i = n!$.
A: 
We obtain
  \begin{align*}
\frac{n!}{x(x+1)\cdots(x+n)}&=\sum_{k=0}^n\frac{A_k}{x+k}\\
n!&=\sum_{k=0}^nA_k\frac{x(x+1)\cdots(x+n)}{x+k}\tag{1}
\end{align*}
  Substituting   $x=-j,0\leq j\leq n$   in (1) we get
  \begin{align*}
n!&=A_j(-j)(-j+1)\cdots   (-1)\cdots1\cdot   2\cdots(n-j)\\
&=A_j(-1)^j  j!(n-j)!\\
\end{align*}
  We get
  \begin{align*}
\color{blue}{A_j}&=(-1)^j\frac{n!}{j!(n-j)!}\\
&\,\,\color{blue}{=(-1)^j\binom{n}{j}\qquad\qquad 0\leq j\leq n}\tag{2}
\end{align*}
  and the claim follows.

Using (1) and (2) we want to show $\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{(k+1)(k+2)}=\frac{1}{n+2}$.

We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k\binom{n}{k}\frac{1}{(k+1)(k+2)}}\\
&=\sum_{k=0}^n(-1)^k\binom{n}{k}\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\\
&=\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{k+1}-\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{k+2}\\
&=\frac{n!}{1\cdot 2\cdots (n+1)}-\frac{n!}{2\cdot 3\cdots (n+2)}\tag{3}\\
&=\frac{1}{n+1}-\frac{1}{(n+1)(n+2)}\\
&\,\,\color{blue}{=\frac{1}{n+2}}
\end{align*}
and the claim  follows.

In (3) we apply (1) and (2) with $x=1$ and $x=2$.
