Binomial Theorem expansion and proving an interesting identity?

Question

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.

Answer 1

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 $$ f(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}g(k)} \quad \Leftrightarrow \quad g(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \binom{n}{k}f(k)} $$

Answer 2

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 $\pm i!(n-i)!A_i = n!$.

Answer 3

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.

Answer 4

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 $x=2$.