Jtdylktuy

Given the following formula
$$
sum^n_{k=0}frac{(-1)^k}{k+x}binom{n}{k},.
$$
How can I show that this is equal to
$$
frac{n!}{x(x+1)cdots(x+n)},?
$$

Induction step:

$$begin{align}
sum_{k=0}^{n+1}&frac{(-1)^k}{x+k}binom{n+1}k=frac1x+frac{(-1)^{n+1}}{x+n+1}+sum_{k=1}^{n}frac{(-1)^k}{x+k}left[binom nk+binom n{k-1}right]
\&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}+sum_{k=1}^{n}frac{(-1)^k}{x+k}binom{n}{k-1}
\&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}-sum_{k=0}^{n-1}frac{(-1)^k}{(x+1)+k}binom{n}{k}
\&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}-frac{n!}{(x+1)(x+2)cdots(x+n+1)}+frac{(-1)^n}{x+n+1}
\&=frac{n!(x+n+1)-n!x}{x(x+1)cdots(x+n+1)}=frac{(n+1)!}{x(x+1)cdots(x+n+1)}
end{align}$$

Consider the (unique) polynomial $p(x)inmathbb{Q}[x]$ of degree at most $n$ such that $p(-k)=1$ for all $k=0,1,2,ldots,n$. Clearly, $p(x)$ is the constant polynomial $1$.

However, using Lagrange interpolation, we have
$$p(x)=sum_{k=0}^n,p(-k),frac{prodlimits_{jin[n]setminus{k}},(x+j)}{prodlimits_{jin[n]setminus{k}},(-k+j)},,$$
where $[n]:={0,1,2,ldots,n}$. This means
$$1=sum_{k=0}^n,frac{prodlimits_{jin[n]setminus{k}},(x+j)}{prodlimits_{jin[n]setminus{k}},(-k+j)}=sum_{k=0}^n,(-1)^k,frac{prodlimits_{jin[n]setminus{k}},(x+j)}{k!,(n-k)!},.$$
Multiplying both sides by $dfrac{n!}{prodlimits_{jin[n]},(x+j)}$ yields
$$frac{n!}{prodlimits_{j=0}^n,(x+j)}=sum_{k=0}^n,(-1)^k,left(frac{n!}{k!,(n-k)!}right),frac{1}{x+k}=sum_{k=0}^n,binom{n}{k},frac{(-1)^k}{x+k},.$$

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newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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}}}
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With $ds{Repars{x} > 0}$:

begin{align}
&bbox[10px,#ffd]{sum_{k = 0}^{n}{pars{-1}^{k} over k + x}{n choose k}} =
sum_{k = 0}^{n}pars{-1}^{k}
pars{int_{0}^{1}t^{k + x - 1},dd t}{n choose k}
\[5mm] = &
int_{0}^{1}t^{x - 1}sum_{k = 0}^{n}
{n choose k}pars{-t}^{k},dd t
\[5mm] = &
int_{0}^{1}t^{x - 1},pars{1 - t}^{n},dd t =
mrm{B}pars{x,n + 1} pars{~mrm{B}: Beta Function~}
\[5mm] = &
{Gammapars{x}Gammapars{n + 1} over Gammapars{x + n + 1}}
phantom{= mrm{B}pars{x,n + 1},,,,,,,,,,,,}
pars{~Gamma: Gamma Function~}
\[5mm] = &
{n! over Gammapars{x + n + 1}/Gammapars{x}} =
{n! over x^{overline{n +1}}} =
bbx{n! over xpars{x + 1}cdotspars{x + n}}
end{align}