Proving that $\sum_{k=0}^{\infty} {x+k-1 \choose k}^{-1}=\frac{x-1}{x-2}, x \in \mathbb{R}_{ >2}$ While doing numerical experiments with series involving binomial coefficients at Mathematica I stumbled up on
$$\sum_{k=0}^{\infty} {x+k-1 \choose k}^{-1}=\frac{x-1}{x-2},
\quad x \in \mathbb{R}_{\ >2}.$$
The question is: How to prove it by hand?
 A: ${x+k-1 \choose k}^{-1}=\dfrac{(x-1)!k!}{(x+k-1)!}$
Introduce $\Gamma$ function:
$\dfrac{\Gamma(x)\Gamma(k+1)}{\Gamma(x+k)}=\dfrac{(x-1)\Gamma(x-1)\Gamma(k+1)}{\Gamma(x+k)}=(x-1)\beta(x-1, k+1)$
where $\beta((x-1, k+1)=\int\limits_0^1t^{x-2}(1-t)^kdt$
Put it back into the sum:
$(x-1)\int\limits_0^1t^{x-2}\sum\limits_{k=0}^\infty (1-t)^kdt=(x-1)\int\limits_0^1t^{x-3}dt=\dfrac{x-1}{x-2}$
A: Note that $${n \choose k}^{-1}=(n+1)\int_{0}^{1} t^k (1-t)^{n-k} dt$$
Then
$$S=\sum_{k=0}^{\infty} {x+k-1 \choose k}^{-1}=\sum_{k=0}^{\infty}  (k+x) \int_{0}^{1} t^k (1-t)^{x-1} dt$$
$$\implies S=\int_{0}^{1} (1-t)^{k-1} \sum_{k=0}^{\infty} (k+x) t^k dt= \int_{0}^{1} (1-t)^{x-1}\left( \frac{t}{(1-t)^2}+\frac{x}{1-t} \right)dt$$
$$\implies S=\int_{0}^{1} [ (x-1)(1-t)^{x-2}+(1-t)^{x-3}]~ dt=\frac{x-1}{x-2}, x >2.$$
A: Another Gamma Approach
$$
\begin{align}
\sum_{k=0}^\infty\binom{x+k-1}{k}^{-1}
&=1+\sum_{k=1}^\infty\frac{k\,\Gamma(k)\,\Gamma(x)}{\Gamma(x+k)}\tag1\\
&=1+\sum_{k=1}^\infty k\int_0^1t^{k-1}(1-t)^{x-1}\,\mathrm{d}t\tag2\\
&=1+\int_0^1\frac{(1-t)^{x-1}}{(1-t)^2}\,\mathrm{d}t\tag3\\[6pt]
&=1+\frac1{x-2}\tag4\\[6pt]
&=\frac{x-1}{x-2}\tag5
\end{align}
$$
Explanation:
$(1)$: pull the $k=0$ term out of the sum
$\phantom{\text{(1):}}$ write in terms of $\Gamma$
$(2)$: use the Beta function
$(3)$: $\sum_{k=1}^\infty kt^{k-1}=(1-t)^{-2}$
$(4)$: integrate
$(5)$: add
