If $r>0$ and $r\notin \mathbb{N}$, is there a simple method to evaluate $ \sum_{n=\lceil r \rceil}^{\infty} {\binom{n}{r}^{-1}}?$

Question

Let $r>0,r\in \mathbb{R}\setminus\mathbb{N}$. Empirically, I have noticed the following relation: $$ \sum_{n=0}^{\lfloor r \rfloor} \frac{1}{\binom{n}{r}} = - \sum_{n=\lceil r \rceil}^{\infty} \frac{1}{\binom{n}{r}}; $$in particular, $\displaystyle{\sum_{n=0}^{\infty} \frac{1}{\binom{n}{r}} =0}$. Note that if $r$ is an integer, the finite sum is not well-defined, although we do have $$ \sum_{j=0}^{k-1} \operatorname{Res} \left(\frac{1}{\binom{z}{k}},z=j\right)= k\cdot\sum_{m=0}^{k-1}\binom{k-1}{m}(-1)^m=0, $$ so in this sense the sum 'cancels'. Mathematica returns the closed-form of $$ \sum_{n=\lceil r \rceil}^{\infty} \frac{1}{\binom{n}{r}}= \frac{\lceil r\rceil }{(r-1) \binom{\lceil r\rceil }{r}}, $$which when $r\in\mathbb{N}$ reduces to this question, but I don't know how to derive that myself. Maybe I'm not fully understanding the answers there but I don't think the same tricks apply when the sum doesn't telescope. So in summary, my questions are:

1. Can someone explain the closed-form?
2. Is there a simple, conceptual reason the finite sum is the negative of the infinite sum?

Answer 1

Here's a computation of the total sum from $n=0$ to $\infty$, which may lead a way to compute the finite sum. Since $$\binom xy\binom yx=\operatorname{sinc}(\pi(x-y)),$$ where $\operatorname{sinc}(x)=\sin(x)/x$, we have $$\frac{1}{\binom xy}=\binom yx\frac{\pi(x-y)}{\sin(\pi(x-y))};$$ in particular, $$\frac{1}{\binom nr}=\binom rn\frac{\pi(r-n)}{\sin(\pi(r-n))}=\pi(r-n)\binom rn\frac{(-1)^n}{\sin \pi r}.$$ So, we want to evaluate $$\frac{\pi}{\sin \pi r}\sum_{n=0}^\infty (-1)^n(r-n)\binom rn.$$ Consider $$f(x)=(1+x)^r=\sum_{n=0}^\infty \binom rnx^n.$$ We have $$\frac{d}{dx}(x^{-r}f(x))=\sum_{n=0}^\infty \binom rn\frac{dx^{n-r}}{x}=\sum_{n=0}^\infty \binom rn(n-r)x^{n-r-1};$$ also, $$\frac{d}{dx}(x^{-r}f(x))=\frac{d}{dx}\left(\frac{1+x}{x}\right)^r=-\frac{r\left(\frac{1+x}{x}\right)^{r-1}}{x^2},$$ and so we have the identity $$\sum_{n=0}^\infty \binom rn(-1)^n(r-n)\binom rn=(-1)^{r+1}\frac{d}{dx}\left(x^{-r}f(x)\right)\bigg|_{x=-1}=0$$ whenever $r>1$.