The limit $\lim_{r\to0}\frac1r\left(1-\binom{n}{r}^{-1}\right)$

I have deduced numerically (See also Wolfram) that we have $$\lim_{r\to0}\frac1r\left(1-\frac1{\binom{n}{r}}\right)=H_n$$ where $H_n$ denotes the $n$th Harmonic number and we define the binomial coefficient by $$\binom{n}{r}=\frac{\Gamma(n+1)}{\Gamma(r+1)\cdot\Gamma(n-r+1)}$$ Has anyone seen this result or similar elsewhere in mathematical literature? Is it possible to analytically prove this result? Would this result be any useful for calculating $H_n$ explicitly (especially for complex arguments)?

Let's use Beta function:

$$\frac1{\binom{n}{r}}=\frac{(n-r)! r!}{n!}=\frac{\Gamma(n-r+1) \Gamma (r+1)}{\Gamma (n+1)}=r B(r,n-r+1)=$$

$$=r \int_0^1 t^{r-1} (1-t)^{n-r} dt$$

We also can write:

$$1=r \int_0^1 t^{r-1} dt$$

Which gives us:

$$\lim_{r\to0}\frac1r\left(1-\frac1{\binom{n}{r}}\right)=\lim_{r\to0} \int_0^1 t^{r-1} \left(1-(1-t)^{n-r} \right) dt=$$

$$=\int_0^1 \frac{1-(1-t)^n}{t} dt=\int_0^1 \frac{1-t^n}{1-t} dt=H_n$$

As per the usual definition of Harmonic numbers.

You could even get more than the limit itself since, for $r$ close to $0$ $$\binom{n}{r}=1+r H_n+\frac{1}{12} r^2 \left(6 \left(H_n\right){}^2-6 \psi ^{(1)}(n+1)-\pi ^2\right)+O\left(r^3\right)$$

$$\frac 1{\binom{n}{r}}=1-r H_n+\frac{1}{12} r^2 \left(6 \left(H_n\right){}^2+6 \psi ^{(1)}(n+1)+\pi ^2\right)+O\left(r^3\right)$$ $$\frac1r\left(1-\frac1{\binom{n}{r}}\right)=H_n-\frac{1}{12} r \left(6 \left(H_n\right){}^2+6 \psi ^{(1)}(n+1)+\pi ^2\right)+O\left(r^2\right)$$