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)? 
 A: 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.
A: You could even get more than the limit itself since, for $r$ close to $0$
$$\binom{n}{r}=1+r H_n+\frac{1}{12} r^2 \left(6 \left(H_n\right){}^2-6 \psi ^{(1)}(n+1)-\pi
   ^2\right)+O\left(r^3\right)$$
$$\frac 1{\binom{n}{r}}=1-r H_n+\frac{1}{12} r^2 \left(6 \left(H_n\right){}^2+6 \psi ^{(1)}(n+1)+\pi
   ^2\right)+O\left(r^3\right)$$
$$\frac1r\left(1-\frac1{\binom{n}{r}}\right)=H_n-\frac{1}{12} r \left(6 \left(H_n\right){}^2+6 \psi ^{(1)}(n+1)+\pi
   ^2\right)+O\left(r^2\right)$$
A: We can write
$$
\eqalign{
  & {1 \over r}\left( {1 - {1 \over {\left( \matrix{ n \cr   r \cr}  \right)}}} \right)
 = {1 \over r}\left( {1 - {{\Gamma \left( {r + 1} \right)\Gamma \left( {n - r + 1} \right)} \over {\Gamma \left( {n + 1} \right)}}} \right) =   \cr 
  &  = \left( {{{\Gamma \left( {n + 1} \right) - \Gamma \left( {r + 1} \right)\Gamma \left( {n + 1 - r} \right)} \over {r\,\Gamma \left( {n + 1} \right)}}} \right) =   \cr 
  &  = {1 \over {\Gamma \left( {n + 1} \right)}}\left( {{{\Gamma \left( {n + 1} \right) - \left( {1 - \gamma \,r + O(r^{\,2} )} \right)\Gamma \left( {n + 1 - r} \right)} \over r}} \right) =   \cr 
  &  = {1 \over {\Gamma \left( {n + 1} \right)}}\left( {{{\Gamma \left( {n + 1} \right) - \Gamma \left( {n + 1 - r} \right)} \over r}
 + \gamma \,\Gamma \left( {n + 1 - r} \right) + O(r)\,\Gamma \left( {n + 1 - r} \right)} \right) \cr} 
$$
and thus
$$
\eqalign{
  & \mathop {\lim }\limits_{r\, \to \,0} \;{1 \over r}\left( {1 - {1 \over {\left( \matrix{
  n \cr 
  r \cr}  \right)}}} \right) = \mathop {\lim }\limits_{r\, \to \,0} {{\left( {{{\Gamma \left( {n + 1} \right) - \Gamma \left( {n + 1 - r} \right)} \over r}} \right)} \over {\Gamma \left( {n + 1} \right)}} + \gamma  =   \cr 
  &  = \psi (n + 1) + \gamma  = H_{\,n} \; \cr} 
$$
A: We can apply L'Hôpital's rule in the $0/0$ indeterminate form case giving
$$\lim_{r\to0}\frac{1-1/\binom{n}{r}}{r}=\lim_{r\to0}\frac{\psi(n-r+1)-\psi(r+1)}{\binom{n}{r}}=\psi(n+1)+\gamma=H_n$$
A: Although $\binom{n}{r}$ is usually only defined for $r\in\mathbb{Z}$, this would be a natural extension for non-integral $r$. With this extension, and this answer, which says that $\frac{\Gamma'(x)}{\Gamma(x)}=-\gamma+H(x-1)$,
$$
\begin{align}
\lim_{r\to0}\frac{\mathrm{d}}{\mathrm{d}r}\binom{n}{r}
&=\lim_{r\to0}\binom{n}{r}\left[\frac{\Gamma'(n-r+1)}{\Gamma(n-r+1)}-\frac{\Gamma'(r+1)}{\Gamma(r+1)}\right]\\
&=\frac{\Gamma'(n+1)}{\Gamma(n+1)}-\frac{\Gamma'(1)}{\Gamma(1)}\\[6pt]
&=(H_n-\gamma)-(H_0-\gamma)\\[12pt]
&=H_n
\end{align}
$$
Thus,
$$
\begin{align}
\lim_{r\to0}\frac1r\left(1-\frac1{\binom{n}{r}}\right)
&=\lim_{r\to0}\frac1{\binom{n}{r}}\frac{\binom{n}{r}-1}{r}\\
&=\lim_{r\to0}\frac1{\binom{n}{r}}\frac{\binom{n}{r}-\binom{n}{0}}{r-0}\\
&=\left.\frac1{\binom{n}{r}}\frac{\mathrm{d}}{\mathrm{d}r}\binom{n}{r}\right|_{r=0}\\[9pt]
&=H_n
\end{align}
$$
