TITLE: The limit $\lim_{r\to0}\frac1r\left(1-\binom{n}{r}^{-1}\right)$
QUESTION [11 upvotes]: 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)?

REPLY [5 votes]: You could even get more than the limit itself since, for $r$ close to $0$
$$\binom{n}{r}=1+r H_n+\frac{1}{12} r^2 \left(6 \left(H_n\right){}^2-6 \psi ^{(1)}(n+1)-\pi
   ^2\right)+O\left(r^3\right)$$
$$\frac 1{\binom{n}{r}}=1-r H_n+\frac{1}{12} r^2 \left(6 \left(H_n\right){}^2+6 \psi ^{(1)}(n+1)+\pi
   ^2\right)+O\left(r^3\right)$$
$$\frac1r\left(1-\frac1{\binom{n}{r}}\right)=H_n-\frac{1}{12} r \left(6 \left(H_n\right){}^2+6 \psi ^{(1)}(n+1)+\pi
   ^2\right)+O\left(r^2\right)$$