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

Question

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)?

Answer 1

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.

Answer 2

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

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

Answer 3

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 $$ \eqalign{ & \mathop {\lim }\limits_{r\, \to \,0} \;{1 \over r}\left( {1 - {1 \over {\left( \matrix{ n \cr r \cr} \right)}}} \right) = \mathop {\lim }\limits_{r\, \to \,0} {{\left( {{{\Gamma \left( {n + 1} \right) - \Gamma \left( {n + 1 - r} \right)} \over r}} \right)} \over {\Gamma \left( {n + 1} \right)}} + \gamma = \cr & = \psi (n + 1) + \gamma = H_{\,n} \; \cr} $$

Answer 4

We can apply L'Hôpital's rule in the $0/0$ indeterminate form case giving $$\lim_{r\to0}\frac{1-1/\binom{n}{r}}{r}=\lim_{r\to0}\frac{\psi(n-r+1)-\psi(r+1)}{\binom{n}{r}}=\psi(n+1)+\gamma=H_n$$

Answer 5

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, $$ \begin{align} \lim_{r\to0}\frac1r\left(1-\frac1{\binom{n}{r}}\right) &=\lim_{r\to0}\frac1{\binom{n}{r}}\frac{\binom{n}{r}-1}{r}\\ &=\lim_{r\to0}\frac1{\binom{n}{r}}\frac{\binom{n}{r}-\binom{n}{0}}{r-0}\\ &=\left.\frac1{\binom{n}{r}}\frac{\mathrm{d}}{\mathrm{d}r}\binom{n}{r}\right|_{r=0}\\[9pt] &=H_n \end{align} $$