Asymptotic of binomial coeficient

Question

I was doing a problem, and I found that I needed to calculate asymptotics for $$ \frac{1}{{n - k \choose k}}$$ Supposing $n = k^2$.

Any help with this would be appreciated, thanks.

Answer 1

There is approximation of ${n\choose k}$ using Entropy function as follows:

$$ {n\choose np}\approx e^{nH(p)} $$ where $H(p)=-p\log p-(1-p)\log(1-p)$. To use this, we can choose $p=\frac{1}{k-1}$ and we get: $$ {k^2-k\choose k}\approx e^{k^2H(\frac{1}{k-1})} $$

Answer 2

$\displaystyle{% {k^{2} - k \choose k}^{-1} = {k!\left(k^{2} - 2k\right)! \over \left(k^{2} - k\right)!}.\quad \mbox{Use Stirling's formula:}\ N! \sim \sqrt{2\pi\,}\,N^{N\ +\ 1/2}\,{\rm e}^{-N}\,,\quad N \gg 1.}$

\begin{align} \left.{k^{2} - k \choose k}^{-1}\right\vert_{k\ \gg\ 1} &\sim {\left[\sqrt{\vphantom{\large A}2\pi\,}\,k^{k\ +\ 1/2}\,{\rm e}^{-k}\right] \left[\sqrt{\vphantom{\large A}2\pi\,}\, \left(k^{2} - 2k\right)^{k^{2} - 2k\ +\ 1/2}\,{\rm e}^{-k^{2}\ +\ 2k}\right] \over \sqrt{\vphantom{\large A}2\pi\,}\,\left(k^{2}\ - k\right)^{k^{2}\ -\ k\ +\ 1/2}\, {\rm e}^{-k^{2}\ +\ k}} \\[3mm]&= {1 \over \sqrt{\vphantom{\large A}2\pi\,}}\, {k^{k + 1/2} \left[k^{2k^{2}\ -\ 4k\ +\ 1}\left(1 - 2/k\right)^{k^{2}\ -\ 2k\ +\ 1/2}\right] \over k^{2k^{2}\ -\ 2k\ +\ 1}\left(1 - 1/k\right)^{k^{2}\ -\ k\ +\ 1/2}} \\[3mm]&\sim {1 \over \sqrt{\vphantom{\large A}2\pi\,}}\,k^{-k + 1/2} {\rm e}^{-k + 3/2} \end{align}

since $\left(~\mbox{for}\ k \gg 1~\right)$: \begin{align} \left(k^{2} - 2k + {1 \over 2}\right)\ln\left(1 - {2 \over k}\right) &\sim -2k + 2 + {1 \over 3k} + {1 \over 3k^{2}} + \cdots \\[3mm] \left(k^{2} - k + {1 \over 2}\right)\ln\left(1 - {1 \over k}\right) &\sim -k + {1 \over 2} - {1 \over 3\,k} - {1 \over 6\,k^{2}} \end{align}

$$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% {k^{2} - k \choose k}^{-1} \sim {{\rm e}^{3/2} \over \sqrt{\vphantom{\large A}2\pi\,}}\, {{\rm e}^{-k} \over k^{k - 1/2}}\,,\quad k \gg 1\quad} \\ \\ \hline \end{array} $$

Answer 3

Stirling yields $$ {k^2\choose k}^{-1}\sim\sqrt{2\pi\mathrm e^3k}\cdot(k\mathrm e)^{-k}. $$