Limit of binomial coefficients

Question

Let $0\leq a_n\leq n$ be a sequence of integers. Under which condition on the $a_n$ does $$\frac{{n-a_n\choose a_n}}{{n\choose a_n}}=\frac{(n-a_n)(n-a_n-1)\dots(n-2a_n+1)}{n(n-1)\dots(n-a_n+1)}$$ (with the convention that ${a\choose b}=0$ for $a<0$) converge to $0$ and $1$? It seems that it converges to $1$ if $a_n$ grows slower than $\sqrt{n}$ whereas it converges to $0$ when $a_n$ grows faster than $\sqrt{n}$. Why is this the case? What if $a_n$ grows as $\sqrt{n}$?

Answer 1

Define $$ A_n = \frac{\binom{n-a_n}{a_n}}{\binom{n}{a_n}} = \prod_{k=0}^{a_n-1} \left(1-\frac{a_n}{n-k}\right) \tag{$\ast$} $$ I'll estimate this just by replacing all the factors in the product with the smallest/largest factor, then using the standard inequalities $1-px\le (1-x)^p\le e^{-px}$ (under suitable conditions).

Fast growth. First suppose $\frac{a_n}{\sqrt n}\to\infty$; we want to show that $A_n\to 0$. If $a_n > \frac n2$ (for a particular $n$), then $n-a_n<a_n$, so $\binom{n-a_n}{a_n} = 0$ and $A_n=0$. If $a_n\le\frac n2$, then $1-\frac{a_n}{n-k}>0$ when $0\le k\le a_n-1$, so the factors on the RHS of ($\ast$) are all positive and we can replace them all with the largest one, getting $$ 0\le A_n \le \left(1-\frac{a_n}{n}\right)^{a_n} \le e^{-a_n^2/n} \to 0 $$

Slow growth. On the other hand, suppose $\frac{a_n}{\sqrt n}\to 0$; we want to show that $A_n\to 1$. For sufficiently large $n$, we have $a_n\le\sqrt n\le\frac n2$, so again the factors on the RHS of ($\ast$) are all positive, and we can replace them all with the smallest one, getting $$ 1\ge A_n \ge \left(1-\frac{a_n}{n-a_n+1}\right)^{a_n} \ge 1 - \frac{a_n^2}{n-a_n+1} \to 1 $$

Medium growth. If $\frac{a_n}{\sqrt n}\to L\in(0,\infty)$, then with minor adjustments the above arguments will give that $$ 1-L^2\le\liminf_{n\to\infty} A_n\le\limsup_{n\to\infty} A_n\le e^{-L^2} $$ If the lower bound is not precise enough for your purposes, perhaps you could use an argument like \begin{align*} \log A_n &\ge a_n\log\left(1-\frac{a_n}{n-a_n+1}\right) \\ &= a_n\left(-\frac{a_n}{n-a_n+1}+O\left(\frac{a_n^2}{(n-a_n+1)^2}\right)\right) \\ &= -\underbrace{\frac{a_n^2}{n-a_n+1}}_{\to L^2} + O\Bigg(\underbrace{\frac{a_n^3}{(n-a_n+1)^2}}_{\to 0}\Bigg) \end{align*} which, now that I think of it, seems like it's probably the right way to go in the first place.

Update: A more explicit version of that last argument can be given using the inequality $$ \log(1+x) \ge \frac{x}{1+x} $$ which is valid for all $x>-1$. (See this question, for example.) Since $\frac{a_n}{\sqrt n}\to L$, we have $a_n<\frac{n+1}2$ eventually and so $-\frac{a_n}{n-a_n+1}>-1$. Thus we can apply the inequality, getting $$ \log A_n \ge a_n\log\left(1-\frac{a_n}{n-a_n+1}\right) \ge \frac{a_n\left(\frac{-a_n}{n-a_n+1}\right)}{1-\frac{a_n}{n-a_n+1}} = -\frac{a_n^2}{n-2a_n+1} \to -L^2 $$

Answer 2

Consider the expression that is very similar to the RHS:

$$ \left( \frac{n - \sqrt{n}}{n} \right)^\sqrt{n}$$

Then, the limit is $1/e$. Replacing $\sqrt{n}$ by some other power of $n$, you can deduce $0$ and $1$ limits you mentioned.