[Math] Growth of binomial coefficient

Question

I am interested in the growth of the binomial coefficient ${n\choose n^a}$ for some fixed $a\in (1/2,1]$. Of course, for $a=1$ the binomial constantly equal to $1$. For $a<1$, computations suggest that
\begin{align}\lim_{n\to\infty}\frac{\exp(n^b)}{{n\choose n^a}}=\begin{cases}0&\text{if }b\leq a\\\infty&\text{if }b>a\end{cases}\end{align} i.e., that ${n\choose n^a}$ grows slightly faster than $\exp(n^a)$ (but for example slower than $\exp(n^a\log n)$). Is there a simple form of the exact growth rate? I find it slightly surprising that the growth rate of ${n\choose n^a}$ increases with $a$ although the binomial coefficient is constantly $1$ for $a=1$.

Answer 1

This should be nice enough: $${n\choose n^a}\sim_{n\to\infty}\left(\frac{n}{n-n^a}\right)^n\left(\frac{n-n^a}{n^a}\right)^{n^a}\frac{1}{\sqrt{2\pi\left(n^a-n^{2a-1}\right)}}. $$

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We have $${n\choose{n^a}}=\frac{n!}{(n-n^a)!\ n^a!},$$ but since $n-n^a$ is not an in integer for all $0<a<1$ (and $n\ne 0,1)$, making use of the $\Gamma$ function write $$\frac{\Gamma(n+1)}{\Gamma(n-n^a+1)\ \Gamma(n^a+1)}. $$ Even so, Stirling's approximation yields $$\frac{\Gamma(n+1)}{\Gamma(n-n^a+1)\ \Gamma(n^a+1)}\sim_{n\to\infty}\frac{\sqrt{2\pi n}(n/e)^n}{\sqrt{4\pi^2 n^a(n-n^a)}\left((n-n^a)/e\right)^{n-n^a}(n^a/e)^{n^a}}=\\ \frac{n^ne^{n-n^a}e^{n^a}}{e^n\sqrt{2\pi(n^a-n^{2a-1})}(n-n^a)^{n-n^a}n^{a\cdot n^a}}=\left(\frac{n}{n-n^a}\right)^n\left(\frac{n-n^a}{n^a}\right)^{n^a}\frac{1}{\sqrt{2\pi\left(n^a-n^{2a-1}\right)}}. $$

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Indeed, we can see now that as $a$ approaches $1$, the binomial coefficient increases. We still need $n\to \infty$ because when $a=1$ the asymptotic formula does not hold, since two fractions in it exist for $a\ne1$, and thus $0!=1$ applies. A significant example: $$\frac{\Gamma(201)}{\Gamma\left(201-200^{0.9}\right) \ \Gamma\left(1+200^{0.9}\right)}\approx 3.9\times 10^{57} $$ but $$\frac{\Gamma(201)}{\Gamma\left(201-200^{0.999}\right) \ \Gamma\left(1+200^{0.999}\right)}\approx 264. $$