Stirling Formula

Question

Find the value of $\lambda$ for this question:

$\dbinom{8n}{4n} \sim \lambda \dfrac{2^{8n}}{\sqrt{n}}$ as $n \to \infty$

I tried using Stirling.

Any help appreciated.

Answer 1

From here, we have $$\dbinom{2m}m \sim \dfrac{4^m}{\sqrt{m\pi}}$$ Taking $m=4n$, we obtain $$\dbinom{8n}{4n} \sim \dfrac{4^{4n}}{\sqrt{4n\pi}} = \dfrac1{2\sqrt{\pi}} \cdot \dfrac{2^{8n}}{\sqrt{n}}$$

Answer 2

A more straightforward approach using factorials and Sterling:

${{8n}\choose{4n}} = \frac{8n!}{4n!4n!}$

We know that $n!$ is approximately $\sqrt{2\pi n}(\frac{n}{e})^n$

So plugging this in for both the numerator and the denominator and you have your result.

Answer 3

More generally, since $n! \sim \sqrt{2\pi n}(n/e)^n$,

$\begin{array}\\ \binom{an}{bn} &=\frac{(an)!}{(bn)!((a-b)n)!}\\ &\sim\frac{\sqrt{2\pi an}(an/e)^{an}}{\sqrt{2\pi bn}(bn/e)^{bn}\sqrt{2\pi (a-b)n}((a-b)n/e)^{(a-b)n}}\\ &=\sqrt{\frac{a}{2\pi b(a-b)n}}\frac{(an)^{an}}{(bn)^{bn}((a-b)n)^{(a-b)n}}\\ &=\sqrt{\frac{a}{2\pi b(a-b)n}}\frac{(a)^{an}}{(b)^{bn}((a-b))^{(a-b)n}}\\ &=\sqrt{\frac{a}{2\pi b(a-b)n}}\left(\frac{a^{a}}{b^{b}(a-b)^{(a-b)}}\right)^n\\ \end{array} $

Setting $a=8$ and $b=4$,

$\begin{array}\\ \binom{8n}{4n} &\sim\sqrt{\frac{8}{2\pi4(4)n}}\left(\frac{8^8}{4^4 4^4}\right)^n\\ &=\sqrt{\frac{1}{4\pi n}}\left(2^8\right)^n\\ &=\sqrt{\frac{1}{4\pi n}}2^{8n}\\ \end{array} $

so $\lambda =\sqrt{\frac{1}{4\pi }} =\frac{1}{2\sqrt{\pi} } $.