Known approximations for binomial coefficients

Question

I keep coming back to this very often, so I am wondering if people are aware of a source that lists approximations of binomial coefficients using Stirling's approximation? I know some regimes are well known, (e.g. ${2n \choose n}$, then there's some unpleasant looking approximations for square roots at this link), but I was wondering if there are nice known answers for some naturally arising regimes (say ${n \choose k}$ for $n=k^i$ and $i>2$, or other dependencies between $n$ and $k$)? Or at least do we know what regimes have nice closed-form expressions?

Answer 1

For rough estimates, I use $(n/e)^n < n! \lt (n/e)^{n+1}$. This is often good enough for showing convergence or divergence.

For more precision, Stirling's approximation $n! \approx \sqrt{2\pi n}(n/e)^n$ is usually good enough.

(The following is taken from earlier work of mine.)

Explicit bounds are (from https://en.wikipedia.org/wiki/Stirling%27s_approximation), if $f(n) =\dfrac{n!}{\sqrt{2\pi n}(n/e)^n}$, then the following inequality holds:

$$1 \lt f(n) \lt e^{1/(12n)} \lt \frac{e}{\sqrt{2\pi}} = 1.0844... $$

As an example of the use of Stirling, you can show that

$\binom{an}{bn} \sim \sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^a}{b^b(a-b)^{a-b}}\right)^n =r(n, a, b) $.

By using the bounds for $n!$ above, you can show that if $u < f(n) < v$, then $\frac{u}{v^2} \lt \dfrac{\binom{an}{bn}}{r(n, a, b)} \lt \frac{v}{u^2} $.

From the simple bounds above, we can take $u = 1$ and $v = \frac{e}{\sqrt{2\pi}} $, so the bounds are $\dfrac{2\pi}{e^2} =0.850... $ and $\dfrac{e}{\sqrt{2\pi}} =1.0844... $.

Therefore $0.850 \lt \dfrac{\binom{an}{bn}}{\sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^a}{b^b(a-b)^{a-b}}\right)^n} \lt 1.085 $.

Answer 2

In the same spirit as @marty cohen but working on a logarithm scale, let $a= k b$ $(k>1)$ to have $$\log \Bigg[\binom{k b n}{b n}\Bigg]=b\Big[k \log (k)-(k-1) \log (k-1)\Big]n-\frac{1}{2} \log \left(2 \pi\frac{ b (k-1) }{k}n\right)-$$ $$\frac {k^2-k+1 }{12 b k(k-1) }\,\frac 1n+\frac{k^6-3 k^5+3 k^4-k^3+3 k^2-3 k+1} { 360b^3 k^3(k-1)^3}\,\frac 1{n^3}+O\left(\frac{1}{n^5}\right)$$ which can be truncated whereever you wish.

Using for example $a=5$, $b=3$ and $n=7$, this truncated series would give $2319959403.8$ instead of the exact $2319959400$