Known approximations for binomial coefficients 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?
 A: 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
$.
A: 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
$2319959400$
