Has anyone seen this formula that I proved? $\lim_{n\to\infty} \frac{2n \choose n}{2n - 2m \choose n - m} = 4^m$ I'm a high-schooler from the Netherlands. I developed the following formula, and I proved it:
$$\lim_{n\to\infty} \frac{2n \choose n}{2n - 2m \choose n - m} = 4^m$$
Has anyone already seen my formula or does somebody know whether it already has been proven or not? I would like to call this 'Groen's first law' :) if it's still unknown.
Thank you and have a nice day!
Jari
 A: Your ‘ first law ’ is just a consequence of Stirling's approximation :
$\dfrac{n^n\sqrt{2\pi n}}{e^n}e^{\frac1{12n+1}}<n!<\dfrac{n^n\sqrt{2\pi n}}{e^n}e^{\frac1{12n}}\;\;,\;\;\forall\,n\in\mathbb N\!=\!\mathbb Z^+.\;\;\color{blue}{(*)}$
From the inequalities $\,(*)\,,\,$ it follows that
$\dfrac{\dfrac{\left(2n\right)^{2n}\sqrt{4\pi n}}{e^{2n}}e^\frac1{24n+1}}{\dfrac{n^{2n}\left(2\pi n\right)}{e^{2n}}e^\frac1{6n}}\!<\!\dfrac{\left(2n\right)!}{\left(n!\right)^2}\!<\!\dfrac{\dfrac{\left(2n\right)^{2n}\sqrt{4\pi n}}{e^{2n}}e^\frac1{24n}}{\dfrac{n^{2n}\left(2\pi n\right)}{e^{2n}}e^\frac2{12n+1}}\;,\;\;\forall\,n\in\mathbb N\;,$
$\displaystyle\dfrac{4^n}{\sqrt{\pi n}}e^{\frac1{24n+1}-\frac1{6n}}<\binom{2n}n<\dfrac{4^n}{\sqrt{\pi n}}e^{\frac1{24n}-\frac2{12n+1}}\;,\quad\forall\,n\in\mathbb N\;\;,$
$\displaystyle\dfrac{4^n}{\sqrt{\pi n}}e^{-\frac1{6n}}<\binom{2n}n<\dfrac{4^n}{\sqrt{\pi n}}\;\;,\;\;\quad\forall\,n\in\mathbb N\;.\quad\color{blue}{(1)}$
Let $\,m\,$ be any integer $\,\big(m\in\mathbb Z\big)\,$.
Let $\;\mathbb N_m^*=\big\{n\,\big|\,n\in\mathbb N\,\land\,n\geqslant m+1\big\}\subseteq\mathbb N\,.$
From the inequalities $\,(1)\,,\,$ it follows that
$\displaystyle\dfrac{4^{n-m}}{\sqrt{\pi\!\left(n\!-\!m\right)}}e^{-\frac1{6\left(n-m\right)}}\!<\!\binom{2n\!-\!2m}{n\!-\!m}\!<\!\dfrac{4^{n-m}}{\sqrt{\pi\!\left(n\!-\!m\right)}}\,,\;\forall\,n\!\in\!\mathbb N_m^*\,.\;\color{blue}{(2)}$
Moreover, from the inequalities $\,(1)\,$ and $\,(2)\,,\,$ it follows that
$\dfrac{\dfrac{4^n}{\sqrt{\pi n}}e^{-\frac1{6n}}}{\dfrac{4^{n-m}}{\sqrt{\pi\!\left(n\!-\!m\right)}}}\!<\!\dfrac{\displaystyle\binom{2n}n}{\displaystyle\binom{2n\!-\!2m}{n\!-\!m}}\!<\!\dfrac{\dfrac{4^n}{\sqrt{\pi n}}}{\dfrac{4^{n-m}}{\sqrt{\pi\!\left(n\!-\!m\right)}}e^{-\frac1{6\left(n-m\right)}}}\,,\;\;\forall\,n\!\in\!\mathbb N_m^*\;,$
$4^m\!\sqrt{1\!-\!\dfrac mn}\,e^{-\frac1{6n}}\!<\!\dfrac{\displaystyle\binom{2n}n}{\displaystyle\binom{2n\!-\!2m}{n\!-\!m}}\!<4^m\!\sqrt{1\!-\!\dfrac mn}\,e^{\frac1{6\left(n-m\right)}},\;\forall\,n\!\in\!\mathbb N_m^*\,.$
Since $\;\lim_\limits{n\to\infty}\left(\!4^m\!\sqrt{1\!-\!\dfrac mn}\,e^{-\frac1{6n}}\!\right)=\lim_\limits{n\to\infty}\left(\!4^m\!\sqrt{1\!-\!\dfrac mn}\,e^{\frac1{6\left(n-m\right)}}\!\right)=4^m\,,\,$
from the previous inequalities, by using Squeeze Theorem, it follows that
$\lim_\limits{n\to\infty} \dfrac{\displaystyle\binom{2n}n}{\displaystyle\binom{2n\!-\!2m}{n\!-\!m}}=4^m\;.$
A: Congratulations! As others have said, this is a straightforward consequence of very classical stuff such as Stirling's approximation $n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n$ or the known asymptotic ${2n \choose n} \sim \frac{4^n}{\sqrt{\pi n}}$ (which also follows from Stirling's but can be proven in other ways). Stirling's approximation is a powerful tool and can be proven in several interesting ways which you may enjoy learning about (some of which do a better or worse job of explaining what the heck that $2 \pi$ is doing there); see, for example, this blog post by Terence Tao.
However since you likely don't know either of these results that is likely not how you proved it. And in fact a much simpler argument is available (which we could guess based on the fact that no square roots or $\pi$s appear in this result), which I'm betting is similar to what you did. Write the quantity we're interested in as $f(n)$, suppressing the dependence on $m$ since it's fixed anyway. We just cancel the factorials:
$$\begin{align*} f(n) = \frac{ {2n \choose n} }{ {2n-2m \choose n-m} } &= \frac{2n! (n-m)!^2}{(2n-2m) n!^2} \\
 &= \frac{2n(2n-1) \dots (2n-2m+1)}{n^2(n-1)^2 \dots (n-m+1)^2} \end{align*}.$$
And now we can divide the numerator and the denominator by $n^{2m}$, which gives
$$f(n) = \frac{2 \left( 2 - \frac{1}{n} \right) \dots \left( 2 - \frac{2m-1}{n} \right)}{\left( 1 - \frac{1}{n} \right)^2 \left( 1 - \frac{2}{n} \right)^2 \dots \left( 1 - \frac{m-1}{n} \right)^2}.$$
Taking $\lim_{n \to \infty}$ gives
$$\lim_{n \to \infty} f(n) = \frac{2^{2m}}{1} = 4^m$$
as desired. This is all that's necessary, no Stirling or anything else.

Working more carefully can give us a more precise asymptotic as follows. First we note that half of the terms in the numerator are even terms of the form $2n - 2k$ and we can divide those by the corresponding terms $n - k$ in the denominator. This gives
$$f(n) = 2^m \frac{(2n-1)(2n-3) \dots (2n - 2m+1)}{n(n-1) \dots (n-m+1)}.$$
Next we'll divide each remaining term in the numerator by the corresponding term in the denominator, which gives
$$f(n) = 2^m \left( 2 - \frac{1}{n} \right) \left( 2 - \frac{1}{n-1} \right) \dots \left( 2 - \frac{1}{n-m+1} \right)$$
or alternatively
$$f(n) = 4^m \left( 1 - \frac{1}{2n} \right) \left( 1 - \frac{1}{2(n-1)} \right) \dots \left( 1 - \frac{1}{2(n-m+1)} \right).$$
To estimate this somewhat mysterious product we can argue as follows. First, if $0 \le x, y \le 1$ we have $(1 - x)(1 - y) = 1 - x - y + xy \ge 1 - x - y$. Inducting on this inequality gives the following result: if $0 \le x_i \le 1$ then
$$(1 - x_1) (1 - x_2) \dots (1 - x_n) \ge 1 - x_1 - x_2 - \dots - x_n.$$
This gives us a lower bound
$$f(n) \ge 4^m \left( 1 - \frac{1}{2n} - \frac{1}{2(n-1)} - \dots - \frac{1}{2(n-m+1)} \right)$$
which is a good bound as long as $m$ is small compared to $n$. In particular with this bound we can now analyze the behavior as $m$ is also allowed to go to infinity but more slowly than $n$. For an upper bound we have the inequality $1 - x \le e^{-x}$ which gives
$$(1 - x_1)(1 - x_2) \dots (1 - x_n) \le \exp \left( - x_1 - x_2 - \dots - x_n \right)$$
which gives an upper bound
$$f(n) \le 4^m \exp \left( - \frac{1}{2n} - \frac{1}{2(n-1)} - \dots - \frac{1}{2(n-m+1)} \right).$$
For both of these bounds we see that it's important to get a bound on the quantity $g(n) = \frac{1}{2n} + \frac{1}{2(n-1)} + \dots + \frac{1}{2(m-1)}$. For $m$ small compared to $n$ we can use a pretty crude bound: bounding based on the smallest resp. largest term of this sum gives
$$\frac{m}{2n} \le g(n) \le \frac{m}{2(n-m+1)}$$
and using these bounds on the upper and lower bounds we wrote down for $f(n)$ gives
$$\boxed{ 4^m \left( 1 - \frac{m}{2(n-m+1)} \right) \le f(n) \le 4^m \exp \left( - \frac{m}{2n} \right) }.$$
Together with the fact that $\exp(-x) \approx 1 - x$ for $x$ small, this tells us that $f(n)$ is approximately $4^m \left( 1 - \frac{m}{2n} \right)$ for large $n$, even if $m$ is also large, as long as it's not too large compared to $n$ (more precisely, as long as $\frac{m}{n} \to 0$). For example we could take $m, n \to \infty$ with $m \sim \sqrt{n}$ and we'd get $f(n) \approx 4^{\sqrt{n}} \left( 1 - \frac{1}{2 \sqrt{n}} \right)$!

There are many more delights to be found in the wild world of asymptotics of combinatorial sequences; for a nice place to start you might try reading Graham, Knuth, and Patashnik's Concrete Mathematics, and if you really want to dive into the deep end you can try dipping your toes into Flajolet and Sedgewick's masterpiece Analytic Combinatorics (decidedly not high school level even in the loosest sense but contains many amazing examples and very much worth reading and rereading over years; I've been slowly working through it for at least a decade). Good luck!
