What is the tightest upper bound we can establish on the central binomial coefficients $ 2n \choose n$ ?


I just tried to proceed a bit, like this:


$ n! > n^{\frac{n}{2}} $


for all $ n>2 $. Thus,


$ \binom{2n}{n} = \frac{ (n+1) \ldots (2n) }{n!} < \frac{\left(\frac{\sum\_{k=1}^n (n+k) }{n}\right)^n }{n^{n/2}} = \frac{ \left( \frac{ n^2 + \frac{n(n+1)}{2} }{n} \right) ^n}{n^{n/2}} = \left( \frac{3n+1}{2\sqrt{n}} \right)^n $


But, I was searching for more tighter bounds using elementary mathematics only (not using Stirling's approximation etc.).



Upper limit on the central binomial coefficient
Here's a way to motivate and refine the argument that
Péter Komjáth attributes to Erdős.


Start by computing the ratio between the $n$-th and $(n-1)$-st
central binomial coefficients:
$$
{2n \choose n} \left/ {2(n-1) \choose n-1} \right.
= \frac{(2n)! \phantom. / \phantom. n!^2}{(2n-2)! \phantom. / \phantom. (n-1)^2}
= \frac{(2n)(2n-1)}{n^2}
= 4 \left( 1 - \frac1{2n} \right).
$$
For large $n$, this ratio approaches $4$, suggesting that
$2n \choose n$ grows roughly as $4^n$.
If the factor $1 - \frac1{2n}$ were $1 - \frac1n = (n-1)/n$,
the growth would be exactly proportional to $n^{-1} 4^n$.
Since $1 - \frac1{2n}$ is (for large $n$) nearly the square root of
$1 - \frac1n$, the actual asymptotic should be proportional to $n^{-1/2} 4^n$.
So we introduce the ratio
$$
r\_n := \left(
 {2n \choose n} \left/ \frac{4^n}{\sqrt n} \right.
 \right)^2
= \frac{n}{16^n} {2n \choose n}^2.
$$
Then
$$
\frac{r\_n}{r\_{n-1}} =
 \left( 1 - \frac1{2n} \right)^2
 \left/ \left( 1 - \frac1n \right) \right.
= \frac{(2n-1)^2}{(2n-2)(2n)} \gt 1.
$$
Thus $r\_{n-1} < r\_n$; and since $r\_1 = (2/4)^2 = 1/4$ we have by induction
$$
r\_1 \lt r\_2 \lt r\_3 \lt r\_4 \lt \cdots \lt r\_n
= \frac12
 \frac{1 \cdot 3}{2 \cdot 2}
 \frac{3 \cdot 5}{4 \cdot 4}
 \frac{5 \cdot 7}{6 \cdot 6}
 \cdots
 \frac{(2n-3)(2n-1)}{(2n-2)(2n-2)}
 \frac{2n-1}{2n}.
$$
Each $r\_{n\_0}$ gives a lower bound on $r\_n$, and thus on $2n\choose n$,
for all $n \geq n\_0$. The OP asked for upper bounds, so consider
$$
R\_n := \frac{2n}{2n-1} r\_n
= \frac{n}{\left(1-\frac 1{2n}\right)16^n} {2n \choose n}^2.
$$
Now $R\_{n+1}/R\_n = (2n-1)(2n+1) \phantom. / \phantom. (2n)^2
= (4n^2-1) \phantom. / \phantom. (4n^2) \lt 1$, so
$$
\frac12 = R\_1 \gt R\_2 \gt R\_3 \gt R\_4 \gt \cdots \gt
R\_{n+1} = 
 \frac12
 \frac{1 \cdot 3}{2 \cdot 2}
 \frac{3 \cdot 5}{4 \cdot 4}
 \frac{5 \cdot 7}{6 \cdot 6}
 \cdots
 \frac{(2n-3)(2n-1)}{(2n-2)(2n-2)}.
$$
It follows that $R\_n \geq r\_{n'}$ for any $n,n'$, so $R\_1=1/2$,
$R\_2=3/8$, $R\_3=45/128$, etc. are a series of upper bounds on every $r\_n$.
Since moreover $r\_n / R\_n = 1 - \frac1{2n} \rightarrow 1$ as
$n \rightarrow \infty$, both $r\_n$ and $R\_n$ converge to a common
limit that is an upper bound on every $r\_n$. If we accept Wallis's
product (which is classical though not as elementary as everything else
in our analysis), then we can evaluate this common limit as $1/\pi$
and thus recover the asymptotically sharp upper bound
${2n \choose n} < 4^n / \sqrt{\pi n}$.


