Upper limit on the central binomial coefficient 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.).
 A: We may prove without using integrals that
$$
\frac1{\sqrt{\pi n}}\geqslant 4^{-n}{2n\choose n}\geqslant \frac{1}{\sqrt{\pi(n+1/2)}}.\quad\quad (1)
$$
(1) is equivalent to $$2n\left(4^{-n}{2n\choose n}\right)^2=\frac12\prod_{k=2}^n \frac{(2k-1)^2}{2k(2k-2)}:=d_n\leqslant \frac2\pi\leqslant c_n\\:=(2n+1)\left(4^{-n}{2n\choose n}\right)^2=\prod_{k=1}^n\left(1-\frac1{4k^2}\right)$$
(the identities are straightforward by induction).
Denote $m:=2n+1$. It is not hard to show that $\sin mx=p_m(\sin x)$ for a polynomial of degree $m$ in $\sin x$. The roots of $p_m$ are $\sin \frac{k\pi}{2n+1}$ for $k=-n,\ldots,n$. Thus $$\sin (2n+1)x=(2n+1)\sin x\prod_{k=1}^n\left(1-\frac{\sin^2 x}{\sin^2 \frac{k\pi}{2n+1}}\right)$$
(the multiple $2n+1$ comes from dividing by $x$ and putting $x=0$).
Put $x=\frac{\pi}{4n+2}$. Using the inequality $\sin tx\leqslant t\sin x$ (which may be proved, for example, by induction in $t=1,2,\ldots$ from  the identity $\sin(t+1)x=\sin tx\cos x+\sin x\cos tx$) for $t=2k$ ($k=1,2,\ldots,n$)
we get
$$
1\leqslant (2n+1)\sin\frac{\pi}{4n+2}\prod_{k=1}^n\left(1-\frac1{4k^2}\right)\leqslant \frac{\pi}{2}c_n,\quad c_n\geqslant \frac{2}\pi. 
$$
Analogously we get
$$
\cos 2nx=\prod_{k=1}^n\left(1-\frac{\sin^2 x}{\sin^2 \frac{\pi(2k-1)}{4n}}\right).
$$
Dividing by $1-\frac{\sin^2 x}{\sin^2 \frac{\pi}{4n}}$ and substituting $x=\frac{\pi}{4n}$ (for computing the LHS at this point use l'Hôpital rule) we get
$$
n\tan\frac{\pi}{4n}\leqslant \prod_{k=2}^n\left(1-\frac1{(2k-1)^2}\right)=\frac1{2d_n},\quad d_n\leqslant \frac1{2n\tan \frac{\pi}{4n}}\leqslant \frac2\pi.
$$
UPD. This is less or more equivalent to Yaglom brothers proof (1953), see their Russian paper (it contains also the derivation of identities $\sum 1/n^2=\pi^2/6$ and $\sum (-1)^{k-1}/(2k-1)=\pi/4$ using these trigonometric things.)
A: Since it gives tighter bounds, I will reproduce my answer from MathSE.

For $n\ge0$, we have (by cross-multiplication)
$$
\begin{align}
\left(\frac{n+\frac12}{n+1}\right)^2
&=\frac{n^2+n+\frac14}{n^2+2n+1}\\
&\le\frac{n+\frac13}{n+\frac43}\tag{1}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}
&=4\frac{n+\frac12}{n+1}\\
&\le4\sqrt{\frac{n+\frac13}{n+\frac43}}\tag{2}
\end{align}
$$
Inequality $(2)$ implies that
$$
\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac13}}{4^n}\text{ is decreasing}}}\tag{3}
$$
For $n\ge0$, we have (by cross-multiplication)
$$
\begin{align}
\left(\frac{n+\frac12}{n+1}\right)^2
&=\frac{n^2+n+\frac14}{n^2+2n+1}\\
&\ge\frac{n+\frac14}{n+\frac54}\tag{4}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}
&=4\frac{n+\frac12}{n+1}\\
&\ge4\sqrt{\frac{n+\frac14}{n+\frac54}}\tag{5}
\end{align}
$$
Inequality $(5)$ implies that
$$
\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac14}}{4^n}\text{ is increasing}}}\tag{6}
$$
Note that the formula in $(3)$, which is decreasing, is bigger than the formula in $(6)$, which is increasing. Their ratio tends to $1$; therefore, they tend to a common limit, $L$.

Theorem $1$ from this answer says
$$
\lim_{n\to\infty}\frac{\sqrt{\pi n}}{4^n}\binom{2n}{n}=1\tag{7}
$$
which means that
$$
\begin{align}
L
&=\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\binom{2n}{n}\\
&=\frac1{\sqrt\pi}\tag8
\end{align}
$$
Combining $(3)$, $(6)$, and $(8)$, we get
$$
\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}}}\tag9
$$
A: You may also prove ${2n\choose n}\sim \frac{4^n}{\sqrt{\pi n}}$ using the combinatorial meaning of ${2n\choose n}$ and the area-of-a-circle definition of $\pi$. That is probably the most elementary we can hope for.
First of all, if $a_n:=\sqrt{n+1/4}{2n\choose n}4^{-n}$ and $b_n:=\sqrt{n+1/2}{2n\choose n}4^{-n}$, we may check that $a_n\leqslant b_n$, $a_n$ increases and $b_n$ decreases, thus $a_n$, $b_n$ have a common positive finite limit which we denote $\alpha$, and $a_n\leqslant \alpha\leqslant b_n$. Since $b_n/a_n=1+O(1/n)$, we get $a_n=\alpha+O(1/n)$, and also $c_n=\alpha+O(1/n)$ where $c_n:=\sqrt{n}{2n\choose n}4^{-n}$.
We should prove that $\alpha=1/\sqrt{\pi}$.
Use the combinatorial identity $$4^{-n}\sum_{k=0}^n k{2k\choose k}(n-k){2(n-k)\choose n-k}=\frac{n(n-1)}8.\quad (\heartsuit)$$
LHS of $(\heartsuit)$ equals
$$
\sum_{k=0}^n \sqrt{k(n-k)}c_kc_{n-k}=\sum_{k=0}^n \sqrt{k(n-k)}\left(\alpha^2+O\left(\frac1k+\frac1{n-k}\right)\right)\\=\alpha^2\sum_{k=0}^n\sqrt{k(n-k)}+O\left(n\sum_{k=0}^n\frac1{\sqrt{k(n-k)}}\right)=\alpha^2\cdot \frac{\pi n^2}8+o(n^2),
$$
since the union of rectangles $[k,k+1]\times [0,\sqrt{k(n-k)}]$ approximates the upper semi-circle with the horizontal diameter $[0,n]$.
Thus $\alpha^2\pi/8=1/8$ and $\alpha=1/\sqrt{\pi}$.
You could use instead of $(\heartsuit)$ the more famous identity
$$
4^{-n}\sum_{k=0}^n {2k\choose k}{2(n-k)\choose n-k}=1
$$
this gives $\alpha^2\int_0^1\frac{dx}{\sqrt{x(1-x)}}=1$ (the integral equals $\pi$ of course, but this is less "elementary" then the area of a circle.)
This argument is similar to J. Wästlund's
(An elementary proof of Wallis' product formula for $\pi$, 2007).
A: Even the asymptotically sharp inequality 
${2n \choose n} < 4^n \left/ \sqrt{\pi n} \right.$
has a short proof:
$$
{2n \choose n} = 
 \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} \cos^{2n} x \phantom. dx < 
 \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} e^{-nx^2} dx <
 \frac{4^n}{\pi} \int_{-\infty}^{\infty} e^{-nx^2} dx =
 \frac{4^n}{\sqrt{\pi n}}.
$$
In the first step, the formula for
$\int_{-\pi/2}^{\pi/2} \cos^{2n} x \phantom. dx$
can be proved by induction via integration by parts,
or using the Beta function.  
The third step is clear, and the last step is the well-known Gaussian integral.
So we need only justify the the second step.
There we need the inequality $\cos x \leq e^{-x^2/2}$, or equivalently
$$
\log \cos x + \frac{x^2}{2} \leq 0,
$$
for $\left|x\right| < \pi/2$, with equality only at $x=0$.
This is true because $\log \cos x +\frac12 x^2$ is an even function of $x$
that vanishes at $x=0$ and whose second derivative
$-\tan^2 x$ is negative for all nonzero $x \in (-\pi/2, \pi/2)$. QED
A: Erdos remarked somewhere the bound
$$
{{2n}\choose{n}}<\frac{4^n}{\sqrt{2n+1}}.
$$
This can be established by induction: 
$$
{{2n+2}\choose{n+1}}=\frac{(2n+1)(2n+2)}{(n+1)(n+1)}{{2n}\choose{n}}
$$
and if we have the bound for $n$, we only have to show 
$$
\frac{2(2n+1)}{(n+1)\sqrt{2n+1}}<\frac{4}{\sqrt{2n+3}}
$$
which reduces to $4n^2+8n+3<4n^2+8n+4$.
A: 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
${2n \choose n} < 4^n / \sqrt{\pi n}$.
A: Noam Elkies notes that there is a quick proof of 
$$\binom{2n}{n} \leq \frac{4^n}{\sqrt{\pi n}}$$
by writing 
$$\binom{2n}{n} = \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} \cos^{2n} x dx$$
and bounding $\cos^2 x \leq e^{-x^2}$.
There is an equally good lower bound by a similar method: 
$$\int_{-\pi/2}^{\pi/2} \cos^{2n} x dx =\int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{(\tan^2 x+1)^{n+1}} = \int_{- \infty}^{\infty} \frac{du}{(1+u^2)^{n+1}} \geq \int_{- \infty}^{\infty} e^{-(n+1) u^2} du $$
so
$$\binom{2n}{n} \geq \frac{4^n}{\sqrt{\pi (n+1)}}.$$
Here the inequality $\tfrac{1}{1+u^2} \geq e^{-u^2}$ follows from the standard bound $e^y \geq 1+y$ for $y\geq 0$.
