<s>[INST] 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.).
 [/INST] 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
$$
\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}}}\tag9
$$
</s>