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.).

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.).