Elementary central binomial coefficient estimates 

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*How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}<\binom{2n}{n}<\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ > 1 ?





*Does anyone know any better elementary estimates?


Attempt.  We have
$$\frac1{2^n}\binom{2n}{n}=\prod_{k=0}^{n-1}\frac{2n-k}{2(n-k)}=\prod_{k=0}^{n-1}\left(1+\frac{k}{2(n-k)}\right).$$
Then we have
$$\left(1+\frac{k}{2(n-k)}\right)>\sqrt{1+\frac{k}{n-k}}=\frac{\sqrt{n}}{\sqrt{n-k}}.$$
So maybe, for the lower bound, we have
$$\frac{n^{\frac{n}{2}}}{\sqrt{n!}}=\prod_{k=0}^{n-1}\frac{\sqrt{n}}{\sqrt{n-k}}>\frac{2^n}{\sqrt{4n}}.$$
By Stirling, $n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, so the lhs becomes
$$\frac{e^{\frac{n}{2}}}{(2\pi n)^{\frac14}},$$
but this isn't $>\frac{2^n}{\sqrt{4n}}$.

Here is a  better estimate of the quantity. I am not going to prove it. Since I know about it, hence I am sharing and will put the reference.
$$\frac{1}{2\sqrt{n}} \le {2n \choose n}2^{-2n} \le \frac{1}{\sqrt{2n}}$$
Refer to Page 590 This is the text book "Boolean functions Complexity" by Stasys Jukna.
