Basic Number Theory Inequality assuming Bertrand's postulate false for some n.

Question

So I am having a hard time doing one problem from George E. Andrews Number Theory book. It is problem $4$ chapter $8$-$3$ if anyone is curious. For it we must assume that Bertrand's Postulate is false for some $n$ (i.e. for some Integer $n$ there is NO prime $p$ such that $n<p<2n$). Also, we should use in our proof one inequality previously postulated. It is the following: $${2n\choose n}\leq\ Q_n \leq\ (2n)^{\pi(2n)}$$ where $Q_n$ is the product of all power of primes not exceeding $2n$ (i.e. the product of all $p^r$ where $r$ is such that $p^r \leq\ 2n <p^{r+1} $) and $\pi(2n)$ is the prime counting function for $2n$. Knowing all this, I need to prove that $${2n\choose n}\leq\ (2n)^{\sqrt {2n}}R_{\frac{2}{3}n} $$ where $R_{\frac{2}{3}n} $ is the product of all primes less than $\frac{2}{3} n $. So my first thought was to use the needed assumption to see that $\pi(2n) = \pi(n)$ and from the first inequality I would just need $$(2n)^{\pi(n)} \leq\ (2n)^{\sqrt {2n}}R_{\frac{2}{3}n} $$ to be true. But this is just false even with our assumption. Take for example $n = 7$ then you would get $14^6 = 7529536 \gt 14^{\sqrt {14}}(2 \cdot\ 3)= 116565.406$ So I am really stuck now because I find that $\frac{2}{3}n$ too specific and I don't know where it might come from for that to be true.

EDIT: so I was thinking on $\phi (7)$ when I did that calculation, the actual result for $n = 7$ should be $14^4 = 38416$ and that is indeed less than $116565.406$. So to the way might be to prove the last inequality but I still don't know how to put there the $R_ {\frac{2}{3}n}$. And Knowing that primes are more dense than squares that ratio should also be enough to make that relation flip, considering the desire result and the exponents on both sides

Answer 1

Once you solve the previous problems in the book you will the idea for this one.

We constructed $Q_n$ by considering every prime $p < 2n : p \space | \binom{2n}{n}$ and finding it's exponent in $\binom{2n}{n}$. Then, the product $Q_n$ is created such that $\binom{2n}{n} | Q_n$

Some points to note (solved in the previous problems in the book):

1. If $p>\sqrt{2n}$, then the exponent of $p$ in $Q_n \leq 1$ .
2. If $ 2n/3<p<n $, then p does not divide $\binom{2n}{n}$
3. Every term $p_i ^ {r_i} \leq 2n$ (definition of $Q_n$)

So, we can go from here

$$ \binom{2n}{n} \leq \prod_{ p \leq \sqrt{2n} } 2n * \prod_{ \sqrt{2n} < p < 2n/3 } p^1 * \prod_{ n < p < 2n} p $$

Note that the term where $ 2n/3 \leq p \leq n $ is not required.

Now as we have no primes from $n$ to $2n$, the third term is 1.

Therefore, we can write:

$$ \binom{2n}{n} \leq \prod_{ p \leq \sqrt{2n} } 2n * \prod_{ \sqrt{2n} < p < 2n/3 } p^1$$

$$ \binom{2n}{n} \leq \prod_{ p \leq \sqrt{2n} } 2n * \prod_{ 1 < p < 2n/3 } p^1$$

$$ \binom{2n}{n} \leq 2n^{\sqrt{2n}} * R_{2n/3}$$