Understanding how to estimate $\pi(x)$ based on Paul Erdos's proof of Bertrand's Postulate

Question

I am reading the 4th Edition of Proofs from the Book. I am not clear on how the proof behind Bertrand's postulate leads to the following statement on page 10 (of my edition):

From (2) one can derive with the same methods that $\prod\limits_{n<p\le 2n}p \ge 2^{\frac{1}{30}n}$ for $n \ge 4000$

Here's (2):

$4^n \le (2n)^{1+\sqrt{2n}} \cdot \prod\limits_{\sqrt{2n}< p \le \frac{2}{3}n} p \cdot \prod\limits_{n < p \le 2n} p$ for $n \ge 3$

After (2) comes the details that for completing the proof. I understand how this leads to the proof. I don't understand how this leads to the estimate for the product of primes

Here are the details presented:

Here's (1):

$\prod\limits_{p \le x}p \le 4^{x-1}$ for all real $x \ge 2$

Assuming that there are no primes between $n$ and $2n$, applying (1) to (2):

Substituting (1) into (2) we get: $$4^n \le (2n)^{1+\sqrt{2n}}4^{\frac{2}{3}n}$$

Then applying $a+1 < 2^a$, the result becomes:

$$2n = (\sqrt[6]{2n})^6 < (\left\lfloor\sqrt[6]{2n}\right\rfloor + 1)^6 < 2^{6\left\lfloor\sqrt[6]{2n}\right\rfloor} \le 2^{6\sqrt[6]{2n}}$$

Then, for $n \ge 50$ ($18 < 2\sqrt{2n}$), combining the last two results gets:

$$2^{2n} \le (2n)^{3(1 + \sqrt{2n})} < 2^{\sqrt[6]{2n}(18+18\sqrt{2n})} < 2^{20\sqrt[6]{2n}\sqrt{2n}} = 2^{20(2n)^{\frac{2}{3}}}$$

While I follow all the details up to this point in the proof (I think), I don't follow how this gets us to:

From (2) one can derive with the same methods that $\prod\limits_{n<p\le 2n}p \ge 2^{\frac{1}{30}n}$ for $n \ge 4000$

If someone could provide details or an explanation about this last result, I would greatly appreciate it.

Answer 1

Okay: I recommend substituting (1) into (2), but not for both products - just for the first product, up to $2n/3$. The resulting inequality will be a lower bound for the product you're trying to find a lower bound for. Then try using the methods from the rest of your post to change the form of that lower bound. Don't worry at first about whether what you'll end up with is exactly what they stated - just work through it once to get a feel for it.

Answer 2

We can put (1) into (2) and we get$$4^{n}\leq\left(2n\right)^{1+\sqrt{2n}}\underset{\sqrt{2n}<p\leq\frac{2}{3}n}{\prod}p\underset{n<p\leq2n}{\prod}p\leq\left(2n\right)^{1+\sqrt{2n}}4^{\frac{2}{3}n-1}\underset{n<p\leq2n}{\prod}p\,\Longrightarrow$$ $$\frac{2^{2n-\frac{4}{3}n-\sqrt{2n}+1}}{n^{1+\sqrt{2n}}}\leq\underset{n<p\leq2n}{\prod}p.$$ Now, if $n$ is large enough (and $n\geq4000$ is large enough) holds$$\frac{1}{n^{1+\sqrt{2n}}}>\frac{1}{2^{n/2}}$$ so$$2^{2n-\frac{4}{3}n-\sqrt{2n}+1-\frac{n}{2}}\leq\underset{n<p\leq2n}{\prod}p$$ and trivially $$2n-\frac{4}{3}n-\sqrt{2n}+1-\frac{n}{2}=\frac{12n-8n-6\sqrt{2n}+6-3n}{6}=\frac{n-6\sqrt{2n}+6}{6}>\frac{n}{30}$$ if $n$ large enough, so we have $$2^{n/30}\leq\underset{n<p\leq2n}{\prod}p.$$