Doubt in proof of Chebysheff theorem

While self studying analytic number theory from Introduction to sieve methods and it's applications by M Ram Murthy and Alina Carmen,I have a doubt in theorem 1.4.1 proof by Chebysheff.

My doubt is -> Author writes $$\frac {(2n) ! } { ( n!) ^2} \leq 2^{2n}$$ and then he writes this step which I am not able to derive - upon taking logarithms $$\theta(2n) - \theta(n) \leq 2n log 2$$

• What is $\theta$? – cangrejo Jan 15 at 12:13
• $\theta(x) = \sum_{p \leq x} ln( p )$ , Its 2nd chebycheff function – Ben Jan 15 at 12:15
• I think there is a mistake. It should be $\theta(2n)-2\theta(n)$. – cangrejo Jan 15 at 12:29
We have $$\frac{(2n)!}{(n!)^2}=\binom{2n}{n}\in \mathbb{N}$$. Therefore $$p|\binom{2n}{n}$$ for all primes $$n and $$\prod_{n. Now use $$\ln$$ to get the desired inequality.