Doubt in proof of Chebysheff theorem

Question

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 $

Can someone please help in how to derive this statement!!

Answer 1

We have $\frac{(2n)!}{(n!)^2}=\binom{2n}{n}\in \mathbb{N}$. Therefore $p|\binom{2n}{n}$ for all primes $n<p<2n$ and $\prod_{n<p<2n}p|\binom{2n}{n}<2^{2n}$. Now use $\ln$ to get the desired inequality.