$\log((2n!)/(n!)^2) =\theta(n)$

Question

I want to show that $\log((2n!)/(n!)^2) =\theta(n)$

I know that $\log(n!)=\theta(n\log n)$

and I know that $\log(2n!)=\theta(2n\log2n)$ can I do this ?

$\log((2n!)/(n!)^2)=\log((2n!)-2\log(n!))=\theta(2n\log2n-2n\log n)=\theta(2n(log2+\log n-\log n))=\theta(2n \log2)$

if not; how can I prove it?

Answer 1

The trouble is $f\sim g+h$ doesn't imply $f-g\sim h$. We need a more detailed approximation, $\ln n!\in n\ln n-n+O(1/n)$. You can then show $\ln\frac{(2n)!}{n!^2}\in 2n\ln 2+O(1/n)$.