 Solution

For every prime number $p$ with $n<p\le 2n$, the numerator $(2n)!$ contains one factor $p$, whereas neither copy of $n!$ contains one. Consequently the product of these prime numbers divides the binomial coefficient $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. Since this binomial coefficient is one positive summand of ${\sum }_{j=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{j}\right)=2^{2n}$, and other summands are positive,For $x\ge 2$, choose $k$ with $2^{k-1}<x\le 2^k$. The Chebyshev function is increasing, so telescoping givesFor $x=1$ the inequality follows from $\theta (1)=0$. Thus $\boxed{\theta (x)<4(\log 2)x}$ throughout the stated range.