An exercise in chapter 1 of A course in Analytic Number Theory by Marius Overholt prompts me to find an upper bound of the binomial coefficient ${2n\choose n}$, to deduce the prime factorisation of ${2n\choose n}$, and hence to show that $\vartheta (x):=\sum_{p\leq x}\ln p =O(x)$.
Therefore, I need to show that $\vartheta (x)$ is bounded by $\ln {2n \choose n}$ in some way, for example, something like $$ \vartheta (2n)\leq C\ln{2n \choose n} \text{ or }\vartheta (n)\leq C\ln{2n \choose n}. $$ However, we have the prime factorisation $$ {2n\choose n}=\frac{\prod_{p,k} p^{[2n/p^k]}}{\prod_{p,k} p^{2[n/p^k]}}=\prod_{p,k} p^{f(p,k)}, $$ where $f(p,k)=1$ if $[2n/p^k]$ is odd , and $f(p,k)=0$ otherwise.
In some cases, this prime factorisation does not contain all primes smaller than $2n$. In fact, when $2n=26$, $$ e^\vartheta(26)=223092870>10400600={26\choose 13}, $$ so the desired bound is very difficult to obtain. In fact, $10400600$ does not even have $3$ as one of its prime factor.
I thus find it very difficult to bound $\vartheta(x)$ above by $\ln {2n\choose n}$, but this seems to be the only reasonable thing to do.
Could anyone give me a hint on this?