Let $S(x)=\sum_{p\leq x} \ln(p)$ where $\sum_{p\leq x}$ denotes a summation over the positive prime numbers that are $\leq x$
Prove that $\forall n \in \mathbb N, S(2n+1)-S(n+1)\leq \ln\left(\binom{2n+1}{n}\right)$
I rewrote $S(2n+1)-S(n+1)$ as $\displaystyle \sum_{n+2\leq p\leq 2n+1}\ln(p)$ and assuming that $n$ is even, we may of course remove every even integer, yielding $$\sum_{n+2\leq p\leq 2n+1}\ln(p) \leq\sum_{k=n+2,k\;\text{odd}}^{2n+1}\ln(k) \leq \ln \frac{(2n+1)!}{(n+1)!(n+2)(n+4)\ldots(2n)}$$
Unfortunately, we don't have $(n+2)(n+4)\ldots(2n) \geq n!$
What should I do ?
After expanding out the factorials, the original inequality can be rewritten as
$$\sum_{n+1<p\leq 2n+1}\ln p + \sum_{k\leq n} \ln k + \leq \sum_{n+1<k\leq 2n+1} \ln k$$
If we let $P(n)$ and $C(n)$ be the set of primes numbers and composite numbers, respectively, in the range $(n+1,2n+1]$, then we can further reduce the inequality to:
$$\sum_{k\leq n} \ln k \leq \sum_{k\in C(n)} \ln k$$
Now, for each prime $p$ and integer $k$, let $a_p(k)$ be the exponent of $p$ in the prime factorization of $k$, and let $P$ be the set of all primes. Expanding each $k$ into its prime factorization, the inequality becomes,
$$\sum_{p \in P} \ln(p){\sum_{k=1}^n a_p(k)} \leq \sum_{p\in P} \ln(p){\sum_{k\in C(n)} a_p(k)}$$
It suffices to show that
$$\sum_{k=1}^n a_p(k) \leq \sum_{k\in C(n)} a_p(k)$$
for all primes $p \leq n$. Or, since $a_p(k)=0$ for $p\leq n$ and $k\in P(n)$, it suffices to show that
$$\sum_{k=1}^n a_p(k) \leq \sum_{k\in C(n)} a_p(k) + \sum_{k\in P(n)} a_p(k) = \sum_{k=n+2}^{2n+1} a_p(k),$$ or that $$ \sum_{k=n+2}^{2n+1} a_p(k) - \sum_{k=1}^n a_p(k) \geq 0$$ This is clear, since the given expression is simply the exponent of $p$ in the prime factorization of $\binom{2n+1}{n}$.