Show $ \frac{1}{n} \sum_{p \in \mathbb P} \left\lfloor \frac{n}{p} \right\rfloor \log p = \log n + O(1)$ Any hints how to prove for $n \in \mathbb N$
$$
 \frac{1}{n} \sum_{p \in \mathbb P} \left\lfloor \frac{n}{p} \right\rfloor \log p = \log n + O(1)
$$
where $\mathbb P$ denotes the set of all primes? As for all primes $p > n$ we have $\lfloor n/p \rfloor = 0$, the sum is finite, i.e. if $p_1, \ldots, p_k$ are all primes $\le n$ we have
$$ 
 \frac{1}{n} \sum_{p \in \mathbb P} \left\lfloor \frac{n}{p} \right\rfloor \log p = \frac{1}{n} \sum_{i=1}^k \left\lfloor \frac{n}{p_i} \right\rfloor \log p_i
  = \frac{1}{n} \sum_{i=1}^k \log\left( p_i^{\lfloor n/p_i \rfloor}\right)
  = \frac{1}{n} \log\left( \prod_{i=1}^k p_i^{\lfloor n/p_i\rfloor} \right)
$$
so I guess we should somehow show that $\log n^n \approx \log\left( \prod_{i=1}^k p_i^{\lfloor n/p_i\rfloor} \right)$, but I have no idea how to proceed, so any hints?
 A: Consider that:
$$ n! = \prod_{p\leq n} p^{\left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^2}\right\rfloor+\left\lfloor\frac{n}{p^3}\right\rfloor+\ldots}$$
hence:
$$ \log(n!) = \sum_{p\leq n}\left\lfloor\frac{n}{p}\right\rfloor \log p + \sum_{k\geq 2}\sum_{p\leq n}\left\lfloor\frac{n}{p^k}\right\rfloor \log p = \sum_{p\leq n}\left\lfloor\frac{n}{p}\right\rfloor \log p + O(n)$$
and the LHS can be estimated through Stirling's approximation.
A: We have $$\frac{1}{n}\sum_{p\leq n}\left\lfloor \frac{n}{p}\right\rfloor \log\left(p\right)=\sum_{p\leq n}\frac{\log\left(p\right)}{p}+O\left(\frac{1}{n}\sum_{p\leq n}\log\left(p\right)\right)
 $$ because $\left\lfloor \frac{n}{p}\right\rfloor =\frac{n}{p}+O\left(1\right)
 $ and by Mertens first theorem we have that $$\sum_{p\leq n}\frac{\log\left(p\right)}{p}=\log\left(n\right)+O\left(1\right)
 $$ and by PNT $$\sum_{p\leq n}\log\left(p\right)=O\left(n\right)
 $$ hence $$\frac{1}{n}\sum_{p\leq n}\left\lfloor \frac{n}{p}\right\rfloor \log\left(p\right)=\log\left(n\right)+O\left(1\right).
 $$
