Proving that and how $ \frac{1}{n}\sum\limits_{p\le n}\lfloor n/p \rfloor - \sum\limits_{p\le n} 1/p $ approaches $0$

Question

Let $p$ denote a generic prime number. By Mertens' second theorem, the sequence $$\sum\limits_{\ p \le n} \frac1p - \log\log n$$ converges to the Meissel-Mertens constant $M\approx 0.2614972$. Now let $\omega(n)$ be the number of distinct prime factors of $n$. Analogously, we have $$\lim_{n\to\infty} \frac{1}{n}\sum\limits_{k\le n} \omega(k) - \log\log n = M,$$ hence combining this, the first result and $$\sum\limits_{k\le n} \omega(k)=\sum\limits_{p\le n} \left\lfloor \frac{n}{p}\right\rfloor, $$ we find $$\bbox[5px,border:2px solid #B2B550]{\lim_{n\to\infty} \frac{1}{n} \sum\limits_{p\le n} \left\lfloor \frac{n}{p}\right\rfloor - \sum_{p\le n} \frac{1}{p}=0. }\tag{$\star$}$$ How could we prove $(\star)$ directly? Is it known if the sequence is negative for all $n>2$, or at least if it changes sign finitely often?

Answer 1

Maybe it's more useful to use $$\sum_{k\leq n}\omega\left(k\right)=\sum_{p\leq n}\left\lfloor \frac{n}{p}\right\rfloor $$ then $$\frac{1}{n}\sum_{k\leq n}\omega\left(k\right)-\sum_{p\leq n}\frac{1}{p}=\frac{1}{n}\sum_{p\leq n}\left\lfloor \frac{n}{p}\right\rfloor -\sum_{p\leq n}\frac{1}{p} $$ $$=\frac{1}{n}\sum_{p\leq n}\frac{n}{p}-\sum_{p\leq n}\frac{1}{p}-\frac{1}{n}\sum_{p\leq n}\left\{ \frac{n}{p}\right\} =-\frac{1}{n}\sum_{p\leq n}\left\{ \frac{n}{p}\right\} \tag{1} $$ and obviously $$0\leq\frac{1}{n}\sum_{p\leq n}\left\{ \frac{n}{p}\right\} \leq\frac{\pi\left(n\right)}{n} $$ so your claim follows. From $(1) $ we have also that the sequence is negative for $n>2 $.