Deje $p$ denotar un genérico de los números primos. Por Mertens' segundo teorema, la secuencia de $$\sum\limits_{\ p \le n} \frac1p - \log\log n$$ converges to the Meissel-Mertens constant $M\aprox 0.2614972$. 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 $$\bbox[5px,border:2px solid #B2B550]{\lim_{n\to\infty} \frac{1}{n} \sum\limits_{p\le n} \left\lfloor \frac{n}{p}\right\rfloor - \sum_{p\le n} \frac{1}{p}=0. }\tag{$\estrella$}$$ How could we prove $(\star)$ directly? Is it known if the sequence is negative for all $n>2$, o, al menos, si cambia de signo finitely a menudo?
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Tal vez sea más útil el uso de $$\sum_{k\leq n}\omega\left(k\right)=\sum_{p\leq n}\left\lfloor \frac{n}{p}\right\rfloor $$ then $$\frac{1}{n}\sum_{k\leq n}\omega\left(k\right)-\sum_{p\leq n}\frac{1}{p}=\frac{1}{n}\sum_{p\leq n}\left\lfloor \frac{n}{p}\right\rfloor -\sum_{p\leq n}\frac{1}{p} $$ $$a=\frac{1}{n}\sum_{p\leq n}\frac{n}{p}-\sum_{p\leq n}\frac{1}{p}-\frac{1}{n}\sum_{p\leq n}\left\{ \frac{n}{p}\right\} =-\frac{1}{n}\sum_{p\leq n}\left\{ \frac{n}{p}\right\} \etiqueta{1} $$ 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 $.
