Two estimations in analytic number theory I want to show the following two estimation:
$$
\sum_{p|n}\frac{\log p}{p}=O(\log\log n)
$$
And
$$
\sum_{p|n}\frac1p=O(\log\log\log n)
$$
First assume $$n=p_1^{\alpha_1}...p_s^{\alpha_s}$$ then the first can be rewritten as:
$$
\sum_{i=1}^s\frac{\log p_i}{p_i}
$$
I think i should bulid some connections between s and n, and could i do the following?
$$
\sum_{i=1}^s\frac{\log p_i}{p_i}=O(\sum_{i=1}^s\frac{1}{i})
$$
If this is true, why?

Take the first as an example: from what we have done, it suffices to estimate:
$$
\sum_{i=1}^s\frac{\log p_i}{p_i}
$$
Classify $p_1,...,p_s$as follows：
$p_1,...,p_{r_0}\leq\log n$，$p_{r_0+1},...,p_s>\log n$, so:
$$
\sum_{i=1}^{r_0}\frac{\log p_i}{p_i}\leq\sum_{p\leq\log n}\frac{\log p}{p}\leq\log\log n+O(1)
$$
We then estimate how large the $r-r_0$ is :
$$
(\log n)^{r-r_0}<p_{r_0+1}...p_s\leq n\to r-r_0\leq\frac{\log n}{\log\log n}
$$
So,
$$
\sum_{i=r_0+1}^s\frac{\log p_i}{p_i}\leq\frac{\log\log n}{\log n}\times\frac{\log n}{\log\log n}=1
$$
I.e
$$
\sum_{i=1}^s\frac{\log p_i}{p_i}=O(\log\log n)
$$
