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?
 A: The only ingredient needed is the elementary (Chebysev estimates) $Ak\log k < q_k < Bk\log k, k \ge 2$ where $q_1=2, q_2=3,...$ is the sequence of primes. This implies:
$\frac{1}{q_k} < \frac{A_1}{k \log k}, k \ge 2$ and $\frac{\log q_k}{q_k} < \frac{A_2}{k}, k \ge 1$.
Also $\frac{\log x}{x}$ is decreasing for $x \ge 3$ so if $p \ge q_k, \frac{\log p}{p} < \frac{A_2}{k}, k \ge 2$
If $n=p_1^{\alpha_1}...p_s^{\alpha_s}$ (in increasing order of the primes) we have:
$\log n=\sum_{k=1}^s\alpha_k \log p_k \ge s\log 2$ since $\alpha_k \ge 1, \log p_k \ge \log 2$, hence :
$s \le C\log n$
Then using that $p_k \ge q_k$ we get that:
$\sum_{p|n}\frac{\log p}{p} \le \sum_{k=1}^{C\log n}\frac{A_2}{k} \le A_3  \log \log n$
$\sum_{p|n}\frac{1}{p} \le \sum_{k=2}^{C\log n}\frac{A_1}{k\log k} \le A_4 \log \log  \log n$, so we are done!
A: 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
$$
\sum_{i=1}^s\frac{\log p_i}{p_i}=O(\log\log n)
$$
