Good bound for $\sum_{p\mid n}\frac{1}{p}$ I am trying to get a good asymptotic bound for $$\sum_{\substack{p\mid n\\ \text{$p$ prime}}}\frac{1}{p}.$$ One way I can think of is $$\sum_{p\mid n}\frac1p\leqslant \sum_{p\leqslant n}\frac{1}{p} \ll \log\log n,$$ by  Mertens' theorem.
Can we do better?
As a follow-up question, if you see a sum like this, is there a way to intuitively guess how large it could be?  Thanks!
 A: $$\sum_{p|n} 1/p \le \sum_{p| \prod_{q\le k_n} q}1/p= \sum_{p\le O(\log n)} 1/p= O(\log\log\log n)$$
where $\prod_{q\le k_n} q$ is the least primorial $\ge n$:
$\prod_{q\le k_n} q\ge \exp(k_n/10)$ gives $k_n\le 10\log(n)$.
A: Similar to how Gronwall deduced
$$
\limsup_{n\to\infty}{\sigma(n)\over e^\gamma n\log\log n}=1
$$
We consider defining $a_n$ such that
$$
\prod_{p\le p_{a_n}}p\le n\le\prod_{p\le p_{a_n+1}}p
$$
Taking logarithms gives
$$
{\vartheta(p_{a_n})\over p_{a_n}}\le{\log n\over p_{a_n}}\le{p_{a_n+1}\over p_m}\cdot{\vartheta(p_{a_n+1})\over p_{a_n+1}}
$$
Now, due to prime number theorem we have $p_n\sim p_{n+1}$ and $\vartheta(x)\sim x$, so that
$$
\log n\sim p_{a_n}\tag1
$$
With these priliminaries, we can begin working on the problem
$$
\sum_{p|n}\frac1p\le\sum_{p\le p_{a_n}}\frac1p=\log\log p_{a_n}+\mathcal O(1)
$$
By (1), we also see that
$$
\sum_{p\le p_{a_n}}\frac1p\sim\log\log\log n
$$
To see how tight this bound is, we set
$$
n_k=(p_1p_2\cdots p_k)^{\lfloor\log p_k\rfloor}
$$
so that
$$
\log n_k=\lfloor\log p_k\rfloor\vartheta(p_k)\sim p_k\log p_k
$$
and
$$
\log\log n_k\sim\log p_k
$$
Plugging these into the original formula, we get
$$
\sum_{p|n}\frac1p=\sum_{p\le p_k}\frac1p\sim\log\log p_k\sim\log\log\log n
$$
Therefore, $\log\log\log n$ is not only an upper bound for $\sum_{p|n}\frac1p$, but also a maximal order for $\sum_{p|n}\frac1p$:
$$
\limsup_{n\to\infty}{1\over\log\log\log n}\sum_{p|n}\frac1p=1
$$
