Good bound for $\sum_{p\mid n}\frac{1}{p}$

Question

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!

Answer 1

$$\sum_{p|n} 1/p \le \sum_{p| \prod_{q\le k_n} q}1/p= \sum_{p\le O(\log n)} 1/p= O(\log\log\log n)$$

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 $k_n\le 10\log(n)$.

Answer 2

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 $\sum_{p|n}\frac1p$:

$$ \limsup_{n\to\infty}{1\over\log\log\log n}\sum_{p|n}\frac1p=1 $$