The estimate of the sum of $\frac{\log p_i}{p_i}$

Question

Let $p_i$ denote the i-th prime, we know the following estimate: $$ \sum_{p\leq x}\frac{\log p}{p}=\log x+O(1) $$

But when I’m trying to solve a problem, i need to get the following estimate: $$ \sum_{i=1}^{n}\frac{\log p_i}{p_i}=O(\log n) $$ Is it also true? Thanks to all your reply!

Answer 1

Evidently, we have

$$ \sum_{i=1}^n{\log p_i\over p_i} =\sum_{p\le p_n}{\log p\over p} =\log p_n+\mathcal O(1) $$

Chebyshev's theorem guarantees the existence of positive constants $A$ and $B$ such that

$$ An\log n<p_n<Bn\log n $$

Taking logarithms gives

$$ \log n+\log\log n+\log A<\log p_n<\log n+\log\log n+\log B $$

This implies $\log p_n\sim\log n$. Consequently, we get

$$ \sum_{i=1}^n{\log p_i\over p_i}=\mathcal O(\log n) $$

Answer 2

You want to say that $$\log p_n=O(\log n)$$

This follows from $$m\le {2m\choose m}=\sum_{p^k\le 2m}\log p (\lfloor \frac{2m}{p^k}\rfloor-2\lfloor \frac{m}{p^k}\rfloor)\le \sum_{p^k\le 2m}\log p$$ $$\le\log(2m) \sum_{p^k\le 2m}1\le \log(2m)^2\pi(2m) $$

Taking $m=p_n/2$ gives the bound $$p_n/2\le \log(p_n)^2 \pi(p_n)= \log(p_n)^2 n $$ Taking the $\log$ of both side gives your result.