$\sum _{p\leq n}\frac{1}{p}=C+\ln \ln n+O\left(\frac{1}{\ln n}\right)$

Question

$\sum _{p\leq n}\frac{\ln p}{p}=\ln n+O(1),n\geq 2,$ where $p$ is a prime number, prove:

$$\sum _{p\leq n}\frac{1}{p}=C+\ln \ln n+O\left(\frac{1}{\ln n}\right)~~~(1)$$

one examination question，I think you're strong in solving this with complete answer in 20/30 minutes.

of course I cannot, give me some hints?

:)

Answer 1

Using Abel summation formula for $f(t)=\dfrac{1}{\ln t}, f:[2,+\infty)\to \mathbb{C}$ and $a_k=\begin{cases} \dfrac{\ln p}{p} ,& k=p \\ 0,& \text{otherwise} \end{cases},$ we find $$\begin{align*}\sum _{p\leq n}\frac{1}{p}&=\frac{1}{\ln n} \sum_{p \leqslant n} \frac{\ln p}{p} + \int_2^{n} \left(\sum_{p \leqslant t} \frac{\ln p}{p}\right)\frac{dt}{t(\ln t)^2}\\ &\overset{\star}{=}1+\mathcal{O}\left(\frac{1}{\ln n}\right)+\int_2^n \left(\ln t - \ln t +\sum_{p \leqslant t}\frac{\ln p}{p}\right)\frac{dt}{t (\ln t)^2}\\ &=1+\mathcal{O}\left(\frac{1}{\ln n}\right)+\int_2^n \frac{dt}{t \ln t}+\int_2^n \left(-\ln t + \sum_{p \leqslant t}\frac{\ln p}{p}\right)\frac{dt}{t (\ln t)^2}\\ &=1+\mathcal{O}\left(\frac{1}{\ln n}\right)+\ln \ln p-\ln\ln 2+\int_2^n \left(-\ln t + \sum_{p \leqslant t}\frac{\ln p}{p}\right)\frac{dt}{t (\ln t)^2}\\ &=C+\ln\ln n+\mathcal{O}\left(\frac{1}{\ln n}\right),\end{align*}$$

where $C\overset{\text{def}}{=}1-\ln\ln 2+\displaystyle\int_2^\infty \left(-\ln t + \sum_{p \leqslant t}\frac{\ln p}{p}\right)\frac{dt}{t (\ln t)^2}.$

We split $\displaystyle\int_2^n \left(-\ln t + \sum_{p \leqslant t}\frac{\ln p}{p}\right)\frac{dt}{t (\ln t)^2}$ in $\displaystyle\int_2^{\infty}$ and $\displaystyle\int_n^{\infty}$. Firt part $\displaystyle\int_2^{\infty}<+\infty$ and $\displaystyle\int_n^{\infty}=\mathcal{O}\left(\int_n^{\infty}\frac{dt}{t(\ln t)^2}\right)=\mathcal{O}\left(\frac{1}{\ln n}\right).$

$\star$ - we used here assumption $\displaystyle\sum _{p\leq n}\frac{\ln p}{p}=\ln n+\mathcal{O}(1),n\geq 2.$

Answer 2

Another approach using $$ \pi(n)=\frac{n}{\log(n)}+O\left(\frac{n}{\log(n)^2}\right)\tag{1} $$ we get, after summation by parts, $$ \begin{align} \sum_{p\le x}\frac1p &=\sum_{n=1}^x\frac{\pi(n)-\pi(n-1)}{n}\\ &=\sum_{n=1}^x\frac{\pi(n)}{n(n+1)}\\ &=\sum_{n=1}^x\left[\frac{1}{\log(n)(n+1)}+O\left(\frac{1}{\log(n)^2(n+1)}\right)\right]\\ &=\log(\log(x))+C+O\left(\frac1{\log(x)}\right)\tag{2} \end{align} $$ where $C$ is the Meissel-Mertens Constant.