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How can the following equation be proven? $\sum\limits_{p \le n; p \in P} \frac{\ln p}{p} \sim \ln(n) + O(1).$

I just wanna understand this sum Sum of reciprocal prime numbers

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dehasi
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  • So what exactly is it that you don't understand? – Gerry Myerson Oct 08 '14 at 10:57
  • look: $\sum\limits_{p \le n; p \in P} \frac{\ln p}{p}$ = $\sum\limits_{x \le n;} \frac{A_1(x)}{x}$. It's ok. And why we can use Shaoiro's theorem and say that it is Ln(x) ?? – dehasi Oct 08 '14 at 11:33

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By partial summation and for Prime Number Theorem in the form $$\pi\left(x\right)=\frac{x}{\log\left(x\right)}+\frac{x}{\left(\log\left(x\right)\right)^{2}}+O\left(\frac{x}{\left(\log\left(x\right)\right)^{3}}\right)$$ you have $$\theta\left(N\right)=\underset{p\leq N}{\sum}\log\left(p\right)=\pi\left(N\right)\log\left(N\right)-\int_{2}^{N}\frac{\pi\left(t\right)}{t}dt=$$ $$=N+O\left(\frac{N}{\left(\log\left(N\right)\right)^{2}}\right)$$ so $$\underset{p\leq N}{\sum}\frac{\log\left(p\right)}{p}=\frac{\theta\left(N\right)}{N}+\int_{2}^{N}\frac{\theta\left(t\right)}{t^{2}}dt=$$ $$=1+O\left(\frac{1}{\left(\log\left(N\right)\right)^{2}}\right)+\int_{2}^{N}\frac{1}{t}dt+O\left(\int_{2}^{N}\frac{1}{t\left(\log\left(t\right)\right)^{2}}dt\right)=$$ $$=\log\left(N\right)+C+o\left(1\right).$$

user180218
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  • Why? $\theta\left(N\right)=\underset{p\leq N}{\sum}\log\left(p\right)=\pi\left(N\right)\log\left(N\right) ...$ why we can do it? – dehasi Oct 08 '14 at 12:41
  • http://en.wikipedia.org/wiki/Abel%27s_summation_formula – user180218 Oct 08 '14 at 12:47
  • I know about Abel's summation formula. Where in $\underset{p\leq N}{\sum}\log\left(p\right)$ $a_n$ and $f(n)$? How we split $\log(p)$? – dehasi Oct 08 '14 at 12:52
  • In $\theta\left(N\right)$ you take $a_{p}=1$ and $f(p)=\log\left(p\right)$. So $A(N)=\underset{p\leq N}{\sum}1=\pi\left(N\right)$. The other summation is similar, you take $A(N)=\theta\left(N\right)$. – user180218 Oct 08 '14 at 12:56