# How to prove $\mathop {\lim }\limits_{x \to \infty } \sum\limits_{{f_x}(p) = 1} {\frac{1}{p}} = \ln 2$ for $p \le x$?

Let ${f_x}(m) = \sum\limits_{\left. p \right|m} {{f_x}(p)}$ be a strongly additive function on positive integer number $m$, where $p$ is a prime number. Set $${f_x}(p) = \left\{ {\begin{array}{*{20}{c}} {0,}\\ {1,}\\ 2, \end{array}} \right.\begin{array}{*{20}{c}} {{\rm{ }}p < \ln \ln x{\rm{ }}\ or \ {\rm{ }}p \ge {{(\ln \ln x)}^4}}\\ {\ln \ln x \le p < {{(\ln \ln x)}^2}}\\ {{{(\ln \ln x)}^2} \le p < {{(\ln \ln x)}^4}} \end{array}.$$ Bekelis (1997) say that $$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{\scriptstyle{\rm{ }}p \le x\atop \scriptstyle{f_x}(p) = 1} {\frac{1}{p}} = \ln 2, \mathop {\lim }\limits_{x \to \infty } \sum\limits_{\scriptstyle{\rm{ }}p \le x\atop \scriptstyle{f_x}(p) = 2} {\frac{1}{p}} = \ln 2.$$ But he does not give a detail proof. How to prove it?

[1]Bekelis, D. (1997). Convolutions of the Poisson laws in number theory. In Analytic and Probabilistic Methods in Number Theory: Proceedings of the Second International Conference in Honour of J. Kubilius, Palanga, Lithuania, 23-27 September 1996 (Vol. 4, p. 283). Walter de Gruyter.

• This follows at once from $\sum_{p\le x} 1/p= \log \log x + B+O(1/\log x)$ which is an elementary result of Mertens. Apr 10 '14 at 14:05

It looks like a partial summation-style problem. So consider the case when $f(p) = 1$

\begin{align} \sum\limits_{\log \log x \leq n < (\log \log x)^2} 1_{\mathbb{P}}(n)n^{-1} &= \pi((\log \log x)^2)(\log \log x)^{-2} - \pi(\log \log x)(\log \log x)^{-1} \\ &+ \int_{\log \log x} ^{((\log \log x)^2} \frac{1_{\mathbb{P}}(t)}{t^2}dt \\ &\sim (\log \log \log x)^{-2} - (\log \log \log x)^{-2} + \int_{\log \log x} ^{((\log \log x)^2} \frac{1}{t\log t}dt\\ &= \log \log ((\log \log x)^2) - \log \log (\log \log x)\\ &= \log \left(\frac{2 \log \log \log x}{\log \log \log x} \right)\\ &= \log 2 \end{align}

It should work similarly when $f(p) = 2$.

There is a well-known Theorem of Mertens (for the proof see the @Stijn's answer): for some constant $C$,

$$\sum_{p\leq x} \frac{1}{p} = \log\log x + C + O(\frac{1}{\log x})\,.$$

It thus follows that

$$\sum_{x \leq p \leq x^2} \frac{1}{p} = \log \log (x^2) - \log \log (x) + O(\frac{1}{\log x})\,.$$

Now $\log\log(x^2) = \log(2\log x) = \log 2 + \log x$, so we find:

$$\sum_{x \leq p \leq x^2} \frac{1}{p} = \log 2 + O(\frac{1}{\log x})\,.$$

The given sums are special cases of this.

• See my comment above. Note that Stijn's answer implicitly uses the prime number theorem. Mertens's result is of course more elementary. Apr 10 '14 at 23:04
• @Lucia: yes; I must admit I didn't read Stijn's answer beyond the expression "partial summation", assuming he was giving the standard proof. Apr 11 '14 at 16:54