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.
 A: 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 $f(p) = 2$.
A: 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.
