1
$\begingroup$

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.

$\endgroup$
1
  • 4
    $\begingroup$ 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. $\endgroup$
    – Lucia
    Apr 10, 2014 at 14:05

2 Answers 2

6
$\begingroup$

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 $f(p) = 2$.

$\endgroup$
0
5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ See my comment above. Note that Stijn's answer implicitly uses the prime number theorem. Mertens's result is of course more elementary. $\endgroup$
    – Lucia
    Apr 10, 2014 at 23:04
  • $\begingroup$ @Lucia: yes; I must admit I didn't read Stijn's answer beyond the expression "partial summation", assuming he was giving the standard proof. $\endgroup$ Apr 11, 2014 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.