7
$\begingroup$

What is a simplest way to get an idea (not rigorous proof) that

$$\sum_{ p \leq x}\frac{1}{p} \approx \log{\log{x}}$$

$\endgroup$
6
  • 2
    $\begingroup$ Two facts: 1) the prime number theorem gives that $p_n \sim n\log(n)$ and 2) we can approximate a sum with an integral $\sum^x \frac{1}{n\log(n)} \sim \int^x \frac{{\rm d}x}{x\log(x)}$. $\endgroup$ – Winther Apr 9 '17 at 8:43
  • 1
    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Wojowu Apr 9 '17 at 8:58
  • $\begingroup$ @ Winther: I meant maybe some tricks with Euler product formula for the Riemann zeta function... $\endgroup$ – Aleksey Druggist Apr 9 '17 at 9:01
  • $\begingroup$ @Wojowu: This is about what I expected $\endgroup$ – Aleksey Druggist Apr 9 '17 at 9:13
  • 1
    $\begingroup$ This paper is closely related to the topic (Idk if it includes a proof inside of the asymptotic behavior). $\endgroup$ – Masacroso Apr 9 '17 at 9:21
4
$\begingroup$

The Von Mangoldt function fulfills the identity $\log(n)=\sum_{d\mid n}\Lambda(d)$, hence $$ \sum_{n\leq x}\log(n) = \sum_{d\leq x}\Lambda(d)\left\lfloor\frac{x}{d}\right\rfloor\tag{1} $$ where the LHS of $(1)$ is $x\log(x)+O(x)$ by Stirling's inequality.
Since $\left\lfloor\frac{x}{d}\right\rfloor=\frac{x}{d}+O(1)$ and $\sum_{d\leq x}\Lambda(d)=O(x)$ from the weak form of the PNT, it follows that $$ \sum_{d\leq x}\frac{\Lambda(d)}{d}=\log(x)+O(1)\tag{2} $$ and since the contribute given by prime powers is bounded by an absolute constant, $$ R(x)=-\log(x)+\sum_{p\leq x}\frac{\log p}{p} = O(1).\tag{3} $$ In particular, by summation by parts, $$\begin{eqnarray*}\sum_{p\leq x}\frac{1}{p}&=&\int_{2}^{x}\frac{dt}{t\log t}+\int_{2}^{x}\frac{R(t)\,dt}{t\log^2 t}+\frac{R(x)}{\log x}\\&=&\log\log(x)+O(1).\end{eqnarray*}\tag{4}$$

$\endgroup$
1
  • 1
    $\begingroup$ Just for remembering : $2^{2n} \ge {2n \choose n} \ge \prod_{n < p \le 2n} p$ so that $\theta(2n)-\theta(n) \le 2n \ln 2$ and $\sum_{p \le 2n} \ln p = \theta(2n) < 4n \ln 2$ $\endgroup$ – reuns Apr 10 '17 at 1:48
2
$\begingroup$

Apart from the answer of Jack D'Aurizio, I would like to add as a complement the following very nice excerpt from "The G.H.Hardy Reader" published by The Mathematical Association of America. It is a compendium of the most significant lectures of Professor G.H.Hardy. Just by chance I am reading exactly the chapter $4$ ("The Indian Mathematician Ramanujan"), and there it is this easy to follow explanation of the topic asked by the OP, so these are the brilliant Professor Hardy's words (any errors in the formulas are my fault, please just let me know):

In the first place, we may start from Euler's identity:

$$\Pi_p\frac{1}{1-p^{-s}}=\frac{1}{(1-2^{-s})(1-3^{-s})(1-5^{-s})\dots}=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\dots=\sum_{n}\frac{1}{n^s}$$

This is true for $s \gt 1$, but both series and product become infinite for $s=1$. It is natural to argue that, when $s=1$, the series an the product should diverge in the same way. Also

$$\log{\Pi\frac{1}{1-p^{-s}}}=\sum\log{\frac{1}{1-p^{-s}}}=\sum\frac{1}{p^s}+\sum{(\frac{1}{2p^{2s}}+\frac{1}{3p^{3s}}+\dots)}$$

and the last series remains infinite for $s=1$. Is is natural to infer that

$$\sum{\frac{1}{p}}$$

diverges like

$$log{(\sum{\frac{1}{n}})},$$

or, more precisely, that

$$\sum_{p \le x} \frac{1}{p} \sim\log{(\sum_{n \le x}\frac{1}{n})}\sim log{\ log{\ x}}$$

for large $x$.

By the way, the book is a little bit expensive for the non-professional ("hobbyist") reader (like myself), but it really worths every cent paid for it. It is a very enjoyable pearl, so I humbly recommend reading it if you have a chance.

$\endgroup$

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.