# Show that $\sum\limits_{p \leq x} \frac{1}{p}$ ~ ${\log\log{x}}$ when ${x \to \infty}$ (here p is a prime)

I saw that some of you were upset over my last question, so I decided to ask a more interesting question:

Show that $\sum\limits_{p \leq x} \frac{1}{p}$ ~ ${\log\log{x}}$ when ${x \to \infty}$
(here the sum goes over all the primes less than or equal to x)

Edit: I've changed the question and wrote it as it appeared in the exam.

• Hint: Google and use Merten's theorem, and you will find that the limit tends to 1. – BlackAdder Jun 25 '13 at 16:01
• @yanbo: the limit does not tend to anything. – user64494 Jun 25 '13 at 16:22
• @user64494 Do you think that there is no limit ? – Robert777 Jun 25 '13 at 16:41

yanbo is right.

Merten's second theorem states that $\lim_{n \to \infty} \big(\sum_{p \le n} \frac1{p}-\ln\ln n \big) =M$ where $M$ is the Meissel–Mertens constant.

Dividing by $\ln\ln n$, we get $\lim_{n \to \infty} \big(\frac1{\ln\ln n}\sum_{p \le n} \frac1{p}-1 \big) =\lim_{n \to \infty} \frac{M}{\ln\ln n} = 0$ (since $\ln \ln n$ goes reluctantly to $\infty$) so $\lim_{n \to \infty} \big(\frac1{\ln\ln n}\sum_{p \le n} \frac1{p} \big) = 1$

• Thanks, I think you're right. – Robert777 Jun 25 '13 at 17:01

Use the Prime Number Theorem, which has as a consequence:

$$p_n \sim n \log{n} \quad (n \to \infty)$$

Then

$$\sum_{k=2}^{n} \frac{1}{p_k} \sim \sum_{k=2}^{n} \frac{1}{k \log{k}} \sim \int_2^n \frac{dt}{t \log{t}} \sim \log{\log{n}} \quad (n \to \infty)$$

• Note that $p_n \sim n \ln n$, so the sum up to $p_n$ is not the same as the sum for $p_i \le n$. However, $\ln \ln n$ grows slowly enough so the results are the same. – marty cohen Jun 25 '13 at 17:30
• @martycohen: good point. Yes, the difference is higher-order, so it does work out. – Ron Gordon Jun 25 '13 at 17:33