0
$\begingroup$

Let's denote $\theta(x)$ as the sum of natural logarithms of primes up to $x$ and denote $\pi(x)$ as the number of primes up to number $x$.Does the inequality

$$\pi(2x)-\pi(x) \geq \frac{\theta(2x)-\theta(x)}{\log(x)}$$

holds where log is natural logarithm?

Edit: Grammar fixes.

$\endgroup$
  • $\begingroup$ Is it donate or denote $\endgroup$ – Darkrai Nov 29 '17 at 18:02
  • $\begingroup$ @Greninja thanks, I edited it. $\endgroup$ – Bright Chancellor Nov 30 '17 at 16:55
1
$\begingroup$

Let $a_n = 1_{n \text{ is prime}}$ then $$\pi(2x)-\pi(x) = \sum_{n \in (x,2x]} a_n \le \sum_{n \in (x,2x]} a_n \frac{\log n}{\log x}= \frac{\theta(2x)-\theta(x)}{\log x}$$

$\endgroup$

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.