1
$\begingroup$

By Bertrand's postulate, we know that there exists at least one prime number between $n$ and $2n$ for any $n > 1$. In other words, we have $$ \pi(2n) - \pi(n) \geq 1, $$ for any $n > 1$. The assertion we would like to prove is that the number of primes between $n$ and $2n$ tends to $\infty$ , if $n \to \infty$, that is, $$ \lim_{n\to\infty} \pi(2n) - \pi(n) = \infty. $$ Do you see an elegant proof?

$\endgroup$
  • $\begingroup$ If a proof using the prime number theorem qualifies, there should be an easy and elegant proof. $\endgroup$ – Peter Mar 19 '20 at 11:01
  • 1
    $\begingroup$ I've shown $\frac{\pi(2n)}{\pi(n)} \to 2$ using PNT. $\endgroup$ – bozcan Mar 19 '20 at 11:07
1
$\begingroup$

By the PNT, we expect

$$\frac{2n}{\ln(2n)}-\frac{n}{\ln(n)}$$

$$=\frac{2n\ln(n)-n\ln(2n)}{\ln(2n)\ln(n)}$$

$$\sim\frac{n}{\ln(n)}$$

$$\to\infty,\;\; \text{as}\;\; n\to\infty$$

primes in this region.

$\endgroup$

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.