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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?

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  • $\begingroup$ 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
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    $\begingroup$ I've shown $\frac{\pi(2n)}{\pi(n)} \to 2$ using PNT. $\endgroup$ – bozcan Mar 19 '20 at 11:07
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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.

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