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Find all the positive integers $m$ such that $$p_{m}≥2m$$

where $(p_{m})$ is the sequence of prime numbers

I have no idea to start.

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Note that $7 < 2\cdot 4$, while $11 > 2\cdot 5$, and every prime after that is odd.

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We have $p_n\ge n\log(n)\ge 2n$ for all $n\ge 8$. This follows from the standard estimates, e.g., see here.

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    $\begingroup$ Isn't the $\log \log n - c$ term important here? $\endgroup$ – Erick Wong Jan 19 '15 at 16:41
  • $\begingroup$ Sorry, you're right, I didn't notice the reference to Rosser 1938. $\endgroup$ – Erick Wong Jan 19 '15 at 17:03
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@Dietrich Burde's answer does the hard part, of proving the high-$m$ behavior.

However, the correct answer to the problem is

$$ m \in \{1 \cup \{n | n \geq 5 \}$$

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    $\begingroup$ Isn't this the complement of the desired set? $\endgroup$ – Erick Wong Jan 19 '15 at 16:43
  • $\begingroup$ No, the question is to find $m$ such that the $m$-th prime is at least as big as $2m$. Not many of those... $\endgroup$ – Mark Fischler Jan 19 '15 at 20:11
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    $\begingroup$ Then your answer is wrong, e.g., for $m=2$ we have $p_2=3$, which is not at least as big as $2m=4$. $\endgroup$ – Dietrich Burde Jan 19 '15 at 20:24
  • $\begingroup$ You are right. I have to change the answer. $m=1$ still works(!) $\endgroup$ – Mark Fischler Jan 19 '15 at 22:55

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