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I am having trouble in solving the following problem.

  • Let $p_{n}$ denote the $n$-th prime. Then prove that $\pi(\sqrt{p_{1}p_{2}\cdots p_{n}})>2n$ for $n \geq 6$.

No idea how to start.

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    @Jonas: You are welcome. Yes, it is an interesting collection and i hope to pose some more problems.2011-02-07

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The case $n=6$ can be verified by hand (you only need to find $13$ primes less than $\sqrt{2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13}$), and once you have it this follows by induction and Bertrand's postulate, because for $n\gt 6$, $p_n\gt 16\Rightarrow \sqrt{p_n}\gt 2^2$.

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    I mean you need to find $13$ primes less than $\sqrt{2\cdot 3\cdots 13}$, because $13\gt 2\cdot 6$. I didn't think about the fact that that might be confusing with $p_6=13$.2011-02-06