$n$-th prime is less than a polynomial of $n$

Asked Apr 02 '21 at 21:03

Active Apr 02 '21 at 21:03

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Let $p_n$ be the $n$-th prime. Is there an elementary way to prove the statement that there exists a polynomial function $P$ such that $$p_n \leq P(n),$$ for all $n \in \mathbb{N}$? (Or, alternatively, for all $n$ from some $N \in \mathbb{N}$ onward?) I know the statement is true, and that $p_n$ "behaves asymptotically like" $n \log n,$ but I'm specifically looking for an elementary proof.

(I'd like to show $\sqrt[n]{p_n} \to 1,$ and since trivially $n \leq p_n,$ the result would follow from the squeeze theorem.)

Thanks in advance!!!!!!

asked Apr 02 '21 at 21:03

Neckverse Herdman

1,270

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See this in particular – diracdeltafunk Apr 02 '21 at 21:15
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By prime number theorem $\log n+\log\log n-1 \leq \frac{p_n}{n} \leq \log n+\log\log n,$ see here, therefore $p_n\le n^2$ for sufficiently large $n.$ See here for more detailed illustrations. – Bumblebee Apr 02 '21 at 21:15
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I was going to answer this but @diracdeltafunk saved me the trouble. You would be best off looking up results showing explicit bounds for $\pi(n)$, the prime counting function. Questions about the size of the the nth prime relative to n are related to $\pi(n).$ – open problem Apr 02 '21 at 21:18
Related. It seems that you can't really avoid a lot of work for this . – Maximilian Janisch Apr 02 '21 at 21:34
Grimm's conjecture comes close if proven as you can turn it into $p_n\leq \frac{n(n-1)}{2}+1$ as a requirement . Grimm's in unsolved though. – Roddy MacPhee Apr 02 '21 at 21:53

$n$-th prime is less than a polynomial of $n$

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