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Consider the function

$$f(n) = \log n /(n\ \log\theta(p_n)),$$

where $\theta$ is the first Chebyshev function and $p_n$ is the $n$-th prime. Does $f$ converge to a constant as $n$ grows to infinity, or does it grow to infinity; and if it converges to a constant, what would that constant be? In either case, if you can answer the question, can you prove it or provide a reference to read about it?

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    $\begingroup$ the function $\theta(n)$ grows like $n\log n$ so $f(n)$ goes to zero more rapidly than $1/n$ for large $n$. $\endgroup$
    – Carlo Beenakker
    Commented Apr 14, 2022 at 17:56
  • $\begingroup$ The reason I am not sure is that if you take $\sigma(N)/(N\ \log\ \theta(n))$ that decreases, but it goes to $e^\gamma/\zeta(2)$, not to 0. And on my computer, it seems to grow at first, at least up to n= 10^8. $\endgroup$
    – EGME
    Commented Apr 14, 2022 at 18:03
  • $\begingroup$ $\sigma(n) < \log n$ $\endgroup$
    – EGME
    Commented Apr 14, 2022 at 18:05
  • $\begingroup$ That was meant to be $\sigma(N) < \log n$, where $N$ is the n-t primorial, but I cannot edit the comment above nor delete it. $\endgroup$
    – EGME
    Commented Apr 14, 2022 at 18:14
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    $\begingroup$ Your function is at most $(\log n)/(n\log 2)$, hence it tends to zero quite rapidly. $\endgroup$
    – GH from MO
    Commented Apr 14, 2022 at 21:50

1 Answer 1

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The prime number theorem suggests that $\theta(x)\sim x$, so we have

$$ f(n)\sim{\log n\over n\log n}=\frac1n\to0 $$

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    $\begingroup$ The statement $\theta(x)\sim x$ is not suggested by the prime number theorem: it is a form of the prime number theorem. Note also that $\log\theta(n)\sim\log n$ is a much weaker statement, e.g. it follows from Chebyshev's estimate $\theta(x)\asymp x$. Finally, as I said in my remark, the trivial bound $\theta(n)\geq\log 2$ already implies that $f(n)$ tends to zero. $\endgroup$
    – GH from MO
    Commented Apr 15, 2022 at 1:59

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