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The following graph shows the proportional error of $n\ln(n)$ for the $n^{th}$ prime.

The proportional error appears to fall towards $0.1$, and not zero, when considering the first $n=10,000,000$ natural numbers.

  1. Is this a correct conclusion, or is this a case of a very very slow asymptotic approach to zero?
  2. Where can I find a proof of the non-zero limit?

In comparison, plots of the proportional error of $\frac{n}{ln(n)}$ and $\text{Li}(n)$ for the number of primes up to $n$, $\pi(n)$, do show it approaching zero more clearly.

enter image description here

Daniel Fischer
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Penelope
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    If you take more terms of the asymptotic expansion cf. e.g. here into account, you can see that $$\frac{p_n}{n\log n} - 1 \sim \frac{\log \log n}{\log n},,$$ and that decreases rather slowly. – Daniel Fischer Jun 08 '20 at 14:34
  • The $0.1$ relative error with your data might be related to Wikipedia's Non-asymptotic bounds on the prime-counting function section which states Dusart proved in $2010$ that $\frac{x}{\log x-1} \lt \pi(x)$ for $x \geq 5393$, and $\pi(x) \lt \frac {x}{\log x-1.1}$ for $x \geq 60184$. The difference in denominators of $0.1$ matches the relative error you got. However, right after this in the Wikipedia article shows $1$ & $1.1$ become $1 - \epsilon$ & $1 + \epsilon$ for large enough $x$. – John Omielan Jun 08 '20 at 17:38

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