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I found formula below$\lim_{n\to\infty}\frac{\operatorname{li^{-1}}(n)}{p_n}=1$ where $\operatorname{li^{-1}}(n)$ is inverse logintegral function and $p_n$ is prime number sequence.

Can anyone prove this formula?

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I would think

$\frac{li^{-1}(n)}{p_n} \sim 1$

li$^{-1}(n) \sim p_n$

li li$^{-1}(n) \sim $ li $ p_n $ or $ n = \pi(p_n)\sim $ li $ p_n$

Working in proper sequence now,

$ n = \pi(p_n)\sim $ li $ p_n$

li$^{-1}(n) \sim p_n$

$\frac{li^{-1}(n)}{p_n} \sim 1$

bearing in mind that $a(n) \sim b(n)$ just means $\lim_{n\to \infty}\frac{a}{b}= 1$

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    @JonasMeyer: I think the second half constitutes a proof. The first half puts us in a position to see it. I could re-edit to remove the process, but I'm not sure it hurts to see that. The process is, in a sense, backward, isn't it?2011-12-31
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Both $li^{-1}(n)$ and $p_n$ are asymptotic to $n\ln n$ (the former by a little integration by parts argument, the latter by the Prime Number Theorem). Therefore their quotient has limit equal to 1.

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    @daniel: thanks for pointing that out, fixed now2012-01-01