# Limit with prime sequence and inverse logintegral

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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I find the logic of this post a little hard to follow. I get $n=\pi(p_n)\sim\mathrm{li}(p_n)\implies \mathrm{li}^{-1}(n)\sim p_n$. – Jonas Meyer Dec 31 '11 at 6:20
I think technically one should begin at the end to reach the beginning line but I hope the edit incorporates your logical point. – daniel Dec 31 '11 at 6:35
daniel: I understand how one might arrive at the point "backwards," but still with the edit it looks like you are arguing backwards, e.g. with the use of "so." A statement like "P, so Q," usually indicates that P is a known statement from which Q follows, but here it seems we want the opposite. – Jonas Meyer Dec 31 '11 at 6:40
@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? – daniel Dec 31 '11 at 6:46

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 now – Greg Martin Jan 1 '12 at 5:19