Limit with prime sequence and inverse logintegral

Question

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?

Answer 1

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 $\lim_{n\to \infty}\frac{a}{b}= 1$

Answer 2

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.