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