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nth $n$th prime: a better approximation

letLet $n$$p_n$ be some integerthe $n$-th prime, then from Wikipedia iI got that :

$p_n \approx n(\ln n + \ln \ln n -1 + \frac{\ln \ln n-2}{\ln n}+\frac{6\ln \ln n-(\ln \ln n)^2-11}{\ln^2 n} $$p_n \approx n \left(\ln n + \ln \ln n -1 + \frac{\ln \ln n-2}{\ln n}+\frac{6\ln \ln n-( \ln \ln n)^2-11}{\ln^2 n} \right)$.

whatWhat is a better approximation that include $O(\frac{(\ln \ln n)^3}{\ln^3 n})$includes $O\big(\frac{(\ln \ln n)^3}{\ln^3 n}\big)$?

nth prime better approximation

let $n$ be some integer then from Wikipedia i got that :

$p_n \approx n(\ln n + \ln \ln n -1 + \frac{\ln \ln n-2}{\ln n}+\frac{6\ln \ln n-(\ln \ln n)^2-11}{\ln^2 n} $.

what is a better approximation that include $O(\frac{(\ln \ln n)^3}{\ln^3 n})$

$n$th prime: a better approximation

Let $p_n$ be the $n$-th prime, then from Wikipedia I got that

$p_n \approx n \left(\ln n + \ln \ln n -1 + \frac{\ln \ln n-2}{\ln n}+\frac{6\ln \ln n-( \ln \ln n)^2-11}{\ln^2 n} \right)$.

What is a better approximation that includes $O\big(\frac{(\ln \ln n)^3}{\ln^3 n}\big)$?

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nth prime better approximation

let $n$ be some integer then from Wikipedia i got that :

$p_n \approx n(\ln n + \ln \ln n -1 + \frac{\ln \ln n-2}{\ln n}+\frac{6\ln \ln n-(\ln \ln n)^2-11}{\ln^2 n} $.

what is a better approximation that include $O(\frac{(\ln \ln n)^3}{\ln^3 n})$