How can I show that: $\pi(n)\simeq\dfrac{n}{10}+\dfrac{5}{\sqrt{n}}+\dfrac{\log{n}+n}{\log_2{n}}$, where $\pi(n)$ is the prime counting function

Question

Progress:

I observed this result:

$\pi(n)\simeq\Bigg\lfloor\dfrac{n}{10}+\dfrac{5}{\sqrt{n}}+\dfrac{\log{n}+n}{\log_2{n}}\Bigg\rfloor$

while studying several inequalities and relations concerning prime gaps and PNT and it tends to be extremely accurate as I tested so far. But how can I prove this one? I would appreciate any help or insight you can provide. Thanks.

Answer 1

Divide both sides by $n$: $$\frac{\pi(n)}n\simeq \frac1{10}+\frac5{n\sqrt{n}}+\frac{\log n + n}{n\log_2n}>\frac1{10}$$ So you say that $\frac{\pi(n)}{n}$ goes to $\frac{1}{10}$ as $n$ goes to infinity. But the PNT says that $\pi(n)\sim\frac{n}{\log n}$, so: $$\lim_{n\to\infty}\frac{\pi(n)}{n}=\lim_{n\to\infty}\frac1{\log n}=0<\frac1{10}$$ This means that, sadly, your result is incorrect.