# $\exists a,b>0$ such that $an\log n<p_n<bn\log n$.

I already know that $\exists a,b>0$ such that $\frac{an}{\log n}<\pi(n)<\frac{bn}{\log n}$, where $\pi(n)$ is the number of primes in the first n natural numbers.

It is intuitive that we can find $a,b>0$ such that $an\log n<p_n<bn\log n$ where $p_n$ is the n$^{th}$ prime. But I do not know how to prove it.

Can it be proved from the previous formula? Or we have to reconstruct the $a$ and $b$?

Any help is appreciated.

-

## 1 Answer

Set $n = p_m$, so that $\pi(n) = m$. The equations become $\frac{a p_m}{\ln p_m} < m < \frac{b p_m}{\ln p_m}$. Now it becomes in converting $\frac{x}{\ln x} <> y$ (where "<>" represents either "<" or ">") into bounds for $x$ in terms of $y$.

I'll do "<" here.

If $\frac{x}{\ln x} < y$, then $x < y \ln x$. If $x < y \ln y$, $\ln x < \ln y +\ln\ln y$ so $x < y (\ln y + \ln \ln y)$.

You can do something similar for ">".

de Bruijn's "Asymptotic Methods in Analysis" has a section on solving $\frac{x}{\ln x} = y$ for $x$.

-