# Inequality with the prime-counting function

• Dec 17th 2008, 09:13 AM
HTale
Inequality with the prime-counting function
Hi there guys, I hope I've posted this in the right section;

I'm going through a paper, and there is one statement which is:

$\displaystyle n^{\pi(n)} < 3^n$

for sufficiently large $\displaystyle n$, where $\displaystyle \pi(n)$ is the prime-counting function defined as the function counting the number of prime numbers less than or equal to $\displaystyle n$. I'm having real difficulty justifying this claim, and I'd be really grateful if someone could help me out.

HTale

• Dec 17th 2008, 11:07 AM
Opalg
Quote:

Originally Posted by HTale
I'm going through a paper, and there is one statement which is:

$\displaystyle n^{\pi(n)} < 3^n$

for sufficiently large $\displaystyle n$, where $\displaystyle \pi(n)$ is the prime-counting function defined as the function counting the number of prime numbers less than or equal to $\displaystyle n$.

Take logs, and it says $\displaystyle \pi(n)<\frac n{\ln n}\ln3$. But ln(3)>1, so the result will follow from the prime number theorem.