Bertrand's postulate in another point of view I was just wondering why can't we use prime number's theorem to prove Bertrand's postulate.We know that if we show that for all natural numbers $n>2, \pi(2n)-\pi(n)>0$ we are done. Why can't it be proven by just showing (By using the prime number's theorem) that for every natural numbers $n>2, \frac{2n}{ln(2n)}-\frac{n}{ln(n)}>0$?

You need a precise estimate of the form $$c_1<\frac{\pi(n)\ln n}{n}<c_2.$$ With that you can derive
$\pi(2n)>c_1\frac{2n}{\ln(2n)}>2c_1\frac{n}{\ln 2+\ln n}$. If you are lucky, you can continue$2c_1\frac n{\ln2+\ln n}>c_2\frac n{\ln n}>\pi(n)$.
However, for this you better have $\frac{2c_1}{c_2} >1+\frac{\ln 2}{\ln n}$. So at least if $c_2<2c_1$, your idea works for sufficiently big $n$.
