Prove: for $x$ large enough, there are more primes in the interval $(1,x]$ than in $(x,2x]$ I want to show that there are more primes in the interval $(1,x]$ than in $(x,2x]$, when $x$ is large enough. Let $\pi$ be the prime counting function, ie. $\pi(x)=\# \{ p \leq x | p \text{ is a prime} \}$. Then the above statement is equivalent to showing:
$\pi (2x)-2\pi(x)=-\log(2)\frac{x}{\log ^2x}+O\left(\frac{x}{\log ^3 x}\right)$
I have a hint saying to use the prime number theorem. I'm trying to use that $\pi(x)= \frac{x}{\log x} + \frac{x}{\log ^2 x} + O \left( \frac{x}{\log ^3 x} \right)$, but can't seem to rewrite it fitting. Maybe I need to use other aspects of the prime number theorem.
Added: It's useful to see alternative ways of solving the problem, but I'm still specifically seeking to show the equation above using PNT.
From Dusart we know that $\frac{x}{\ln x}\left(1+\frac{1}{\ln x} \right) &lt; \pi(x )&lt;\frac{x}{\ln x}\left(1+\frac{1.3}{\ln x} \right) $ for all $x \geq 600$ and so you need to prove that $\pi(2x) -2 \pi(x) \leq \frac{2x}{\ln 2x} \left(1+\frac{1.3}{\ln 2x} \right)-2\frac{x}{\ln x}\left(1+\frac{1}{\ln x} \right) &lt;&lt; 0  $
Or Equivalently $\frac{1}{\ln 2x} \left(1+\frac{1.3}{\ln 2x} \right) &lt;&lt; \frac{1}{\ln x}\left(1+\frac{1}{\ln x} \right) =&gt; \ln x (1+\frac{1.3}{\ln 2x}) &lt; (\ln x +\ln 2)(1+\frac{1}{\ln x}) =&gt; \ln x +\frac{1.3 \ln x}{\ln 2x} &lt; ln x +1.3&lt; \ln x+ 1 + \ln 2 &lt; \ln x+1+\ln 2 + \frac{\ln 2}{\ln x} $
Checking for smaller cases the inequality is true for all $x\geq 11$
There is no need for error estimates.  Use the PNT in the form $\pi(x) \sim x/\log x$. That implies $\pi(2x)/\pi(x) \sim (2x/\log(2x))/(x/\log x) = 2(\log x)/\log(2x)$.  Now convince yourself that this tends to $2$ as $x\to \infty$, so the difference
$$
\pi(2x) - \pi(x) = \pi(x)\left(\frac{\pi(2x)}{\pi(x)} - 1\right)
$$
is large for large $x$.  Similar reasoning shows for each $c &gt; 1$ that $\pi(cx) - \pi(x)$ is large for large $x$.