Let $\psi(x) := \sum_{n\leq x} \Lambda(n)$ where $\Lambda(n)$ is the Von-Mangoldt function. I want to show that if $ \lim_{x \rightarrow \infty} \frac{\psi(x)}{x} =1 $ then also $\lim_{x\rightarrow \infty} \frac{\pi(x) \log x }{x}=1.$
I tried to play a little bit with $\psi$, what I want to show is that:
$\left| \frac{\pi(x) \log x}{x} -1 \right| \leq \left| \frac{\psi(x)}{x} -1 \right| \rightarrow 0$
So I tried to develop $\psi$ a little bit, but I got astray.
So I have $ \frac{\psi(x)}{x} -1 = \sum_{p^k \leq x , k \geq 1} \frac{\log p}{x} -1 = \frac{1}{x}\left(\sum_{p\leq x} \log p + \sum_{p^2\leq x} \log p + ...+ \sum_{p^k \leq x, p^{k+1} >x} \log p \right) -1 $ and I want to estimate its aboslute value from below, but I don't have any idea?
Any hints?
Thanks.