Prove that $\frac{\pi(2x)\text{log}\, x}{x}$ has a constant upper bound.

Question

I know that \begin{equation*} \frac{\pi(2x)\text{log}(x)}{x} = \sum_{p < 2x} \frac{\text{log} \, x}{x}, \end{equation*} but other than that I'm not sure how to proceed. Any hints would be appreciated, thanks!

Answer 1

Suppose that $x\geq 2$. By Chebyshev's result, there is a $c>0$ such that $$ \pi (2x) < c\frac{{2x}}{{\log (2x)}} = 2c\frac{x}{{\log x + \log 2}} <2c\frac{x}{{\log x}} , $$ whence $$ \frac{{\pi (2x)\log x}}{x} < 2c. $$ Obviously this result extends to $x>0$.