Showing that the density of primes is zero from the inequality $\prod_{p\leq n} p\leq 4^{n-1}$

You can do it without summation by parts or Stieltjes integrals. The number of primes up to $n$ is $\sum\nolimits_{p \le n} 1$. Divide this sum into two parts and estimate the parts trivially and then by using your inequality: \begin{align*} \sum\limits_{p \le n} 1 & = \sum\limits_{p \le n} {\log p\frac{1}{{\log p}}} = \sum\limits_{p \le \sqrt n } {\log p\frac{1}{{\log p}}} + \sum\limits_{\sqrt n < p \le n} {\log p\frac{1}{{\log p}}} \\ & \le \sum\limits_{p \le \sqrt n } {\log p\frac{1}{{\log 2}}} + \sum\limits_{\sqrt n < p \le n} {\log p\frac{1}{{\log \sqrt n }}} \\ & = \frac{1}{{\log 2}}\sum\limits_{p \le \sqrt n } {\log p} + \frac{1}{{\log \sqrt n }}\sum\limits_{\sqrt n < p \le n} {\log p} \\ & \le \frac{1}{{\log 2}}\sum\limits_{p \le \sqrt n } {\log p} + \frac{1}{{\log \sqrt n }}\sum\limits_{p \le n} {\log p} \\ & \le \frac{{\sqrt n \log 4}}{{\log 2}} + \frac{{n\log 4}}{{\log \sqrt n }} = 2\sqrt n + 4\log 2\frac{n}{{\log n}}. \end{align*} Can you finish from here?