\begin{equation}\label{eq1}
\sum_{ p \le x } \frac{ \log p}{p} = \log x + R \quad\text{ with }\quad | R | \le 2
\end{equation}
I use Mertens method to prove the variant
\begin{equation}\label{eq2} \sum_{ n \le x } \frac{ \Lambda(n)}{n} = \log x + R \quad \text{ with } \quad -1 \le R \le 2
\end{equation}
Here, \( \Lambda \) is the von Mangoldt function. Equation \eqref{eq2} is thus similar to \eqref{eq1}, the sum is over prime powers instead of primes. It turns out that it is easier to prove \eqref{eq2} than \eqref{eq1}, because including the prime powers actually reduces the amount of estimates one has to make. The proof of \eqref{eq2} serves as a light version of the proof of \eqref{eq1} and gives insight into how the proof of \eqref{eq1} is organized.
\end{equation}
I use Mertens method to prove the variant
\begin{equation}\label{eq2} \sum_{ n \le x } \frac{ \Lambda(n)}{n} = \log x + R \quad \text{ with } \quad -1 \le R \le 2
\end{equation}
Here, \( \Lambda \) is the von Mangoldt function. Equation \eqref{eq2} is thus similar to \eqref{eq1}, the sum is over prime powers instead of primes. It turns out that it is easier to prove \eqref{eq2} than \eqref{eq1}, because including the prime powers actually reduces the amount of estimates one has to make. The proof of \eqref{eq2} serves as a light version of the proof of \eqref{eq1} and gives insight into how the proof of \eqref{eq1} is organized.