Variation on a theme by Mertens

One of the formulas that Mertens proves in his paper is

$$\begin{equation}\label{eq1} \sum_{ p \le x } \frac{ \log p}{p} = \log x + R \quad\text{ with }\quad | R | \le 2 \end{equation}$$

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.

The proof starts with the formula
$$\begin{equation*} \log n = \sum_{d | n } \Lambda(n) \end{equation*}$$
From this follows
$$\begin{align*} \sum_{ n \le x }\log n &= \sum_{ n \le x }\Lambda(n)  \Big\lfloor \frac{x}{n} \Big\rfloor \\ & =  \sum_{ n \le x }\Lambda(n) \frac{x}{n}  -  \sum_{ n \le x }\Lambda(n)  \Big\{ \frac{x}{n}\Big \} \end{align*}$$
thus
$$\begin{equation*} \sum_{ n \le x } \frac{ \Lambda(n)}{n} = \frac{1}{x} \sum_{ n \le x }\log n + \frac{1}{x} \sum_{ n \le x }\Lambda(n)  \Big\{ \frac{x}{n}\Big \} \end{equation*}$$

The first term on the right hand side is easily estimated from the inequality
$$\begin{equation*}  x \log x - x + 1 \le \sum_{ n \le x }\log n \le  (x+1) \log x - x + 1 \end{equation*}$$
This inequality is well known; it follows easily from comparing the sum with the integral. Hence

$$\begin{equation*} \frac{1}{x} \sum_{ n \le x }\log n = \log x + R_1 \end{equation*}$$
with
$$\begin{equation*}- 1 + \frac{1}{x} \le R_1 \le \frac{\log x}{x} - 1 + \frac{1}{x} \end{equation*}$$
and therefor
$$\begin{equation*}  - 1\le R_1 \le 0 \end{equation*}$$

The estimation of the second term $R_2 = \frac{1}{x} \sum_{ n \le x }\Lambda(n)  \Big\{ \frac{x}{n}\Big \}$ proceeds as follows. It is clear that

$$\begin{equation*} 0 \le R_2 \le  \frac{1}{x} \sum_{ n \le x }\Lambda(n) \end{equation*}$$
One only needs an upper bound on the Chebyshev function $\psi(x) = \sum_{ n \le x }\Lambda(n)$. In the first section of his paper, Mertens proves that $\psi(x) < 2 x$, hence$0 \le R_2 \le 2$, and the result $\eqref{eq2}$ is proven.

• The proof of $\psi(x) < 2 x$ uses tighter bounds on $\log n!$ than the ones used in this post. So, all in all, one still needs tight bounds on $\log n!$ to prove $\eqref{eq2}$.
• Better upper bounds on $\frac{\psi(x)}{x}$ are known, they thus lead directly to a better upper bound in $\eqref{eq2}$.
• This post is also a solution to exercise 3.1.7 in Murty, but with an explicit constant instead of $O(1)$

Here is a graph illustrating $\eqref{eq2}$

In the graph above, the blue dots are the function $\sum_{ n \le x } \frac{ \Lambda(n)}{n} - \log x$ and the black lines are the bounds.