\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.

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.

\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}

## No comments:

## Post a Comment