## Sunday, May 31, 2015

### Variation on a theme by Mertens

One of the formulas that Mertens proves in his paper is
\label{eq1} \sum_{ p \le x } \frac{ \log p}{p} = \log x + R \quad\text{ with }\quad | R | \le 2

I use Mertens method to prove the variant

\label{eq2} \sum_{ n \le x } \frac{ \Lambda(n)}{n} = \log x + R \quad \text{ with } \quad -1 \le R \le 2

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.