Quick query: How to demonstrate that $\sum\limits_{d > x} {\frac{{\mu (d)\,\log d}}{{{d^2}}}} \,\, = \,\,O\left( {\frac{{\log x}}{x}} \right)$

Question

I am completing a proof regarding the asymptotic formula for $\sum\limits_{n \le x} {\frac{{\phi (n)}}{{{n^2}}}} $

The only part of the proof I am struggling with involves processing the term $\sum\limits_{d > x} {\frac{{\mu (d)\,\log d}}{{{d^2}}}} $ and showing that

$$\sum\limits_{d > x} {\frac{{\mu (d)\,\log d}}{{{d^2}}}} \,\, = \,\,O\left( {\frac{{\log x}}{x}} \right)$$

I have seen some solutions online, but the approach in all the ways I have seen don't seem logical.

One such analysis argues that

$$\left| {\sum\limits_{d > x} {\frac{{\mu (d)\,\log d}}{{{d^2}}}} } \right|\,\,\, \le \,\,\sum\limits_{d > x} {\frac{{\log x}}{{{d^2}}}} \le \,\,\log x\sum\limits_{d > x} {\frac{1}{{{d^2}}}} \,\, = \,\,O\left( {\frac{{\log x}}{x}} \right)$$

but how on earth can it be reasoned that

$$\left| {\sum\limits_{d > x} {\frac{{\mu (d)\,\log d}}{{{d^2}}}} } \right|\,\,\, \le \,\,\sum\limits_{d > x} {\frac{{\log x}}{{{d^2}}}} $$

when it is clear that, because of the summation, $\log d > \log x$?

Maybe there is a trick removing the Möbius function, because $\mu (d)$ can take on the values $ - 1 $, $0$ or $1$ , but I cannot see it.

Comments much appreciated!

Answer 1

Since $x\to \frac{\log x}{x^2}$ is eventually decreasing, it should be $$\begin{align}\left| {\sum\limits_{d > x} {\frac{{\mu (d)\,\log d}}{{{d^2}}}} } \right|&\leq \sum_{d>x} \frac{\log d}{d^2} \leq \frac{\log x}{x^2} + \int_x^\infty \frac{\log t}{t^2}\,dt= \frac{\log x}{x^2} + \left[ -\frac{\log t +1}{t}\right]_x^\infty \\&=\frac{\log x}{x^2} +\frac{\log x}{x}+\frac{1}{x}=O\left( {\frac{{\log x}}{x}} \right). \end{align}$$