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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!

Ken_M_W
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1 Answers1

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

Robert Z
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  • Thanks @RobertZ that is a really sound and clear line of argument. Much appreciated. – Ken_M_W Jul 24 '21 at 14:13
  • Actually @RobertZ the lower bound is quite useful to see the approach! I checked the previous edit. – Ken_M_W Jul 24 '21 at 16:45
  • @Ken_M_W I am glad you find it useful – Robert Z Jul 24 '21 at 16:48
  • Hello again @RobertZ . I have been playing with your solution approach a little more. One thing that struck me, and I cannot get my head around, is the condition in the problem I have stated that $x$ is a real number, not just an integer, and that we are summing $x>d$. If $x$ happens to be an integer, $m$, then our dummy variable $d$ sums from $m+1$ to $\infty$. I have been drawing rectangles under monotonic decreasing graphs to get geometric clarity, and I cannot seem to get a satisfactory result if the summation begins with $\left\lfloor x \right\rfloor + 1$. – Ken_M_W Jul 24 '21 at 19:16
  • @RobertZ I will carry on playing around .... it might yet "click" in my head, but can you see the issue I am juggling with? Thanks again :) – Ken_M_W Jul 24 '21 at 19:17
  • @Ken_M_W The same idea is explained here: https://en.wikipedia.org/wiki/Integral_test_for_convergence – Robert Z Jul 24 '21 at 19:30
  • @RobertZ thank you for the tip about "accepted" answers. Also, I had been playing around with integral test ideas to try and see how the inequality works. I will carry on playing around and report back. I can see how it works up to a point, but I can't yet "feel" it! Thanks again. Maybe a good night's sleep will do the trick. It often does! – Ken_M_W Jul 24 '21 at 21:02
  • @RobertZ I can see why $\sum\limits_{d > x} {\frac{{\log d}}{{{d^2}}}} \le \frac{{\log x}}{{{x^2}}} + \int\limits_x^\infty {\frac{{\log t}}{{{t^2}}}{\rm{d}}t} $ . A bit of manipulation with my diagram and it became (painfully) obvious! – Ken_M_W Jul 24 '21 at 22:23
  • @RobertZ But I am suspicious of the result that $\sum\limits_{d > x} {\frac{{\log d}}{{{d^2}}}} \ge \int\limits_x^\infty {\frac{{\log t}}{{{t^2}}}{\rm{d}}t} $ . For some real value $x$, if $x$ is allowed to approach $\left\lfloor x \right\rfloor$ then this is equivalent to the claim being $\int_x^{x + 1} {\frac{{\log t}}{{{t^2}}}{\rm{d}}t} \ge \sum\limits_{d = x + 1}^\infty {\frac{{\log d}}{{{d^2}}}} , - ,\int_{x + 1}^\infty {\frac{{\log t}}{{{t^2}}}{\rm{d}}t} $, which looks ludicrous on my diagram... – Ken_M_W Jul 24 '21 at 22:23
  • Let us continue this discussion in chat. – Ken_M_W Jul 24 '21 at 22:24