There is a well-known question that seeks the asymptotic behaviour of this function, for $x\geq 2$: $$\sum_{n\leq x} \frac{\phi(n)}{n^2}.$$ See, for example, Apostol "Introduction to Analytic Number Theory" question 6 on page 71.

I have found three posts on stackexchange related to this question, namely post A, post B, and post C. Using methods available to someone who has just read Chapter 3 of Apostol's text, the best solution to this question would seem to be apatch's response to post A.

The method boils down to showing the following identity $$\sum_{d>x}\frac{\mu(d)\log d}{d^2} = O\left(\frac{\log x}{x}\right).$$

This is where I get unstuck. The first step is clearly $$\left|\sum_{d>x}\frac{\mu(d)\log d}{d^2}\right| \leq \sum_{d>x}\frac{\log d}{d^2}$$ but then what do you do with what remains?

One solution I found simply jumps from here to $O(\frac{\log x}{x})$, which seems non-trivial. Apatch's solution in post A does the following: $$\sum_{d>x}\frac{\log d}{d^2}=\sum_{d>x}\frac{\log d}{d^\frac{1}{2}}.\frac{1}{d^\frac{3}{2}}<\frac{\log x}{x^\frac{1}{2}}\sum_{d>x}\frac{1}{d^\frac{3}{2}}$$ but I don't see how can you justify the inequality, since $\frac{\log x}{\sqrt{x}}$ only reaches its maximum until around $x\approx 7.39$, which is well above the $x>2$ condition stated in the question.

Eric Naslund's solution to post C states that $$\sum_{d> x}\frac{\mu(d)\log d}{d^2}=O(1/x)$$ but this less elementary result is presumably even harder to derive.

Can anyone identify the "correct" answer, and explain why?

  • 2
    $\begingroup$ $\int_x^\infty \frac{\log t}{t^2}\,dt$ should yield the conclusion. If nobody has done then, I'll compute after dinner. $\endgroup$ – Daniel Fischer Nov 18 '13 at 17:23


$$ \sum_{d \geq x} \frac{\log d}{d^2} \leq \frac{\log x}{x^2} + \int_x^\infty \frac{\log t}{t^2}\,dt. $$

  • $\begingroup$ Thanks! I'm guessing this is what Apostol intended readers to use: it is suitably elementary (and I'm now wondering why I didn't think of it). $\endgroup$ – rbrignall Nov 18 '13 at 22:35

One way to show that

$$\sum_{n>x}\frac{\mu(n)\log n}{n^{2}}=O\left(\frac{\log x}{x}\right)$$

is to split into diadic intervals and use the bound $\sum_{n> y}\frac{1}{n^2}\ll \frac{1}{y}$. We have that

$$\sum_{n>x}\frac{\mu(n)\log n}{n^{2}}\leq\sum_{k=1}^{\infty}\sum_{2^{k-1}x<n\leq2^{k}x}\frac{\log n}{n^{2}} $$

$$\leq \sum_{k=1}^{\infty}\log(2^{k}x)\sum_{2^{k-1}x<n\leq2^{k}x}\frac{1}{n^{2}} $$

$$\ll \sum_{k=1}^{\infty}\log(2^{k}x)\frac{1}{2^{k-1}x}$$

$$=\frac{\log x}{x}\sum_{k=1}^{\infty}\frac{1}{2^{k-1}}+\frac{1}{x}\sum_{k=1}^{\infty}\frac{k\log2}{2^{k-1}}\ll\frac{\log x}{x}.$$

In my answer that you mentioned, I was thinking of this elementary bound, and forgot the $\log x$ term. However, the result I quoted is not false - we just need to take into account the cancellation resulting from the $\mu(n)$ term. Letting $M(x)=\sum_{n\leq x} \mu(n)$, integration by parts tells us that

$$\sum_{n>x}\frac{\mu(n)\log n}{n^{2}}=\int_{x}^{\infty}\frac{\log t}{t^{2}}d\left(M(t)\right)$$

$$=\frac{M(x)\log x}{x^{2}}+\int_{x}^{\infty}\frac{M(x)(2\log x-1)}{x^{3}}dx.$$ By the prime number theorem, there exists $c>0$ such that $$M(x)\ll xe^{-c\sqrt{\log x}},$$ and so we see that there is a constant $c_1>0$ such that $$\sum_{n>x}\frac{\mu(n)\log n}{n^{2}}\ll \frac{1}{x}e^{-c\sqrt{\log x}}.$$

  • $\begingroup$ Hi, thanks, that diadic approach is cute. Also, apologies: I didn't mean to suggest what you said in your earlier post was necessarily wrong (though I'm guessing Apostol did not have this particular argument in mind). $\endgroup$ – rbrignall Nov 18 '13 at 22:24
  • $\begingroup$ Can you give a hint or so how the prime number theorem gives the bound $O(x e^{-c\sqrt{\log x}})$ for the Mertens function? I don't see how to derive it :( $\endgroup$ – Daniel Fischer Nov 19 '13 at 11:23
  • $\begingroup$ @DanielFischer: I am quoting the quantitative version of the prime number theorem due to Hadamaard and De La Vallee Poussin. The bounds for $\pi(x)-\text{li}(x)$ and $M(x)$ will be within a logarithmic factor of each other, and so you cannot deduce the statement I wrote solely from the fact that $\pi(x)-\text{li}(x)=o\left(\frac{x}{\log x}\right)$. You need to use the stronger bound $$\pi(x)-\text{li}(x)=O\left(xe^{-c\sqrt{\log x}}\right),$$ which was originally prove in 1897. $\endgroup$ – Eric Naslund Nov 19 '13 at 13:46
  • $\begingroup$ Thanks. Now I'll go attempt to figure out the relation between $\pi(x)-\operatorname{li}(x)$ and $M(x)$. $\endgroup$ – Daniel Fischer Nov 19 '13 at 13:49
  • $\begingroup$ @DanielFischer: I strongly encourage you to do so! It is a very important connection, and it is said that phrasing things in terms of the randomness that the mobius function is the best way to understand primes. $\endgroup$ – Eric Naslund Nov 19 '13 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.