3

This question is from my number theory assignment and I was not able to solve it. I have been following Analytic Number Theory by Dekonick and Luca.

Let $\tau(n)$ be the number of divisors of n. Show that $\sum_{n\leq x , gcd(3,n)=1} \frac{ \tau(n)}{n} = \frac{ 2(\log x)^2} {9} +O(\log x)$.

As $\tau(n) =\sum_{d|n} 1$, gcd(3,n)=1 means that n of the form 3m+1, 3m+2 ($m\in \mathbb{Z}$).

I can't change n to n=qd and then writing $\sum_{n\leq x} d(n) = \sum_{q,d , qd \leq x} 1$ because summation is only over 3n+1 , 3n+2.

So, Can you please tell how should I approach this problem ?

Thanks!

TravorLZH
  • 6,718
  • I think it would be better if you think $gcd(n,3)=1$ as all $n$ which aren't the divisors of $n$. –  Nov 07 '21 at 18:34
  • And also $\tau(n)=\sum_{k=1}^{n}\lfloor \frac{n}{k} \rfloor$ –  Nov 07 '21 at 18:35
  • this is similar to one of your other questions: you need to know about two things: about how the partial sums of a sequence are related to the dirichlet series of that sequence, and about euler products. – tomos Nov 07 '21 at 19:18

1 Answers1

6

It would be interesting if we consider the following generalized problem:

$$ F(x)=\sum_{\substack{n\le x\\(n,q)=1}}{\tau(n)\over n} $$

To approach this, we consider studying the Dirichlet series associated to $F(x)$:

\begin{aligned} \hat F(s) &=\sum_{\substack{n\ge1\\(n,q)=1}}{\tau(n)\over n^s}=\prod_{p\nmid q}\left(1-{1\over p^s}\right)^{-2} \\ &=\zeta^2(s)\prod_{p|q}\left(1-{1\over p^s}\right)^2 \end{aligned}

This indicates that if we define

$$ f(n)= \begin{cases} \tau(n)/n & (n,q)=1 \\ 0 & \text{otherwise} \end{cases} $$

and

$$ G(s)=\sum_{n\ge1}{g(n)\over n^s}=\prod_{p|q}\left(1-{1\over p^{s+1}}\right)^2 $$

Then we have

$$ f(n)=\sum_{n=ab}g(a){\tau(b)\over b} $$

and

$$ F(x)=\sum_{n\le x}f(n)=\sum_{a\le x}g(a)\sum_{b\le x/a}{\tau(b)\over b} $$

This means our task gets transformed into evaluating the inner sum. For convenience, set $z=x/a$, so that Stieltjes integration gives

\begin{aligned} \sum_{b\le z}{\tau(b)\over b} &=\sum_{m\le z}\frac1m\sum_{n\le z/m}\frac1n=\sum_{m\le z}\frac1m\left[\log\frac zm+\mathcal O(1)\right] \\ &=\sum_{m\le z}{\log z\over m}-\sum_{m\le z}{\log m\over m}+\mathcal O(\log z) \\ &=\frac12(\log z)^2+\mathcal O(\log z) \\ &=\frac12(\log x)^2+\mathcal O\left\{\log x(\log a)^2\right\} \end{aligned}

Plugging this result back in, we get

$$ F(x)=\frac12(\log x)^2\sum_{a\le x}g(a)+\mathcal O\left\{\sum_{a\le x}|g(a)|(\log a)^2\log x\right\} $$

Because for every fixed $q$, $G(s)$ converges absolutely everywhere. Therefore using the identity $G(1)=\varphi(q)^2/q^2$, we see that for every fixed $q>0$, there is

$$ \sum_{\substack{n\le x\\(n,q)=1}}{\tau(n)\over n}={\varphi(q)^2\over2q^2}(\log x)^2+\mathcal O(\log x) $$

Plugging in $q=3$ gives exactly what the OP asked for.

TravorLZH
  • 6,718
  • Where that the square on the $\log a$ term come from? – Daniel Flores Nov 12 '21 at 16:46
  • @Daniel It appears when you approximate $\sum_m\log m/m$ by the integral of $\log x/x$. – TravorLZH Nov 13 '21 at 06:15
  • also isn't it necessary to prove that $\sum_{n > x} g(a) = O\left( \frac{1}{\log x} \right)$ to get the error term to be $O(\log x)$? Is there an easy of proving this? – Daniel Flores Nov 14 '21 at 20:02
  • @Daniel Since $\chi_{>x}(n)\le(\log n/\log x)^2$, we can see that the sum you mentioned is of $\mathcal O{(\log x)^{-2}}. – TravorLZH Nov 15 '21 at 00:40
  • @Daniel By the way, you can see a lot of treatments like this in §3.4 of Montgomery & Vaughan's Multiplicative Number Theory I: Classical Theory. – TravorLZH Nov 15 '21 at 00:42
  • thank you found in the book where they proved that $\sum_{n > x}g(n) \ll 1/\log x$ – Daniel Flores Nov 15 '21 at 00:54
  • @Daniel My pleasure. I was astonished when I see such operation for the first time few months ago. It was this example that always pushes me to study the Euler product of Dirichlet series before actually doing the calculations. – TravorLZH Nov 15 '21 at 03:48
  • @Travor How did you deduced that $\hat{F(s)} $ is equal to the infinite product :$\prod_{p\nmid q}\left(1-{1\over p^s}\right)^{-2}$? –  Dec 25 '21 at 12:34
  • @Evan Use the Euler product formula. i.e. if $f(n)$ is a multiplicative function then $$\sum_{n\ge1}{f(n)\over n^s}=\prod_p\sum_{r\ge0}{f(p)\over p^{rs}}$$ – TravorLZH Dec 25 '21 at 14:52
  • @TravorLZH I understand this theorem but how exactly $\sum_{n\geq 1, (n,q)=1} \frac{ \tau(n)}{n^s} =\prod_{p \nmid q} (1- 1/{p^s})^{-2}$. Can you elaborate on proof of this? –  Jun 04 '22 at 15:07
  • Let $\chi(n)$ be the characteristic function for $(n,q)=1$, then we have $\sum_{(n,q)=1}{f(n)\over n^s}=\prod_p\sum_{r\ge0}{\chi(p^r)f(p^r)\over p^{rs}}=\prod_{p\nmid q}\sum_{r\ge0}{f(p^r)\over p^{rs}}$. The right hand side follows from the fact that $\chi(p^r)=0$ when $p$ divides $q$. – TravorLZH Jun 05 '22 at 05:37