3
$\begingroup$

Let $c_{q}(n)=\displaystyle \sum_{\substack{a=1,...,q\\ (a,q)=1}} e\left(\frac{an}{q}\right)$ be the Ramanujan's sum and consider the following series:

$$\displaystyle\sum_{q>r} \frac{\mu^{2}(q)c_{q}(n)}{\varphi^{2}(q)}$$

It's possible to find an upper bound explicitly in $n$ and $r$?

I want to find a form such as $\frac{n}{\varphi(n)}\frac{C}{r}$, with $C>0$.

I try to do this by myself, but in all my attemps I reduce me to find an explicit bound for a sum that is convergent but not absolutely convergent and so doesn't have an Euler's product (that in fact diverge).

Thanks in advance for any suggestion.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Your sum is bounded by $$C\frac{\log (2+r)}{r} \frac{n \tau(n)}{\varphi(n)},$$ for some absolute constant $C$.

Proof. Using that $c_q(n) = \sum_{d|(n,q)} d \mu(q/d)$, and reversing the order of summation, we get that your sum is bounded by $$\sum_{d|n} \frac{|\mu(d)| d}{\varphi^2(d)} \sum_{\ell > \frac{r}{d}} \frac{|\mu(\ell)|}{\varphi^2(\ell)}.$$ Next I claim that $$\sum_{\ell \geq N} \frac{1}{\varphi^2(\ell)} \ll \frac{\log (2+N)}{N}.$$ Taking this for granted, the bound on the sum becomes $$\ll \frac{\log(2+r)}{r} \sum_{d|n} \frac{|\mu(d)| d^2}{\varphi^2(d)} = \frac{\log(2+r)}{r} \prod_{p|n} (1 + \frac{p^2}{(p-1)^2}), $$ and $$\prod_{p|n} (1 + \frac{p^2}{(p-1)^2}) = \prod_{p|n} (2 + \frac{2}{p} + O(p^{-2})) \ll \frac{n \tau(n)}{\varphi(n)}.$$ Finally, to bound the sum over $\ell$, we use Rankin's trick (we may assume $N \geq 1$) giving $$\sum_{\ell \geq N} \frac{|\mu(\ell)|}{\varphi(\ell)^2} \leq \sum_{\ell \geq N} \frac{\ell^{1-\varepsilon}}{N^{1-\varepsilon}} \frac{|\mu(\ell)|}{\varphi(\ell)^2} \leq \sum_{\ell \geq 1} \frac{\ell^{1-\varepsilon}}{N^{1-\varepsilon}} \frac{|\mu(\ell)|}{\varphi(\ell)^2}. $$ Here $\varepsilon > 0$ will be chosen to be $\varepsilon = \frac{1}{\log (2+N)}$. Now this final sum can be expressed as an Euler product giving $$\sum_{\ell \geq 1} \frac{\ell^{1-\varepsilon}}{N^{1-\varepsilon}} \frac{|\mu(\ell)|}{\varphi(\ell)^2} = \frac{1}{N^{1-\varepsilon}} \prod_{p} (1 + \frac{p^{1-\varepsilon}}{(p-1)^2}) \ll \frac{\zeta(1+\varepsilon)}{N^{1-\varepsilon}} \ll \frac{\log (2+N)}{N},$$ which completes the proof.

Improve this answer
$\endgroup$
5
  • 3
    $\begingroup$ From the general philosophy "$\phi(m)\asymp m$ on average", we should be able to get rid of the $\log r$ term. Indeed, $\sum_{\ell\le x} \frac{\ell^2}{\phi(\ell)^2}$ can be shown to be $O(x)$ by standard multiplicative-function methods, and then $\sum_{U\le\ell\le2U} \frac1{\phi(\ell)^2} \ll \frac1U$ follows easily and implies $\sum_{U\ge N} \frac1{\phi(\ell)^2} \ll \frac1N$. $\endgroup$
    – Greg Martin
    Mar 31, 2017 at 16:51
  • 1
    $\begingroup$ Thanks for your answer! ..but this is not what I was looking for. In fact, you give an estimate for the sum of the absolute values of the general term, and this is easy, as you showed. I yet known the bound with $\frac{\tau(n)n}{\varphi(n)}$, but this is to much big for my purposes. As I said above, I want a bound on the sum itself, that probably is of the form $\frac{C}{r}\frac{n}{\varphi(n)}$. Using your method, one reduce to find a bound for $\sum_{l>\frac{r}{d}}\frac{\mu(l)}{\varphi(l)^{2}}$. $\endgroup$
    – The Number Theorist
    Mar 31, 2017 at 21:39
  • 1
    $\begingroup$ @GregMartin In fact one can prove that $\sum_{l>r}\frac{\mu^{2}(l)}{\varphi^{2}(l)}\leq \frac{6.7345}{r}$. $\endgroup$
    – The Number Theorist
    Mar 31, 2017 at 21:41
  • 3
    $\begingroup$ Note that the Dirichlet series $\sum_{\ell=1}^\infty \frac{\mu(\ell)\ell^2}{\phi(\ell)^2} \ell^{-s}$ is $\frac1{\zeta(s)^2}$ times a series that converges absolutely down to $\Re(s)>\frac12$. We want the tail of this series at $s=2$. It should be possible, using contour integration a la the proof of the PNT, to get a bound $\sum_{\ell>N} \frac{\mu(\ell)}{\phi(\ell)^2} \ll \frac1N \exp(-\sqrt{c\log N})$ for a small $c>0$. $\endgroup$
    – Greg Martin
    Mar 31, 2017 at 23:56
  • $\begingroup$ @GregMartin It's true! I agree with you! In fact, I'm trying to use this method. $\endgroup$
    – The Number Theorist
    Apr 1, 2017 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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