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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.

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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.

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    $\begingroup$ 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 '17 at 16:51
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    $\begingroup$ 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 '17 at 21:39
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    $\begingroup$ @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 '17 at 21:41
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    $\begingroup$ 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 '17 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 '17 at 7:57

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