Bounding a series involving Ramanujan's sum 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.
 A: 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.
