Expressing $\sum_{k=1}^\infty\frac{\mu(k)}{k}\cos\left(2\pi\frac{n}{k}\right)$ in terms of $n$ For integers $n\geq 1$, let $\mu(n)$ the Möbius function. See the definiton of this arithmetic function, for example from this MathWorld.

Question. I am wondering if is it possible get a good approximation of $$\sum_{k=1}^\infty\frac{\mu(k)}{k}\cos\left(2\pi\frac{n}{k}\right)$$ in terms of a fixed integer $n\geq 1.$ Thanks in advance.

I know Abel's summation formula and the Taylor series for the cosine function, but I don't know if it is a good idea to get a good calculations. Also I don't know if series of this type was in the literature (then feel free to refers the literature).
 A: The original series equals
$$ \sum_{m\geq 1}\frac{(2\pi)^{2m}n^{2m}(-1)^m}{(2m)!}\sum_{k\geq 1}\frac{\mu(k)}{k^{2m+1}}=\sum_{m\geq 1}\frac{(2i\pi n)^{2m}}{(2m)!\,\zeta(2m+1)}$$
and if $\zeta(2m+1)$ is replaced by $1$ that simply becomes $-1+\cosh(2\pi i n)=0$.
If $\frac{1}{\zeta(2m+1)}$ is replaced by $1-\frac{1}{2^{2m+1}}$ we get the more accurate approximation $\frac{1}{2} \left(-1-\cosh(\pi i n)+2\cosh(2\pi i n)\right)=\frac{1-(-1)^n}{2}$, strongly depending on the parity of $n$. 
In general it is pretty difficult to prove accurate approximations for the given function that hold uniformly for any $n$ in some range (for instance, $n\in[N,2N]$): due to the presence of the Moebius function the given series essentially encodes the remainder term of a sieve. These kind of objects usually have an erratic behaviour.
A: Well, you don't need Abel summation, and for numerical calculations, the Taylor series of $\cos$ is sufficient, if $n$ is not a very large number. Just substitute the Taylor series, and observe $$\sum^\infty_{k=1}\frac{\mu(k)}{k}=0,$$ that's equivalent to the prime number theorem, and the reason why your series converges at all. In general, you have $$\sum^\infty_{k=1}\frac{\mu(k)}{k^m}=\frac{1}{\zeta(m)}.$$
For large $n,$ this is tricky, though, and if you find a really good asymptotic expression, you might become famous. 
