If I am right it is standard to prove that $$\int_0^{2\pi}\sum_{n=1}^\infty\frac{\mu(n)}{n}\left(\frac{\cos(nx)}{n}\right)^2dx=\frac{\pi}{\zeta(3)}\tag{1}$$ using uniform convergence, where $\mu(n)$ denotes the Möbius function (the definition for example from this MathWorld), and $\zeta(s)$ the Riemann's zeta. I wondered this question when I was refreshing the first example (Fourier) of page 63 from [1].
I am interested to know more about the function in the integrand.
Question. Can you prove an interval $\mathcal{I}\subset[0,2\pi]$ where $$f(x):=\sum_{n=1}^\infty\frac{\mu(n)}{n}\left(\frac{\cos(nx)}{n}\right)^2\tag{2}$$ is positive or well negative? Many thanks.
I think that determine with accuracy the roots of such function, and thus the subintervals (of $[0,2\pi]$) where our function is negative or positive is very difficult, thus I am asking about what work can be done.
My question was inspired in the plot for a toy model of our function $(2)$, see this code
plot sum MoebiusMu(n)/n (cos(nx)/n)^2, from n=1 to 1000, 0<x<2pi
instead of plot sum 1/n (cos(nx)/n)^2, from n=1 to 1000, 0<x<2pi or plot sum (-1)^n/n (cos(nx)/n)^2, from n=1 to 1000, 0<x<2pi using Wolfram Alpha online calculator.
References:
[1] Jack D’Aurizio, Superior Mathematics from an Elementary point of view, course notes, University of Pisa (2017-2018).
An example of such subinterval can be found quite easily. By the Weierstrass M-test the series converges absolutely and uniformly in $\left[0,2\pi\right]$, then $f\left(x\right)$ is continuous. So take $x=\pi.$ Then $$f\left(\pi\right)=\sum_{n\geq1}\frac{\mu\left(n\right)}{n^{3}}\cos^{2}\left(n\pi\right)=\sum_{n\geq1}\frac{\mu\left(n\right)}{n^{3}}=\frac{1}{\zeta\left(3\right)}>0$$ then by the sign preserving property for continuous functions exists and open interval $I$ containing $\pi$ such that $$ f\left(x\right)>0,\,\forall x\in I.$$