I have been trying to solve $$y(x)=\sum_{k=1}^{\infty} \frac{|\cos(kx)|}{k}$$ however, this is proving to be more difficult than I had hoped, and cannot seem to figure this out.
What I have figured out so far are the simpler sums: $$y_1(x)=\sum_{k=1}^{\infty} \frac{|x|^k}{k} = \int_{0}^{x} \frac{\text{sgn}(x')}{1-|x'|}dx' = -\ln(1-|x|)$$ $$y_2(x)=\sum_{k=1}^{\infty} \frac{\cos(k x)}{k} = -\frac{1}{2}\ln(2(1-\cos (x))) $$
I have tried using the same method for the absolute value of cosine that I did for the regular cosine sum, which is just -
$$\sum_{k=1}^{\infty} \frac{\cos(kx)}{k} = \Re \left(\sum_{k=1}^{\infty}\frac{e^{ikx}}{k}\right) $$ but since $$|\Re(z)| \neq \Re(||z||)$$ for a complex z, I'm not too sure how to finish this and would appreciate a hint.
Thanks.
This sum diverges for any real number $x$. Here's a sketch of the proof:
Assume first that $\frac x{2\pi}$ is irrational. In this case, the multiples of $x$ are uniformly distributed (mod $2\pi$). In particular, asymptotically $25\%$ of the multiples of $x$ are between $-\frac\pi4$ and $\frac\pi4$ (mod $2\pi$). For all these multiples, the corresponding term in the sum defining $y(x)$ is at least $\frac1{k\sqrt2}$. And if $K$ is any set of integers with positive density, then $\sum_{k\in S} \frac1k$ diverges (since the other terms are all nonnegative).
On the other hand, $\frac x{2\pi}$ is rational, say with denominator $d$, then every $d$th term in the sum equals $\frac1k$, and so the sum diverges for the same reason as above.