A closed form for a series

Question

This maybe duplicated, please let me know if it is so. I would like to find a closed form for the following sum $$ \sum_k \frac {1}{k^2}\cos (kx)$$ Any suggestion would be helpful.

Answer 1

This is simply a Bernoulli polynomial $\;B_2(x)\;$ up to a constant $\pi^2$ for $\,x \in (0,2\pi)$ :

\begin{align} S&:=\sum_{k=1}^\infty \frac {\cos (kx)}{k^2}\\ &=\pi^2\;B_2\left(\frac x{2\pi}\right)\\ &=\pi^2\;\left(\left(\frac x{2\pi}\right)^2-\frac x{2\pi}+\frac 16\right)\\ &=\frac{\pi^2}6-\frac{\pi x}2+\frac{x^2}4\\ \end{align}

On the other side if you replace the $\cos$ function by a $\sin$ you would get a non elementary Clausen function $Cl_2$. For some (nontrivial) intuition about all this you may see this answer or this thread or this answer.