A closed form for a series 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.

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.
