Is $\sum_{k=-\infty}^{+\infty}\dfrac{\cos((4k+1)x)}{(4k+1)^n}$ a polynomial in $x$ for $n\in \mathbb{N}$?

Question

My guess is that $$\sum_{k=-\infty}^{+\infty}\dfrac{\cos((4k+1)x)}{(4k+1)^n}$$ might be a polynomial on $(0,\pi/2)$ with real coefficients in $x$ for $n\in \mathbb{N}$.
Since $\cos(0)=1$ the constant term of a possible polynomial is $$\sum_{k=-\infty}^{+\infty}\dfrac{1}{(4k+1)^n}$$ a listing of these numbers is found on the second page of this preprint. For the other coefficients except the constant term i numerically get \begin{align} \sum_{k=-\infty}^{+\infty}\dfrac{\cos((4k+1)x)}{(4k+1)^1}=&+\dfrac{\pi}{4}\\ \sum_{k=-\infty}^{+\infty}\dfrac{\cos((4k+1)x)}{(4k+1)^2}=&-\dfrac{\pi}{4}x+\dfrac{\pi^2}{8}\\ \sum_{k=-\infty}^{+\infty}\dfrac{\cos((4k+1)x)}{(4k+1)^3}=&-\dfrac{\pi}{8}x^2+\dfrac{\pi^3}{32} \end{align} I guessed this from looking at $K_n(u,0)$ where $K_n(u,v)$ is defined in part $3$ of the preprint and from Mercer's theorem. Since $K_n(u,v)$ is defined by recursively integrating a constant function it may be a piecewise-polynomial surface.

Answer 1

Let $P_n$ be the sum of the series. If $n\ge4$ the series can be derivated term by term at least twice, obtaining $$ P_n''=-P_{n-2}. $$ Since $P_1$, $P_2$ and $P_3$ are polynomials, $P_n$ is a polynomial for all $n\in\Bbb N$.

Answer 2

$\sum_{k\in\mathbb{Z}}\frac{\cos((4k+1)x)}{4k+1}$ is the Fourier series of a rectangle wave that equals $\frac{\pi}{4}$ over the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, while $\sum_{k\in\mathbb{Z}}\frac{\cos((4k+1)x)}{(4k+1)^2}$ is the Fourier series of a triangle wave that equals $\frac{\pi^2}{8}-\frac{\pi}{4}x$ over the interval $(0,\pi)$. These facts can be deduced by computing the Fourier series from the mentioned piecewise-polynomial functions, or by manipulating the Fourier series of Bernoulli polynomials. By integrating with respect to $x$ twice the claim follows by induction.